Commutative Algebra 6

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.


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.


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. ♦


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).

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s