# Injective and Surjective Maps

Proposition.

Let $f : (V\subseteq \mathbb A^n) \to (W \subseteq \mathbb A^m)$ be a morphism of closed sets, with corresponding $f^* : k[W] \to k[V]$.

• $f^*$ is injective if and only if $f(V) \subseteq W$ is dense.
• $f^*$ is surjective if and only if $f$ is an embedding of V as a closed subspace of W.

Reminder

A continuous map $f:X \to Y$ of topological spaces is called an embedding if f is injective and the topology of subspace of X is induced via the subspace topology from Y. For a non-example, the map

$f : (-1, 0] \cup (1, 2) \to \mathbb R, \ f(x) = \begin{cases} x+1, \ &\text{if } x \le 0,\\ x, \ &\text{if } x > 0,\end{cases}$

is injective and continuous but not an embedding.

Proof

First claim: saying $f^*$ is injective is equivalent to: for any regular $g:W\to \mathbb A^1$, if $g\circ f = 0$ then $g=0$. This is equivalent to: if g vanishes on f(V), then it vanishes on the whole of W.

• If $\overline {f(V)} = W$ this holds since g is continuous.
• Conversely, if $W' := \overline{f(V)} \subsetneq W$, then $I(W') \supsetneq I(W)$ so there exists $f \in I(W') - I(W)$, i.e. f vanishes on W’ (and hence f(V)) but not W.

Second claim: $f^*$ is surjective if and only if it factors through

$k[W] \stackrel \pi\longrightarrow k[W]/\mathfrak a \stackrel \phi\longrightarrow k[V]$

for some ideal $\mathfrak a\subseteq k[W]$ and isomorphism $\phi$, where $\pi$ is the canonical map. Since k[V] is a reduced ring, such an $\mathfrak a$ must be a radical ideal. So the above homomorphisms correspond to: $V \stackrel {\cong} \longrightarrow W' \hookrightarrow W$ for some closed subset $W'\subseteq W$. ♦

## Example

Take the map $f : \{(x, y)\in \mathbb A^2 : xy = 1\} \to \mathbb A^1$, where $f(x, y) = x$. Algebraically

$f^* : k[T] \to k[X, Y]/(XY - 1) \cong k[X, \frac 1 X],\quad T \mapsto X.$

[Image edited from GeoGebra plot.]

• The image of f is $\mathbb A^1 - \{0\}$ which is dense in $\mathbb A^1$; correspondingly $f^*$ is injective.
• Even though f is injective, its image is not closed. Correspondingly $f^*$ is not surjective.

We will devote the remainder of this article to two rather difficult examples. Since the reasoning is a bit involved, this is all we will cover for now.

# Case Study A

Let $k = \mathbb C$. We take the following set

$V = \{(t^3, t^4, t^5) \in \mathbb A^3 : t \in \mathbb C\}$.

### Problem: is this a closed subset? What is the ideal I(V)?

For the first question, we claim that V is cut out by the equations $X^4 - Y^3, X^5 - Z^3$ and $Y^5 - Z^4$. Indeed if xyz are complex values and $x^4 = y^3$, $x^5 = z^3$ and $y^5 = z^4$, then either x = 0 (in which case y = z = 0), or we set t := y/x to obtain

$x = (y/x)^3 = t^3,\ y =tx = t^4,\ z = (y/x)^5 = t^5.$

The next question is tricky; if $\mathfrak a = I(V)$, then we just saw that $X^4 - Y^3$, $X^5 - Z^3$ and $Y^5 - Z^4$ are in $\mathfrak a$. A bit of experimentation gives us more elements

$X^3 - YZ, Y^2 - XZ, Z^2 - X^2 Y \in \mathfrak a.$

Let $\mathfrak b \subseteq \mathfrak a$ be the ideal generated by these 3 elements; we will show that $\mathfrak b = \mathfrak a$. Now let $B := k[X,Y,Z]/\mathfrak b$; we claim that B is spanned by $k[X]$, $k[X]Y$ and $k[X]Z$ as a k-vector space.

• Since $Y^2 \equiv XZ \pmod {\mathfrak{b}}$ and $Z^2 \equiv X^2 Y \pmod {\mathfrak b}$, we can replace any monomial modulo $\mathfrak b$ by $X^m Y^j Z^k$ where j = 0, 1 and k = 0, 1.
• Finally since $YZ \equiv X^3 \pmod {\mathfrak b}$ we are done.

The morphism $f:\mathbb A^1 \to V$ where $f(t) = (t^3, t^4, t^5)$ corresponds to the homomorphism

$f^* : k[X,Y,Z]/\mathfrak a \longrightarrow k[T], \quad X \mapsto T^3, Y \mapsto T^4, Z \mapsto T^5.$

Since $\mathfrak b\subseteq \mathfrak a$, we obtain the composition

$B = k[X,Y,Z]/\mathfrak b\longrightarrow k[X,Y,Z]/\mathfrak a \longrightarrow k[T]$

which maps $k[X]$, $k[X]Y$ and $k[X]Z$ into subspaces $k[T^3]$, $k[T^3]T^4$ and $k[T^3]T^5$ of $k[T]$. These spaces are linearly independent over k so the composed map is injective. Hence, the first map is injective and we see that $\mathfrak b = \mathfrak a$.

Questions to Ponder

1. Is $\mathfrak a$ a prime ideal of k[XYZ]? Equivalently, is k[V] an integral domain?
2. Can $\mathfrak a$ be generated by two or less elements?

# Case Study B

Let us take the subring $A := k[S^4, S^3 T, ST^3, T^4]$ of k[ST]. Consider the k-algebra homomorphism:

$\phi : k[W, X, Y, Z] \longrightarrow A, \quad W \mapsto S^4, X \mapsto S^3 T, Y \mapsto ST^3, Z \mapsto T^4.$

### Problem: describe the kernel $\mathfrak a$ of this map.

Step 1. Note that the map $k[W, Z] \to K[W,X,Y,Z]/\mathfrak a$ is an injection.

Indeed, the map corresponds to $k[S^4, T^4] \hookrightarrow k[S^4, S^3 T, ST^3, T^4]$ and the homomorphism $k[W, Z] \to k[S^4, T^4]$ is an isomorphism. Now $k[S^4, T^4]$ has basis given by $S^{4i} T^{4j}$ for all $i, j\ge 0$. Let us compute a small set of monomial representatives $m_i$ such that

$k[S^4, S^3 T, ST^3, T^4] = \sum_i k[S^4, T^4] m_i$

as vector spaces.

• Degree 1 : we have $S^4, S^3 T, ST^3, T^4$. For these we take

$m_1 = 1,\ m_2 = S^3 T, \ m_3 = ST^3$.

• Degree 2 : we have $S^6 T^2, S^4 T^4, S^2 T^6$. But then $S^4 T^4 \in k[S^4, T^4]$ so we leave out this term and take

$m_3 = S^6 T^2,\ m_4 = S^2 T^6$.

• Degree 3 :  we have $S^9 T^3, S^7 T^5, S^5 T^7, S^3 T^9$. But $S^7 T^5, S^3 T^9 \in k[S^4, T^4] m_2$ and $S^5 T^7, S^9 T^3 \in k[S^4, T^4] m_3$ so there are no new terms added.

Thus for $B = k[S^4, T^4]$ we have shown

$A = B + B\cdot S^3 T + B\cdot ST^3 + B\cdot S^6 T^2 + B\cdot S^2 T^6.$

The occurring monomials can be plotted as follows:

Note that the point (2, 2) is conspicuously absent.

Step 2. We write down as many relations as we can in the ideal. Some clear ones are

$WZ - XY,\ W^2 Y - X^3,\ XZ^2 - Y^3,\ X^2 Z - WY^2 \in \mathfrak a.$

Do these generate the whole ideal? Let $\mathfrak b\subseteq \mathfrak a$ be the ideal generated by these four elements.

Strategy. We will show that $k[W, X, Y, Z]/\mathfrak b \to A$ is injective.

For that, we pick any $f(W,X,Y,Z) \in k[W,X,Y,Z]$.

• Replacing $Y^3$ with $XZ^2$ and $X^3$ with $YW^2$ modulo $\mathfrak b$, we obtain f in $\sum k[W, Z]X^i Y^j$ where the sum is over $0 \le i \le 2$, $0\le j \le 2$.
• Replacing $XY$ with $WZ$ modulo $\mathfrak b$, the sum is now over $(i, j) = (0, 0), (0, 1), (0, 2), (1, 0), (2, 0)$.
• Finally replacing $WY^2$ with $X^2 Z$, we have $f \equiv g \pmod {\mathfrak b}$ where

$g \in k[W, Z] + k[W, Z] X + k[W, Z] X^2 + k[W, Z] Y + k[Z] Y^2.$

Now suppose $f\in \mathfrak a$, then $g\in \mathfrak a$ so $g(S^4, S^3 T, ST^3, T^4) = 0$. By looking at the monomials occuring, this can only happen for g = 0. This proves:

$\mathfrak a = \mathfrak b = (WZ - XY,\ W^2 Y - X^3,\ XZ^2 - Y^3,\ X^2 Z - WY^2).$

Questions to Ponder

1. Is $\mathfrak a$ a prime ideal of $k[W,X,Y,Z]$?
2. Can $\mathfrak a$ be generated by three elements or less?

# Morphisms in Algebraic Geometry

Next we study the “nice” functions between closed subspaces of $\mathbb A^n$.

Definition.

Suppose $V\subseteq \mathbb A^n$ and $W\subseteq A^m$ are closed subsets. A morphism $f:V\to W$ is a function which can be expressed as:

$f(v_1, \ldots, v_n) = (f_1(v_1, \ldots, v_n), f_2(v_1, \ldots, v_n), \ldots, f_m(v_1, \ldots, v_n))$

for some polynomials $f_1, f_2, \ldots, f_m \in k[X_1, \ldots, X_n]$. We also say f is a regular map.

Example

1. Let $V \subseteq \mathbb A^n$ be any closed subset. Regular maps of the form $V \to \mathbb A^1$ are given by polynomials $f \in k[X_1, \ldots, X_n]$.
2. Take $V = \mathbb A^1$ and $W = \{(x, y, z) \in \mathbb A^3 : x^2 + y^2 = z^2\}$. Define $f:V\to W$ by $t \mapsto (t^2 - 1, 2t, t^2 + 1)$. We write this as $x = t^2 - 1, y = 2t, z = t^2 + 1$.
3. Take $V = \mathbb A^1$ and $W = \{(x, y) \in \mathbb A^2 : y^2 = x^3\}$. Define $f:V\to W$ by $t\mapsto (t^2, t^3)$. We write this as $x = t^2, y = t^3$.

[Example 3: image edited from GeoGebra plot.]

The first example, although basic, is of huge importance.

Definition.

regular function for a closed subset $V\subseteq \mathbb A^n$ is a morphism of the form $V\to \mathbb A^1$.

The of such functions is called the coordinate ring of V and is denoted by $k[V]$.

Note

Each regular function is given by $f\in k[X_1, \ldots, X_n]$ and two polynomials f, g induce the same function on V if and only if

$(\forall P \in V, f(P) = g(P)) \iff (\forall P \in V, (f-g)(P) = 0) \iff f-g \in I(V).$

Thus we have $k[V] \cong k[X_1, \ldots, X_n]/I(V)$, which gives k[V] its ring structure. To describe this ring structure in terms of regular functions, we have:

$\text{morphisms } f, g : V \to \mathbb A^1 \implies \begin{cases} (f+g) : V\to \mathbb A^1, \ &P \mapsto f(P) + g(P) \in k, \\ (f\cdot g) : V\to \mathbb A^1, \ &P \mapsto f(P)g(P)\in k. \end{cases}$

# Morphisms in Algebra

General morphisms $f:V\to W$ can now be described in the language of their coordinate rings.

Proposition.

Let $V\subseteq \mathbb A^n$ and $W\subseteq \mathbb A^m$ be closed. There is a bijection between:

• morphisms $f:V\to W$;
• ring homomorphisms $f^* : k[W] \to k[V]$ which are linear over k.

Note

The second condition can be rephrased as: $k[W] \to k[V]$ is a homomorphism of k-algebras. In a later article, we will cover algebras over a ring in greater detail.

Proof

Given a morphism $f:V\to W$, upon composing with a regular function $g : W\to \mathbb A^1$, we obtain a regular function $g\circ f : V\to \mathbb A^1$ of V. This gives a ring homomorphism $k[W] \to k[V]$ which is clearly linear over k.

Conversely, suppose $\phi : k[W] \to k[V]$ is a ring homomorphism which is linear over k. Let $\overline X_1, \ldots, \overline X_m$ be the images of  $X_1, \ldots, X_m$ in $k[W] = k[X_1, \ldots, X_m]/I(W)$. Let $f_i = \phi(\overline X_i) \in k[V]$ which is a regular function $V\to \mathbb A^1$. We claim that the function

$f : V \to \mathbb A^m, \quad f(P) := (f_1(P), \ldots, f_m(P))$

has image in W. Indeed for any $g\in I(W)\subseteq k[X_1, \ldots, X_m]$ we have

$g\circ f = g(f_1, \ldots, f_m) = g(\phi(\overline X_1), \ldots, \phi(\overline X_m)) = \phi(g(\overline X_1, \ldots, \overline X_m)) = \phi(\overline{g(X_1, \ldots, X_m)})$

which is 0 since $\overline g =0$ in k[W]. This creates a bijection. ♦

Note

In order to swap $g(\phi(\ldots)) = \phi(g(\ldots))$ we require $\phi$ to be a ring homomorphism linear over k. E.g. if $g = \alpha X_1^2 +\beta X_2$ with $\alpha, \beta\in k$ then

$g(\phi(\overline X_1), \phi(\overline X_2)) = \alpha\phi(\overline X_1)^2 + \beta \phi(\overline X_2) = \phi(\alpha \overline X_1^2 + \beta \overline X_2) = \phi(g(\overline X_1, \overline X_2)).$

The following properties are obvious:

Lemma.

• For any closed set V, we have $(1_V)^* = 1_{k[V]}$.
• For any morphisms $f:V \to V', g:V' \to V''$ of closed sets we have

$(g\circ f)^* =f^* \circ g^* : k[V''] \longrightarrow k[V].$

Proof

The first is clear. For the second, pick any $h\in k[V'']$, a regular map $V''\to \mathbb A^1$. Then

$f^*(g^*(h)) = f^*(h\circ g) = (h \circ g)\circ f = h\circ (g\circ f) = (g\circ f)^*(h)$. ♦

## Examples

Let us interpret the earlier examples as homomorphisms $k[W] \to k[V]$.

Example 2. We have:

$f : \mathbb A^1 \to \{(x, y, z) \in \mathbb A^3 : x^2 + y^2 = z^2\}, \quad t \mapsto (t^2 - 1, 2t, t^2 + 1)$.

We wrote this as $x = t^2 - 1, y = 2t, z = t^2 + 1$ for a reason, for f corresponds to:

$f^* : k[X, Y, Z]/(X^2 + Y^2 - Z^2) \longrightarrow k[T], \quad X \mapsto T^2 - 1, Y \mapsto 2T, Z \mapsto T^2 + 1.$

Example 3. Similarly

$f : \mathbb A^1 \to \{(x,y) \in \mathbb A^2 : y^2 = x^3\}, \quad t\mapsto (t^2, t^3)$

was written as $x = t^2, y = t^3$. Algebraically,

$f^* : k[X,Y]/(Y^2 - X^3) \longrightarrow k[T], \quad X \mapsto T^2, Y \mapsto T^3.$

# Properties

Since we defined a topology on closed sets, the morphisms should be continuous.

Proposition.

A morphism $f : (V\subseteq \mathbb A^n) \to (W\subseteq \mathbb A^m)$ is a continuous map with respect to the Zariski topology on both sets.

Proof

Let W’ be a closed subset of W; we need to show $f^{-1}(W')$ is closed in V (equivalently, in $\mathbb A^n)$. Now W’ can be written as

$W' = \{Q \in W: \text{ for each } g \in S', g(Q) = 0\}$

for some subset $S' \subseteq k[X_1, \ldots, X_m]$. It follows that

$f^{-1}(W') = \{P \in V : \text{ for each } g \in S', g(f(P)) = 0\}$

is cut out from V by equations $\{g\circ f : g\in S'\}$. Hence $f^{-1}(W')$ is closed. ♦

## Isomorphisms

Definition.

Closed subsets $V \subseteq \mathbb A^n$ and $W\subseteq \mathbb A^m$ are said to be isomorphic if there exist regular maps $f : V\to W$ and $g:W\to V$ such that $g\circ f = 1_V$ and $f\circ g = 1_W$.

This implies:

$f^* : k[W] \to k[V],\ g^* : k[V] \to k[W], \quad f^*\circ g^* = 1_{k[V]}, \ g^* \circ f^* = 1_{k[W]}.$

Hence isomorphism of the closed subsets corresponds to isomorphism of the underlying k-algebras!

Example

Let us take example 3 from above, where $f : \mathbb A^1 \to \{(x,y) \in \mathbb A^2 : y^2 = x^3\}$ is defined by $f(t) = (t^2, t^3)$. Note that f is bijective on the points, and it corresponds to:

$f^* : k[X,Y]/(Y^2 - X^3) \longrightarrow k[T], \quad X \mapsto T^2, Y \mapsto T^3.$

The image of $f^*$ is not surjective since it does not contain T. We have thus learnt:

There exist bijective regular maps $V\to W$ which are not isomorphisms.

# More General Correspondence

Putting it together, we obtain the following bijective correspondences:

• The top correspondence was the original one.
• The left correspondence follows from point-set topology.
• The right correspondence follows from the correspondence between ideals of $A/\mathfrak a$ and ideals of $A$ containing $\mathfrak a$.
• The correspondence preserves radical ideals because $\mathfrak b$ is a radical ideal of A if and only if $A/\mathfrak b$ is a reduced ring; now apply $(A/\mathfrak a)\, /\, (\mathfrak b/\mathfrak a) \cong A/\mathfrak a$.

Summary.

In other words, we have a bijection between radical ideals of the coordinate ring k[V] and closed subsets of V. This enables us to look at V and its coordinate ring k[V], ignoring the ambient affine space it sits in.

# More Concepts in Algebraic Geometry

Recall that we have a bijection between radical ideals of $A = k[X_1, \ldots, X_n]$ and closed subsets of $\mathbb A^n_k$.

The bijection reverses the inclusion so $V(\mathfrak a) \subseteq V(\mathfrak b)$ if and only if $\mathfrak a \supseteq \mathfrak b$. Not too surprisingly, operations on ideals translate to operations on the corresponding closed subsets.

Proposition.

Suppose closed subsets $S, T, S_i \subseteq \mathbb A^n_k$ correspond to radical ideals $\mathfrak a, \mathfrak b, \mathfrak a_i \subseteq A$.

• $\cap_i S_i$ corresponds to $r(\sum_i \mathfrak a_i)$.
• $S\cup T$ corresponds to $\mathfrak a \cap \mathfrak b = r(\mathfrak {ab})$.

Proof

For the first claim:

• Since $\mathfrak a_j \subseteq r(\sum_i \mathfrak a_i)$ for any j, we have $S_j = V(\mathfrak a_j) \supseteq V(r(\sum_i \mathfrak a_i))$ and thus $\cap_i S_i \supseteq V(r(\sum_i \mathfrak a_i))$.
• Conversely, if $P\in \cap S_i$, then any $f\in r(\sum_i \mathfrak a_i)$ gives $f^n \in \sum_i \mathfrak a_i$ for some n > 0 so $f^n$ is a finite sum of terms $g_j \in \mathfrak a_{i_j}$. Each such $g_j(P) = 0$ so $f(P)^n = 0$ and thus $f(P) = 0$. We have shown: $\cap_i S_i = V(r(\sum_i \mathfrak a_i))$.
• Finally, since $r(\sum_i \mathfrak a_i)$ is a radical ideal, we are done.

Second claim:

• First, we show $S\cup T = V(\mathfrak a \cap \mathfrak b)$.
• Since $\mathfrak a \cap \mathfrak b \subseteq \mathfrak a$ we have $S = V(\mathfrak a) \subseteq V(\mathfrak a \cap \mathfrak b)$. Likewise $T\subseteq V(\mathfrak a \cap \mathfrak b)$ and we have shown ⊆.
• Conversely, if $P\in \mathbb A^n_k$ lies outside S and T then there exist $f\in \mathfrak a$ and $g\in \mathfrak b$ such that $f(P), g(P) \ne 0$ and thus $(fg)(P) \ne 0$. Since $fg \in \mathfrak a \cap\mathfrak b$ this gives $P\not\in V(\mathfrak a \cap \mathfrak b)$.
• Since intersection of radical ideals is radical, we have $r(\mathfrak a \cap \mathfrak b) = \mathfrak a \cap \mathfrak b$. And since $r(\mathfrak {ab}) = r(\mathfrak a \cap \mathfrak b)$ from here, we are done. ♦

# Prime and Maximal Ideals

Next, we wish to find the closed subsets corresponding to maximal and prime ideals of A.

Proposition.

Maximal ideals correspond to singleton subsets of $\mathbb A^n_k$.

Proof

For a singleton set {P}, consider the evaluation map:

$e_P : A = k[X_1, \ldots, X_n] \longrightarrow k, \quad f \mapsto f(P)$

which is a ring homomorphism. This is clearly surjective so its kernel $\mathfrak m_P$ is a maximal ideal. By definition $I(\{P\}) = \mathfrak m_P$.

Conversely, for a maximal ideal $\mathfrak m \subset A$, since $m\ne A$ its corresponding subset $S \ne \emptyset$ by the Nullstellensatz. Thus it contains some P. From $P\in S$ we get $\mathfrak m = V(S) \subseteq V(\{P\}) = \mathfrak m_P$. By maximality of $\mathfrak m$ equality holds so S = {P}. ♦

Exercise.

Prove that if $P =(v_1, \ldots, v_n)$ then $\mathfrak m_P = (X_1 - v_1, \ldots, X_n - v_n)$.

Prime ideals are a little trickier, so we will have to introduce a new topological concept.

Definition.

A topological space X is said to be irreducible if it is non-empty, and any non-empty open subset of X is dense in X. Otherwise we say it is reducible.

We are getting repetitive, but empty spaces are excluded from the class of irreducible spaces for the same reason 1 is not prime. Later, we will see that closed subsets can be uniquely “factored” as a union of irreducible closed subsets.

Lemma.

The following are equivalent for any non-empty topological space X.

1. X is irreducible.
2. Any two non-empty open subsets of X must intersect.
3. If closed subsets $C, C'\subseteq X$ have union X, then C = X or C’ = X.

Note

All criteria for irreducibility are useful at some point of time so it pays to take heed. We will study such spaces in greater detail at a later time.

Proof

The equivalence between 2 and 3 is straightforward. For equivalence of 1 and 2, use the fact that a subset of a topological space is dense if and only if every non-empty open subset intersects it. ♦

Proposition.

The prime ideals of A correspond to the irreducible closed subspaces of $\mathbb A^n_k$.

Proof

Suppose V is irreducible and $fg\in A- I(V)$. Then $C := \{P \in V : f(P) = 0\}$ and $C' := \{P \in V : g(P) = 0\}$ are closed subsets of V and $C, C'\ne V$. Thus $C\cup C' \ne V$ so we have $fg \not\in I(V)$.

Conversely suppose $V = V(\mathfrak p)$ for a prime ideal $\mathfrak p \subset A$. Let $C, C'\subseteq V$ be closed subsets (of V, and hence of $\mathbb A^n_k$) with union V. Write

$C = V(\mathfrak a), C' = V(\mathfrak a')$ for radical ideals $\mathfrak a \supseteq \mathfrak p, \mathfrak a' \supseteq \mathfrak p.$

Thus $V = C \cup C'$ corresponds to the ideal $\mathfrak a \cap \mathfrak a'$ and we have $\mathfrak a \cap \mathfrak a' = \mathfrak p$. It remains to prove the following, which we will leave as an easy exercise. ♦

Exercise

Suppose $\mathfrak a, \mathfrak a', \mathfrak p$ are ideals of any ring A with $\mathfrak p$ prime. If $\mathfrak a, \mathfrak a' \supseteq \mathfrak p$ and $\mathfrak a \cap \mathfrak a' = \mathfrak p$, then $\mathfrak a = \mathfrak p$ or $\mathfrak a' = \mathfrak p$.

# Simple Example

We will work through a simple example step-by-step.

Let $k = \mathbb C$ and consider the closed subset V cut out by $f = X^2 + Y^2 - 2$ and $g = X-Y$. Geometrically this gives the points of intersection between $X^2 + Y^2 = 2$ (a circle) and $X = Y$ (a line). Clearly, we get two points (1, 1) and (-1, -1).

[Graph plotted by GeoGebra.]

Let us verify this algebraically.

Let $\mathfrak a = (X^2 + Y^2 - 2, X - Y) \subset \mathbb C[X, Y]$. We have $V(\mathfrak a) = V$ so it remains to show that $\mathfrak a$ is a radical ideal. We prove this by applying the following to the quotient ring $\mathbb C[X, Y]/\mathfrak a$.

Lemma

A ring A is said to be reduced if (0) is a radical ideal in A. Prove that an ideal $\mathfrak a\subseteq A$ is radical if and only if $A/\mathfrak a$ is reduced.

First note that for any ring A and $a\in A$, we have an evaluation map $e : A[X] \to A$ taking $f(X) \mapsto f(a)$. The kernel of this map is precisely (X – a); indeed it clearly contains (X – a), conversely any $f(X)\in A[X]$ can be written as $f(X) = (X-a)g(X) + r$ with $r\in A$; if $f(a) = 0$ we have r = 0. Hence we have $A[X]/(X-a) \cong A$ where the isomorphism takes X to a.

With that in mind we get:

$\mathbb C[X, Y]/(X - Y) \cong \mathbb C[Y] \Rightarrow \mathbb C[X, Y]/(X^2 + Y^2 - 2, X - Y) \cong \mathbb C[Y]/(2Y^2 - 2)$

because the first isomorphism takes X to Y. This corresponds to a set of two points {-1, +1} in the affine line $\mathbb A^1$. Now by Chinese Remainder Theorem, we have

$\mathbb C[Y]/(Y^2 - 1) \cong \mathbb C[Y]/(Y-1) \times \mathbb C[Y]/(Y+1) \cong \mathbb C\times \mathbb C$

which is a reduced ring. Hence $\mathbb C[X,Y]/\mathfrak a$ is a reduced ring and $\mathfrak a$ is a radical ideal.

The set of two points is reducible so its corresponding ideal $\mathfrak a = (X^2 + Y^2 - 2, X-Y)$ is not prime. Indeed, we saw that $Y^2 - 1\in \mathfrak a$, but $Y+1, Y-1\not\in \mathfrak a$.

## Slight Variation

Let us now consider the set cut out by $f = X^2 + Y^2 - 2$ and $g = X+Y - 2$. Geometrically, we now have only one point of intersection.

[Graph plotted by GeoGebra.]

Let $\mathfrak a = (X^2 + Y^2 - 2, X+Y - 2)$ now. We have

\begin{aligned}\mathbb C[X, Y]/(X^2 + Y^2 - 2, X + Y - 2) &\cong \mathbb C[Y]/((2 - Y)^2 + Y^2 - 2)\\ &= \mathbb C[Y]/(2Y^2 -4Y + 2).\end{aligned}

This ring is non-reduced so $\mathfrak a$ is not a radical ideal. In fact, $r(\mathfrak a) = (Y-1, X+Y-2) = (X-1, Y-1)$ which corresponds to the geometric picture.

But the story is not quite over!

Note that upon taking the radical of $\mathfrak a$, we are actually losing information on the multiplicity of intersection. Since $\mathfrak a$ satisfies $\dim_{\mathbb C} (A/\mathfrak a) = 2$, this suggests that the intersection multiplicity is 2 here. Indeed we can construct a theory of intersection via considering general ideals instead of merely radical ideals.

## Note

This seems like a tremendous amount of work for such simple geometric examples. But it looks tedious only because we took pains to explicitly justify every step. As you progress, the above computations will eventually seem too trivial to even contemplate.

In the following articles, we will consider a much harder example when we have more tools at our disposal.

# Algebraic Geometry Concepts

We have decided to introduce, at this early point, some basics of algebraic geometry in order to motivate the later concepts.

In summary, algebraic geometry studies solutions to polynomial equations over a field.

First we consider a case which has applications to number theory.

## Example

Let us compute all integer solutions to $a^2 + 2b^2 = 3c^2$. For c = 0, we only have the trivial solution; for $c\ne 0$, we get $(\frac a c)^2 + 2(\frac b c)^2 = 3$ so we need to solve $x^2 + 2y^2 = 3$ in the field of rational numbers.

By observation we obtain our first solution P = (1, 1). For any other point $Q = (x,y)$ with coordinates in ℚ, let us take the gradient of the line PQ, given by $m = \frac{y-1}{x-1}$ which is a rational number. Thus each Q gives us a rational number m.

[ Image edited from Geogebra plotter.]

Conversely, suppose we are given $m\in \mathbb Q$. The line through P = (1, 1) with gradient m intersects the ellipse in another point Q = (x, y). Now x and y are rational, because x is the root of a quadratic equation with rational coefficients; thus the sum of the two roots is rational; since one of the root is 1, the other root is also rational.

E.g. take $m = \frac 3 2$. The line through P of gradient m is then $(y-1) = \frac 3 2 (x-1)$, or $y = \frac 3 2 x - \frac 1 2$. Substituting this into the equation of the ellipse then gives

$x^2 + 2\left( \frac 3 2 x - \frac 1 2\right)^2 = 3 \implies \frac{11}2 x^2 - 3x - \frac 5 2 = 0$

which has two roots with sum $\frac 6 {11}$. Since one of the roots is 1, the other is $x=\frac 6{11} - 1 = -\frac 5 {11}$. And from $y = \frac 3 2 x - \frac 1 2$ we get $y = -\frac{13}{11}$. This gives the solution $a = -5, b=-13, c = 11$ to the original equation $a^2 + 2b^2 = 3c^2$.

Hence, we have shown that all solutions to our equation can be obtained in this way. This solution generalizes to all quadratic forms in three variables, e.g. $a^2 + ab + b^2 = 3c^2$. The case for cubic forms is much more complicated and involves elliptic curves, which is a huge topic.

Exercises

Write down parametric solutions for $a^2 + 2b^2 = 3c^2$ and $a^2 + ab + b^2 = 3c^2$.

# The Affine Space

Let k be a fixed field.

Definition.

The affine n-space $\mathbb A_k^n$ over the field k is the set of all n-tuples $(x_1, \ldots, x_n)$ with each $x_i \in k$.

If the base field is implicit, we often just write $\mathbb A^n$.

Set-theoretically, $\mathbb A_k^n$ is just $k^n$. However, we have chosen a different notation because one often mentally associates $k^n$ with an n-dimensional vector space over k.

Let $A = k[X_1, \ldots, X_n]$ be the ring of polynomials in variables and coefficients in k. Each $f\in A$ can be interpreted as a function on $\mathbb A_k^n$. For example,

\left.\begin{aligned} f = X^2 - Y^3 + 2Z \in \mathbb C[X, Y, Z]\\P = (2, 1, 0) \in \mathbb A_{\mathbb C}^3 \end{aligned}\right\} \implies f(P) = 2^2 - 1^3 + 2\cdot 0 = 3.

Definition.

Given a collection $S \subseteq A$ of polynomials, let

$V(S) = \{ P \in \mathbb A_k^n : f(P) = 0 \text{ for all } f\in S\}.$

A subset of $\mathbb A_k^n$ of this form is called a closed subset.

In words, we say that V(S) is the subset of $\mathbb A^n_k$ carved out by the polynomials in S. For example if n = 3 and $S \subset \mathbb R[X, Y, Z]$ comprises of $f = X^2 + Y^2 - 1$ and $g = Z$, the resulting V(S) is a unit circle on the XY-plane.

Proposition.

The collection of all closed subsets in $\mathbb A_k^n$ forms a topology for $\mathbb A_k^n$. Specifically, we have the following.

• $V(\{0\}) = \mathbb A_k^n$ and $V(\{1\}) = \emptyset$.
• For any collection of subsets $S_i \subseteq A$, we have $\cap_i V(S_i) = V(\cup_i S_i)$.
• Given $S, T\subseteq A$ we have $V(S) \cup V(T) = V(ST)$, where $ST = \{fg : f\in S, g\in T\}$.

Proof

The first property is obvious.

For the second, $P \in \cap_i V(S_i)$ is equivalent to: for each i, $P\in V(S_i)$, which is equivalent to: for each i and $f \in S_i$, we have $f(P) = 0$, which is finally equivalent to: for each $f\in \cup_i S_i$ we have $f(P) = 0$.

Finally clearly $V(S) \subseteq V(ST)$: if P satisfies $f(P) = 0$ for each $f\in S$, then certainly $(fg)(P) = f(P)g(P) = 0$ for all $fg\in ST$. Similarly $V(T) \subseteq V(ST)$ so we have proven ⊆ in the claim.

For the reverse inclusion, suppose P lies outside V(S) or V(T); thus there exists $f\in S, g\in T$ such that $f(P) \ne 0$, $g(P) \ne 0$. But this means $fg\in ST$ satisfies $(fg)(P) \ne 0$.♦

Definition.

The above topology is called the Zariski topology on the affine n-space.

Exercise

Prove that the Zariski topology on $\mathbb A^1_k$ is the cofinite topology (i.e. a subset is closed if and only if it is finite, or the whole space).

Let $k = \mathbb R$. Prove that $\{(x, e^x) : x\in \mathbb R\}$ is not a closed subset of $\mathbb A^2_{\mathbb R}$.

# Ideals of Polynomial Ring

In the previous section, a subset S of the polynomial ring A gives us a subset of the affine space. We now define the reverse.

Definition.

Given any subset $V\subseteq \mathbb A^n_k$, let $I(V)$ be the set of all $f\in A$ such that $f(P) = 0$ for all $P\in V$.

Immediately we have:

Lemma.

For any subset V of $\mathbb A^n_k$, $I(V)$ is a radical ideal of the ring of polynomials A.

Proof

Clearly $0\in I(V)$. Next if $f, g\in I(V)$ then for any $P\in V$ we have $(f-g)(P) = f(P) - g(P) = 0$. Similarly if $f\in I(V)$ and $g\in A$, then for any $P\in V$ we have $(fg)(P) = f(P)g(P) = 0$ so we have $fg\in I(V)$.

Finally, if $f \in A$ satisfies $f^n \in I(V)$ then for any $P\in V$ we have $f(P)^n = 0$ and so $f(P) = 0$. This gives $f\in I(V)$. ♦

Hence we have the following diagram.

# Properties of the Correspondence

Clearly the above does not provide a bijection, but it will when we restrict the sets on both sides.

Proposition.

Take any $S, S' \subseteq A$ and $V, V' \subseteq \mathbb A^n_k$.

• $V \subseteq V' \implies I(V) \supseteq I(V')$.
• $S \subseteq S' \implies V(S) \supseteq V(S')$.
• $S \subseteq IV(S)$.
• $V\subseteq VI(V)$.
• $VIV(S) = V(S)$.
• $IVI(V) = I(V)$. [Apologies for the awkward notation!]

Note

The fifth claim says that if V is a closed subset of $\mathbb A^n_k$, then VI takes V back to itself.

Proof

We will prove only half of the claims, since the remaining are similar.

First claim: if $V\subseteq V'$, then for any $f \in I(V')$, any $P\in V$ also lies in V’ so $f(P) = 0$ and we have $f \in I(V)$.

Third claim: if $f\in S$, we take any $P\in V(S)$; by the definition of V(S) we have $f(P) = 0$. Hence f lies in I(V(S)).

Fifth claim: we have $VI(V(S)) \supseteq V(S)$ by the fourth claim. On the other hand since $IV(S) \supseteq S$ by the third claim we have $VIV(S) \subseteq V(S)$ by the second claim. ♦

Now we are very close to achieving a bijective correspondence, but we need a final additional condition.

Theorem (Nullstellensatz)

Suppose k is an algebraically closed field (e.g. $k = \mathbb C$). The above correspondence gives a bijection between:

• closed subsets $V\subseteq \mathbb A^n_k$,
• radical ideals of $A = k[X_1, \ldots, X_n]$.

Exercise

Find a counter-example for the real field ℝ.

Unfortunately, we have to defer the proof till much later. For now, we will have to contend with exploring some examples.

## Commutative Algebra 2

In this installation, we will study more on ideals of a ring A.

Definition.

If $\mathfrak a\subseteq A$ is an ideal, its radical is defined by

$r(\mathfrak a) := \{ x \in A: x^n \in \mathfrak a \text{ for some } n>0\}.$

To fix ideas, again consider the case $A = \mathbb Z$ again. For the ideal (m) where $m = p_1^{e_1}\ldots p_k^{e_k}$, each $e_i > 0$, its radical is simply (m’) where $m' = p_1 \ldots p_k$ with the repeated exponents removed. Thus one thinks of the radical as “retaining only the prime factors”.

Our first result is:

Lemma.

The radical of an ideal $\mathfrak a$ is also an ideal.

Proof

Let $\mathfrak b = r(\mathfrak a)$. Suppose $x, y \in \mathfrak b$; pick m, n > 0 such that $x^m, y^n \in \mathfrak a$.

As before, $(x+y)^{m+n}$ is a sum of terms $Cx^i y^j$ with $i+j = m+n$ and so each term is a multiple of $x^m$ or $y^n$. Thus $(x+y)^{m+n} \in \mathfrak a$ so $x+y \in \mathfrak b$.

Also for any $z\in A$ we have $(xz)^m = x^m z^m \in \mathfrak a$. Hence $xz \in \mathfrak b$. Finally since $0\in \mathfrak b$, we see that $\mathfrak b$ is an ideal of A. ♦

The following properties relate the radical of an ideal to earlier constructions.

Proposition.

Let $\mathfrak a, \mathfrak b$ be ideals of A, and $(\mathfrak a_i)$ be any collection of ideals of A.

• $r(r(\mathfrak a)) = r(\mathfrak a)$.
• $r(\sum_i \mathfrak a_i) = r(\sum_i r(\mathfrak a_i))$.
• $r(\mathfrak a \cap \mathfrak b) = r(\mathfrak {ab}) = r(\mathfrak a) \cap r(\mathfrak b)$.

Proof

Since $\mathfrak a \subseteq r(\mathfrak a)$ we have proven ⊇ of the first claim. Conversely, if $x \in r(r(\mathfrak a))$ then $x^n \in \mathfrak a$ for some n > 0 and so $(x^n)^m = x^{mn} \in \mathfrak a$ for some mn > 0. Thus $x\in r(\mathfrak a)$.

For second claim, since $\mathfrak a_i \subseteq r(\mathfrak a_i)$, ⊆ is obvious. Conversely, if x lies in the RHS then $x^n \in \sum_i r(\mathfrak a_i)$ for some n > 0, and so $x^n = y_1 + \ldots + y_k$ with $y_j \in r(\mathfrak a_{i_j})$. Without loss of generality, there is an m > 0 such that $y_j^m \in \mathfrak a_{i_j}$ for each j = 1, …, k (take m large enough). Then

$(x^n)^{mk} = (y_1 + \ldots + y_k)^{mk}$ = sum of terms of the form $y_1^{e_1} \ldots y_k^{e_k}$ with $e_1 +\ldots + e_k = mk.$

In each term, we have $e_j \ge m$ for some j, hence the term is a multiple of $y_j^m \in \mathfrak a_{i_j}$. Thus $(x^n)^{mk} \in \sum_i \mathfrak a_i$ and x lies in the LHS.

Finally for the last claim.

• Since $\mathfrak{ab} \subseteq \mathfrak a \cap \mathfrak b$ the second term is contained in the first.
• Since $\mathfrak a \cap \mathfrak b \subseteq \mathfrak a$ we have $r(\mathfrak a \cap \mathfrak b) \subseteq r(\mathfrak a)$ and similarly $r(\mathfrak a \cap \mathfrak b)\subseteq r(\mathfrak b)$ so the first term is contained in the third.
• Finally if $x\in r(\mathfrak a) \cap r(\mathfrak b)$ there exist mn > 0 such that $x^m \in \mathfrak a, x^n \in\mathfrak b$. Then $x^{m+n} \in \mathfrak{ab}$ so the third term is contained in the second. ♦

It is not true that $r(\cap \mathfrak a_i) = \cap_i r(\mathfrak a_i)$ for any class of ideals $\mathfrak a_i\subseteq A$. For example, take $A = \mathbb Z$ and $\mathfrak a_n = (2^n)$ for $n = 1, 2, \ldots$. Then $r(\mathfrak a_n) = (2)$ so

$r(\cap_{n\ge 1} \mathfrak a_n) = r((0)) = (0), \quad \cap_n r(\mathfrak a_n) = \cap_n (2) = (2).$

Definition.

An ideal $\mathfrak a \subseteq A$ is called a radical ideal if $r(\mathfrak a) = \mathfrak a$.

Note that for any ideal $\mathfrak a$, $r(\mathfrak a)$ is a radical ideal.

### Exercise.

1. Prove that a prime ideal is radical.

2. Decide which of the following is true. Find counter-examples for the false claims.

• If $\mathfrak a, \mathfrak b$ are radical ideals, so is $\mathfrak a + \mathfrak b$.
• If $\mathfrak a, \mathfrak b$ are radical ideals, so is $\mathfrak a \cap \mathfrak b$.
• If $\mathfrak a, \mathfrak b$ are radical ideals, so is $\mathfrak a\mathfrak b$.
• If $(\mathfrak a_i)$ is a collection of radical ideals, so is $\sum_i \mathfrak a_i$.
• If $(\mathfrak a_i)$ is a collection of radical ideals, so is $\cap_i \mathfrak a_i$.

[For a counter-example to the first claim, take the ring A = ℤ[X], the ring of polynomials with integer coefficients.]

# Division of Ideals

Finally, we wish to divide ideal $\mathfrak a$ by $\mathfrak b$.

Definition.

Let $\mathfrak a,\mathfrak b\subseteq A$ be ideals. Write

$(\mathfrak a : \mathfrak b) := \{ x \in A: x\mathfrak b\subseteq \mathfrak a\}.$

Here, the notation $x\mathfrak b$ means $\{xy : y\in \mathfrak b\}$; note that this is an ideal of A. As a convenient mnemonic for the definition (whether it is $x\mathfrak b \subseteq \mathfrak a$ or $x \mathfrak a \subseteq \mathfrak b$), just recall that in the ring ℤ we have (mnℤ : nℤ) = mℤ.

Lemma.

The set $\mathfrak c := (\mathfrak a : \mathfrak b)$ is an ideal of A.

Proof

Clearly $0 \in \mathfrak c$ since $0\mathfrak b = (0)$.

Next suppose $x, y \in \mathfrak c$, so $x\mathfrak b, y\mathfrak b \subseteq \mathfrak a$. Then $(x-y)\mathfrak b \subseteq x\mathfrak b + y\mathfrak b \subseteq \mathfrak a$.

Finally if $x\in \mathfrak c$, so $x\mathfrak b\subseteq \mathfrak a$, then any $z\in A$ gives us $xz \mathfrak b \subseteq z\mathfrak a \subseteq \mathfrak a$ since $\mathfrak a$ is an ideal of A. ♦

Finally, we go through some basic properties of ideal division.

Proposition

Let $\mathfrak a, \mathfrak b, \mathfrak c$ be ideals of A, and $(\mathfrak a_i), (\mathfrak b_i)$ be any collection of ideals of A.

• $(\mathfrak a : \mathfrak b)\mathfrak b \subseteq \mathfrak a$.
• $((\mathfrak a : \mathfrak b) : \mathfrak c) = (\mathfrak a : \mathfrak {bc})$.
• $(\cap_i \mathfrak a_i : \mathfrak b) = \cap_i (\mathfrak a_i : \mathfrak b)$.
• $(\mathfrak a : \sum_i \mathfrak b_i) = \cap_i (\mathfrak a : \mathfrak b_i)$.

Proof

First claim: if $x \in (\mathfrak a : \mathfrak b)$, $y\in\mathfrak b$ then by definition $xy\in \mathfrak a$. Hence $\mathfrak a$ also contains any finite sum of $x_i y_i$ with $x_i \in (\mathfrak a : \mathfrak b)$, $y_i \in \mathfrak b$.

Second claim: for $x\in A$,

$x \in ((\mathfrak a : \mathfrak b) : \mathfrak c) \iff x\mathfrak c \subseteq (\mathfrak a : \mathfrak b) \iff (x\mathfrak c)\mathfrak b \subseteq \mathfrak a \iff x \in (\mathfrak a : \mathfrak {cb}).$

Third claim: for $x \in A$,

$x\in (\cap_i \mathfrak a_i : \mathfrak b) \iff x\mathfrak b \subseteq \cap_i \mathfrak a_i \iff (\forall i, x\mathfrak b\subseteq \mathfrak a_i) \iff (\forall i, x \in (\mathfrak a_i : \mathfrak b)).$

Fourth claim: $x \in (\mathfrak a : \sum_i \mathfrak b_i) \iff x(\sum_i \mathfrak b_i) \subseteq \mathfrak a \iff (\forall i, x\mathfrak b_i \subseteq \mathfrak a),$ where the second equivalence follows from: $x\sum_i \mathfrak b_i = \sum_i (x\mathfrak b_i)$ and for any collection of ideals $\mathfrak b_i$, $\sum_i \mathfrak b_i \subseteq \mathfrak c \iff (\forall i, \mathfrak b_i \subseteq \mathfrak c)$. ♦

### Note

In the next article, we will be looking at some basic ideas in algebraic geometry to motivate many of our subsequent concepts. The concept of a radical ideal is of paramount importance there. We will not be seeing much of ideal division for a while, until we encounter invertible ideals.

## Commutative Algebra 1

Recall that we defined three operations on ideals: intersection, sum and product. We can take intersection and sum of any collection of ideals (even infinitely many of them), but we can only define the product of finitely many ideals.

As we mentioned earlier, ideals are a generalization of elements of the ring. In the case of $A = \mathbb Z$ with ideals $\mathfrak a = (m), \mathfrak b = (n)$, we have

$\mathfrak a \mathfrak b = (mn), \quad \mathfrak a + \mathfrak b = (\gcd(m,n)), \quad \mathfrak a \cap \mathfrak b = (\mathrm{lcm}(m,n)), \quad \mathfrak a\subseteq \mathfrak b \iff n | m.$

Thus, philosophically, we have the following correspondences in mind

• product of ideals ↔ product,
• sum of ideals ↔ gcd,
• intersection of ideals ↔ lcm,
• reverse containment ↔ divides.

Armed with the above picture, the following properties become quite natural.

Proposition 1.

Given any ideals $\mathfrak a, \mathfrak b$ and collection of ideals $\mathfrak b_i$ of the ring $A$, we have:

• $\mathfrak a(\sum_i \mathfrak b_i) = \sum_i (\mathfrak{ab}_i)$.
• $\mathfrak a\mathfrak b \subseteq \mathfrak a \cap \mathfrak b$.
• $(\mathfrak a \cap \mathfrak b)(\mathfrak a + \mathfrak b) \subseteq \mathfrak{ab}$.

Proof

For the first claim, ⊆ follows from: if $x\in \mathfrak a$ and $y \in \sum_i \mathfrak b_i$, then

$y = z_1 + \ldots + z_k$ for some $z_1 \in \mathfrak b_{i_1}, \ldots, z_k \in \mathfrak b_{i_k}$.

Then $xy = \sum_{j=1}^k xz_j$ where each $xz_j \in \mathfrak {ab}_{i_j}$. Hence $xy \in \sum_i (\mathfrak{ab}_i)$.

For the other containment ⊇, note that since $\mathfrak b_i \subseteq \sum_i \mathfrak b_i$ we have $\mathfrak {ab}_i \subseteq \mathfrak a\sum_i \mathfrak b_i$ and hence $\sum_i \mathfrak{ab}_i \subseteq \mathfrak a\sum_i \mathfrak b_i$.

Second claim: if $x\in \mathfrak a, y\in \mathfrak b$, then $xy \in \mathfrak a$ since it is a multiple of $x$, and similarly it lies in $\mathfrak b$ since it is a multiple of $y$. Thus $xy\in \mathfrak a\cap \mathfrak b$ and so $\mathfrak a\cap \mathfrak b$ must contain all finite sums $x_1 y_1 + \ldots + x_k y_k$ with $x_i \in \mathfrak a$ and $y_i \in \mathfrak b$.

Last claim: $(\mathfrak a \cap \mathfrak b)(\mathfrak a + \mathfrak b) = (\mathfrak a \cap \mathfrak b)\mathfrak a + (\mathfrak a \cap \mathfrak b)\mathfrak b \subseteq \mathfrak b\mathfrak a + \mathfrak a \mathfrak b = \mathfrak {ab}$. ♦

# Coprime Ideals

Suppose $m,n\in \mathbb Z$ are coprime. The following properties of $\mathbb Z$ are familiar to us.

• (Bezout’s theorem) There exist integers a, b such that $am+bn = 1$.
• Any common multiple of m and n is a multiple of mn.
• (Chinese Remainder Theorem) For any c mod m and d mod n, there is a unique a mod mn such that $a\equiv c \pmod m$ and $a\equiv d \pmod n$.
• Any powers $m^k$, $n^j$ are also coprime.

Let us generalize this to the setting of ideals.

Definition.

Let $A$ be a ring. We say that ideals $\mathfrak{a}, \mathfrak{b} \subseteq A$ are coprime if $\mathfrak{a} + \mathfrak{b} = (1) = A$.

Note

It is clear that ideals $\mathfrak a, \mathfrak b\subseteq A$ are coprime if and only if there exist $x\in \mathfrak a, y\in\mathfrak b$ such that $x+y = 1$.

Also, if $\mathfrak m$ is a maximal ideal of $A$, then it is coprime to any ideal $\mathfrak a$ not contained in it, because $\mathfrak m + \mathfrak a$ is an ideal which strictly contains $\mathfrak m$, so by the maximality of $\mathfrak m$, $\mathfrak m + \mathfrak a = A$.

The following two results give us a recipe for producing more coprime pairs of ideals given existing ones. Philosophically the results make sense if we imagine coprime to mean “not sharing any factor”.

Proposition 2.

If ideals $\mathfrak a, \mathfrak b\subseteq A$ are coprime, then so are any powers $\mathfrak a^m, \mathfrak b^n$.

Proof

Pick $x\in \mathfrak a, y\in\mathfrak b$ such that $x+y = 1$. Taking this to the (m+n)-th power, the LHS is a sum of m+n+1 terms, each of the form $Cx^i y^j$ where C is an integer, $i, j\ge 0$ are integers such that $i+j=m+n$. Since either $i\ge m$ or $j \ge n$, this term is a multiple of $x^m$ or $y^n$ so it lies in $\mathfrak a^m$ or $\mathfrak b^n$. Hence the whole sum lies in $\mathfrak a^m + \mathfrak b^n$ and $1 \in \mathfrak a^m + \mathfrak b^n$. ♦

Proposition 3.

If $\mathfrak a, \mathfrak b, \mathfrak c\subseteq A$ are ideals such that $(\mathfrak a, \mathfrak b)$ is coprime and $(\mathfrak a,\mathfrak c)$ is coprime, then $(\mathfrak a, \mathfrak {bc})$ is coprime.

Proof

Find $x\in \mathfrak a, y\in \mathfrak b$ such that $x+y = 1$. Also find $x'\in \mathfrak a, z\in \mathfrak c$ such that $x'+z = 1$. Then

$1 = (x+y)(x'+z) = \overbrace{x(x'+z) + x'y}^{\in \mathfrak a}+ yz \in \mathfrak a + \mathfrak {bc}$

and we are done. ♦

Corollary.

If $\mathfrak a, \mathfrak b_1, \ldots, \mathfrak b_n\subseteq A$ are ideals such that $\mathfrak a$ and $\mathfrak b_i$ are coprime for each $1 \le i \le n$, then $\mathfrak a$ and $\prod_{i=1}^n \mathfrak b_i$ are coprime.

Proof

Repeatedly apply proposition 3: since $(\mathfrak a, \mathfrak b_1)$ and $(\mathfrak a, \mathfrak b_2)$ are coprime pairs, so is $(\mathfrak a, \mathfrak b_1 \mathfrak b_2)$. Together with the fact that $(\mathfrak a, \mathfrak b_3)$ is coprime, we see that $(\mathfrak a, \mathfrak b_1 \mathfrak b_2 \mathfrak b_3)$ is coprime, etc. ♦

# Consequences

Finally we discuss the consequences given two coprime ideals of a ring A. Immediately we obtain Bezout’s theorem from the definition of coprimality: if $\mathfrak a,\mathfrak b$ are coprime, there exist $x\in \mathfrak a,y \in\mathfrak b$ such that $x+y = 1$. Next we have:

Proposition 4.

If $\mathfrak a,\mathfrak b\subseteq A$ are coprime ideals, then $\mathfrak{ab} = \mathfrak a \cap \mathfrak b$.

Proof

By proposition 1, $\mathfrak a \cap \mathfrak b \supseteq \mathfrak{ab} \supseteq (\mathfrak a \cap \mathfrak b)(\mathfrak a + \mathfrak b) = (\mathfrak a \cap \mathfrak b)(1) = \mathfrak a \cap \mathfrak b.$ so equality holds throughout. ♦

Corollary.

If $\mathfrak a_1, \ldots, \mathfrak a_n$ are pairwise coprime ideals of A, then $\mathfrak a_1 \cap \ldots \cap \mathfrak a_n = \mathfrak a_1 \ldots \mathfrak a_n$.

Note

In general, a collection of items is said to satisfy pairwise X if any two of them satisfy X.

Proof

When n = 1, clear. For $n\ge 2$, suppose it holds for n – 1 so that $\mathfrak a_1 \cap\ldots \cap \mathfrak a_{n-1}= \mathfrak a_1 \ldots \mathfrak a_{n-1}$. By corollary to proposition 3, $\mathfrak a_n$ and $\mathfrak a_1 \ldots \mathfrak a_{n-1}$ are coprime. By proposition 4 we have

$\mathfrak a_1 \ldots \mathfrak a_{n-1} \mathfrak a_n = (\mathfrak a_1 \cap \ldots \cap \mathfrak a_{n-1}) \mathfrak a_n = \mathfrak a_1 \cap \ldots \cap \mathfrak a_{n-1} \cap \mathfrak a_n.$

Chinese Remainder Theorem (CRT).

If $\mathfrak a, \mathfrak b\subseteq A$ are coprime ideals, then the natural map

$A/(\mathfrak{ab}) \longrightarrow (A/\mathfrak a) \times (A/\mathfrak b), \qquad a + \mathfrak{ab} \mapsto (a + \mathfrak a, a + \mathfrak b)$,

is an isomorphism.

Proof

The kernel of the map is clearly $\mathfrak a\cap \mathfrak b$ which is $\mathfrak {ab}$ by proposition 4. To prove that the map is surjective, pick $x\in \mathfrak a, y\in \mathfrak b$ such that $x+y = 1$. Now for any $(a + \mathfrak a, b + \mathfrak b)$ in the RHS, let $z := ay + bx \in A$. Since $x\equiv 1 \pmod {\mathfrak{b}}$ and $y \equiv 1 \pmod {\mathfrak{a}}$ we have

$z \equiv ay \equiv a \pmod{\mathfrak{a}}, \qquad z \equiv bx \equiv b \pmod{\mathfrak{b}}$

so $z + \mathfrak{ab}$ maps to $(a+\mathfrak a, b + \mathfrak b)$. ♦

As before, this immediately generalizes to the following.

Corollary.

If $\mathfrak a_1, \ldots, \mathfrak a_n \subseteq A$ are pairwise coprime ideals, then the natural map

$A/(\mathfrak a_1 \ldots \mathfrak a_n) \longrightarrow (A/\mathfrak a_1) \times \ldots \times (A/\mathfrak a_n)$

is an isomorphism.

Proof

For n = 1, OK. For n > 1, by corollary to proposition 3, $\mathfrak a_n$ and $\mathfrak a_1 \ldots \mathfrak a_{n-1}$ are coprime so CRT gives us

$A/(\mathfrak a_1 \ldots \mathfrak a_n) \cong A/(\mathfrak a_1\ldots \mathfrak a_{n-1}) \times A/\mathfrak a_n$

via the natural map. Now proceed inductively. ♦

The following is an important special case.

Example.

Let $\mathfrak m_1, \ldots, \mathfrak m_n$ be distinct maximal ideals of the ring A. Then their powers $\mathfrak m_1^{k_1}, \ldots, \mathfrak m_n^{k_n}$ are pairwise coprime for any $k_1, \ldots, k_n \ge 1$. Hence we have:

\begin{aligned}\mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n} &= \mathfrak m_1^{k_1} \cap \ldots \cap \mathfrak m_n^{k_n}, \\ A/(\mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}) &\cong A/\mathfrak m_1^{k_1} \times \ldots \times A/\mathfrak m_n^{k_n}.\end{aligned}

## Commutative Algebra 0

We’re starting a new series on commutative algebra. This has been in the works for way too long, and eventually we just decided to push ahead with it anyway. Most of the articles will be short, and we’ll try to illustrate the concepts with examples and diagrams whenever possible. Hopefully this will turn out well.

Since the notes are rather disorganized, repetitions will undoubtedly occur. For instance, we may introduce concept X in an earlier installation only to reintroduce it later as a special case of a more general machinery. However, we strive at clarity in explaining the concepts. Of course, the reader will be the best judge of how successful we are.

This is a summary of things you should already be familiar with, so we will proceed rather quickly.

# Basic Concepts

We assume you have some basic knowledge about rings and ideals. Here, all rings are assumed to be commutative with 1. For a ring $A$, any ideal $\mathfrak a\subseteq A$ gives us a ring quotient $A/\mathfrak a$. Ideals also allow us to talk about “modulo” relations.

Definition.

If $\mathfrak a\subseteq A$ is an ideal, then for $x,y\in A$, we write $x\equiv y\pmod {\mathfrak a}$ if their difference $x-y\in \mathfrak a$.

Note that this generalizes the concept of $x \equiv y \pmod n$ in modular arithmetic. Thanks to the definition of an ideal, we immediately get:

• If $x \equiv y \pmod {\mathfrak a}$ and $x' \equiv y' \pmod {\mathfrak a}$, then

$x+y \equiv x' + y' \pmod {\mathfrak a}, \quad xy \equiv x'y' \pmod {\mathfrak a}.$

• If $x\equiv y\pmod {\mathfrak a}$, then for all $n\ge 1$ we have $x^n \equiv y^n \pmod {\mathfrak a}$.

Proof : exercise.

This illustrates a general maxim here: that ideals generalize the concept of individual elements of a ring. We will see many more examples later. The most startling case of this is the example of a Dedekind domain, where we have unique factorization of ideals into prime ideals, instead of unique factorization of elements into primes.

## Subrings

Also, let us properly define a subring.

Definition.

A subring of the ring $A$ is a subset $B\subseteq A$ satisfying:

• $1 \in B$;
• $(B, +)$ is a subgroup of $(A, +)$;
• if $x, y \in B$, then $xy \in B$.

Note

The trivial ring $A = \{0\}$ cannot be a subring of any other ring except itself, because of the first condition. However, it can be a ring quotient since for any ring $A$, the ideal $\mathfrak a = A$ gives a trivial ring for the quotient $A/\mathfrak a$.

# Ideals

Recall that we have the following construction of ideals.

Definition

Let $A$ be any ring.

• If $\{\mathfrak a_i\}$ is a collection of ideals of $A$, their intersection $\cap_i \mathfrak a_i$ is also an ideal of $A$.
• If $\mathfrak a, \mathfrak b\subseteq A$ are ideals, their product $\mathfrak{ab}$ is the set of all finite sums $x_1 y_1 + \ldots + x_k y_k$, where $x_1, \ldots, x_k \in \mathfrak a$ and $y_1, \ldots, y_k \in \mathfrak b$.

Beginning students in algebra are often a bit bewildered by the presence of finite sums in the product of two ideals. But the definition is sensible, for the product satisfies

$\mathfrak {ab} = \mathfrak {ba}, \quad (\mathfrak{ab})\mathfrak c = \mathfrak a(\mathfrak {bc}), \quad A\mathfrak a = \mathfrak a.$

For this reason, we often write the ideal $A$ (whole ring) as $(1)$. Aesthetically this is pleasing for we get $\mathfrak a (1) = \mathfrak a$.

# Generating an Ideal

The fact that ideals are closed under arbitrary intersections is important, for we can take any subset $S\subseteq A$ and generate the smallest ideal containing it.

Definition

Consider the collection $\Sigma$ of all ideals $\mathfrak a\subseteq A$ containing $S$. Note that $\Sigma$ is non-empty since $A\in \Sigma$. The intersection of all $\mathfrak a\in \Sigma$ gives

$(S) := \bigcap_{\mathfrak a \in \Sigma} \mathfrak a,$

the ideal generated by set $S$. If $S = \{a_1, a_2, \ldots, a_n\}$, we also denote $(S)$ by $(a_1, a_2, \ldots, a_n)$.

Note

We will use the computer science notation `:=’ to refer to definition. Thus $A := B$ means the notation $A$ on the LHS is defined to be $B$ on the RHS.

An ideal generated by one element is easy to describe:

$(a) = \{ ab : b\in A\}.$

More generally, one sees that $(S)$ is given by the set of all finite sums of the form $a_1 b_1 + a_2 b_2 + \ldots + a_n b_n$ where $a_1, \ldots, a_n \in S$ and $b_1, \ldots, b_n \in A$. We can only restrict ourselves to finite sums since we are dealing with algebra here and infinite sums are not well-defined. [One can consider infinite sums in, say, functional analysis, but that is another story for another day.]

## Sums of Ideals

Now suppose $\mathfrak a_i$ is a collection of ideals of $A$. If $S = \cup_i \mathfrak a_i$, the ideal generated by S is called the sum of the ideals $\mathfrak a_i$. It is clear that the sum can also be defined as follows.

Definition.

The ideal $\sum_i \mathfrak a_i$ is the collection of all finite sums $x_1 + x_2 + \ldots + x_n$, where

$x_1 \in \mathfrak a_{i_1}, \ x_2 \in \mathfrak a_{i_2}, \ \ldots\ , x_n \in \mathfrak a_{i_n}.$

As expected, multiplication is distributive over addition.

# Integral Domains and Fields

Next, we have the following.

Definition.

An element $a$ of ring $A$ is a zero-divisor if there exists $b\in A$, $b\ne 0$ such that $ab=0$.

Note that the trivial ring $A = \{0\}$ has no zero-divisor by definition. Otherwise for non-trivial ring $A$, the element $0\in A$ is a zero-divisor since $0\cdot 1 = 0$ and $1\ne 0$Be forewarned that the trivial ring can be a pitfall for the unwary.

Definition.

The ring $A$ is an integral domain (or just domain) if it is non-trivial, and the only zero-divisor in $A$ is $0$.

Note that we have explicitly excluded the case of the trivial ring. Examples of integral domains include

• $\mathbb Z \subset \mathbb Q \subset \mathbb R \subset \mathbb C$;
• $\mathbb Z[i] = \{a + bi : a, b\in\mathbb Z\}$, where $i =\sqrt{-1}$.

Clearly any subring of an integral domain is an integral domain.

Definition.

An element $a$ of a ring $A$ is a unit if there exists $b\in A$ such that $ab=1$.

Note that the set of units of a ring A forms a group under multiplication. We denote this by $U(A)$ and call it the unit group of A. For example, $U(\mathbb Z) = \{+1, -1\}$ and $U(\mathbb Z[i]) = \{+1, +i, -1, -i\}$. Finally, we define:

Definition.

The ring $A$ is said to be a field if it is non-trivial and any non-zero element of $A$ is a unit.

For example, $\mathbb Q\subset \mathbb R\subset \mathbb C$ are all fields and $\mathbb Z$ is not a field. The following is standard, whose proof we skip.

Theorem.

• Any field is an integral domain.
• Any finite integral domain is a field.

Note

A ring A is a field if and only if it has exactly two ideals: 0 and A. [The trivial ring has only one ideal.]

# Prime and Maximal Ideals

Given an ideal $\mathfrak a$ of ring $A$, one would like to know when $A/\mathfrak a$ is an integral domain or a field.

Theorem.

• $A/\mathfrak a$ is an integral domain if and only if $\mathfrak a\ne A$ and

$x, y\in A, \ xy \in \mathfrak a \implies x \in \mathfrak a \text{ or } y\in \mathfrak a.$

• $A/\mathfrak a$ is a field if and only if $\mathfrak a\ne A$ and, for any ideal $\mathfrak b\subseteq A$ containing $\mathfrak a$, we have $\mathfrak b = \mathfrak a$ or $\mathfrak b = A$.

Note

For the first case, we say that $\mathfrak a$ is a prime ideal of A. For the second case, we say that $\mathfrak a$ is a maximal ideal of A. By definition, the whole ring $A = (1)$ is neither prime nor maximal. This should not come as a surprise, since in our definition of prime numbers, we explicitly exclude 1.

We will not prove the above theorem, but the proof of the first result is immediate. For the second result, we use the fact that a field has exactly two ideals, and that the ideals of the ring quotient $A/\mathfrak a$ are in bijection with ideals $\mathfrak b$ of $A$ containing $\mathfrak a$.

Pictorially, the lattices of ideals correspond as follows (where the arrows represent inclusion):

Furthermore, the prime and maximal ideals coincide, i.e. $\mathfrak b/\mathfrak a$ is a prime (resp. maximal) ideal of the ring $A/\mathfrak a$ if and only if $\mathfrak b$ is a prime (resp. maximal) ideal of the ring $A$. For a more concrete example, here is the correspondence between ideals of $\mathbb Z/(30)$ and ideals of $\mathbb Z$ containing $(30)$.

Thus, taking the quotient $A/\mathfrak a$ corresponds to “chopping off” ideals from lower branches in the lattice.