# More Concepts in Algebraic Geometry

As before, k denotes an algebraically closed field.

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 1.

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 proposition 1 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 2.

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 A

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 1.

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 3.

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

Proof

Suppose V is irreducible and $f, g\in A- I(V)$. 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 B

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 2.

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

Proof

Exercise. ♦

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.

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### 7 Responses to Commutative Algebra 4

1. Vanya says:

It would be useful to mention that k is algebraically closed at the beginning of the post.

• limsup says:

Good point. Done.

2. Vanya says:

Could you explain how to obtain the radical of the ideal”a” in the secret example.

• Vanya says:

“Second example

• limsup says:

By observation. Note that $X-1, Y-1\in r(\mathfrak a)$. Also $(X-1, Y-1)$ is a maximal ideal so $r(\mathfrak a)$ cannot contain anything more.

3. Vanya says:

The first part of the proof of the statement “The prime ideals of A correspond to the irreducible closed subspaces of \mathbb A^n_k.” does not seem entirely right. Should there be a comma between f and g? like “Suppose V is irreducible and $f, g \not \in I(V)$. Then the closed subsets …’

• limsup says:

Thanks good catch!