# Injective and Surjective Maps

Proposition 1.

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 $g \in I(W') - I(W)$, i.e. g 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?

This entry was posted in Advanced Algebra and tagged , , , , . Bookmark the permalink.

### 4 Responses to Commutative Algebra 6

1. Vanya says:

What do you mean by degree in the second example?

• limsup says:

Ya I was a little vague in that part. We have the ring $k[S^4, S^3 T, ST^3, T^4]$ there. Think of it as $k[A, B, C, D]$ where $A = S^4, B = S^3 T$ etc. So by degree 1, we mean degree-1 monomials in $A, B, C, D$. Degree 2 just means degree-2 monomials, i.e. $S^7 T, S^6 T^2, etc$.

2. Vanya says:

In the proof of the proposition, in the second bullet, there is a name clash between function f and element f \in I(W’) – I(W).

• limsup says:

Thanks. Changed it to $g$. I’ve been using too many $f$‘s.