The Ideal given by the kernel of map $ f \in R \mapsto f(0,0) \in \mathbb C[X,Y]$ is not finitely generated for this particular polynomial subring

Question

In the following problem, $R$-module means left $R$-module and $R$ is a ring.

I have already proven these facts that may or not be needed:

-Show that an $R$-module $ M$ is finitely generated if and only if there exists a surjective homomorphism of $R$-modules $R^n → M$, for some $n$.

-Every quotient of a finitely generated module is again fnitely generated.

I am trying to prove the following:

Now set $R \subseteq \mathbb C[X,Y]$ be the subring of all polynomials $f \in \mathbb C[X,Y]$ that can be written as $f = g(X)+X ·h(X,Y)$. (Equivalent: a polynomial $f = \sum a_{ij}X^iY^j$ is in $R$ if and only all coeffcients $a_{0,j}=0 $ for $j > 0$). $R $ is a subring of $\mathbb C[X,Y]$.

a. Let $I \subset R$ be the kernel of the evaluation map $R \rightarrow C$ given by$ f \mapsto f(0,0)$. Show that the ideal $ I$ is not finitely generated.

b. Conclude that in general a submodule of a finitely generated module need not be finitely generated

This is my try:

a) First of all, a remark: Is it a typo that its written $f = g(X)+X ·h(X,Y)$? Shouldn't it be $f(X,Y) = g(X)+X ·h(X,Y)$? And is there any particular reason for using capital letters?

Let $ev(f):=f(0,0)$ denote the evaluation map

$I= \text{Ker (ev)}=\{f \in R: ev(f)=f(0,0)=0\}=\{ f(x,y)=g(x)+x·h(x,y): g(0)=0\}$

I guess I can proceed by absurd suppose that I is finitely generated. then $\exists$ polynomials $f_1, f_2,...,f_n \in I$ such that $\forall f \in I, f=\sum_{i=1}^{n} r_if_i$ with $r_i \in R$

b) I guess I could try to use any of the previous 3 propositions, but I don't really see how

I am stuck here, I have no idea what to do next. Any help is appreciated BTW Still waiting for a less convoluted solution

Answer 1

Here is an outline of a possible approach for proving $I$ is not finitely generated:

• First, show that for each $f \in R$, in fact you can write $f$ in the form $c + X g(X, Y)$ where $c \in \mathbb{C}$ and $g \in \mathbb{C}[X, Y]$. Also, $I$ is the ideal of polynomials of this form where $c = 0$.
• Now for the meat of the argument: define a function $\psi : I \to \mathbb{C}[Y]$ which takes $X g(X, Y) \in I$ to $g(0, Y)$. In other words, $\psi(f)$ takes the coefficients $a_{1,0}, a_{1,1}, a_{1,2}, a_{1,3}, \ldots$ of $X, XY, XY^2, XY^3, \ldots$ in $f$ and returns $a_{1,0} + a_{1,1} Y + a_{1,2} Y^2 + a_{1,3} Y^3 + \cdots$.
• Show that $\psi$ respects addition. Also, if $\lambda = c_0 + X g_0(X, Y) \in R$ and $f = X g_1(X, Y) \in I$, then $\psi(\lambda f) = c_0 \psi(f) = \lambda(0, 0) \psi(f)$.
• Now, start as you did and suppose, for sake of contradiction, that $I$ were finitely generated, i.e. suppose there were $f_1, \ldots, f_n \in I$ such that every $f \in I$ can be written as $\sum_{i=1}^n \lambda_i f_i$ for some $\lambda_i \in R$. Then by the above, $\psi(f) = \sum_{i=1}^n \lambda_i(0,0) \psi(f_i)$, which is in the $\mathbb{C}$-linear span of $\{ \psi(f_1), \psi(f_2), \ldots, \psi(f_n) \}$ (where $\mathbb{C}[Y]$ is a vector space over $\mathbb{C}$).
• From here, you should be able to derive a contradiction.

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As for part (b): Can you think of some particular $R$-module which $I$ is contained in? For that $R$-module, is it finitely generated?

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(This argument is a concrete form of a special case of the general trick I hinted at in the comments, where we reduce questions about generators of a module $M$ to questions about a spanning set of a vector space $M / JM$, where $J$ is some ideal such that $R / J$ is a field. In this particular case, we use $M = I$ and $J = I$, and consider the vector space $I / I^2$ over the field $R / I \simeq \mathbb{C}$. Then in this case, it turns out $I / I^2 \simeq \mathbb{C}[Y]$ where the isomorphism is closely related to the map $\psi$ defined above.)