Show $I(V(f))=(f)$ when $f$ is irreducible.

Question

This is a question from Perrin's text, and it goes like this: Let $k$ be algebraically closed. Let $F\in k[x,y]$ be an irreducible polynomial. Assume that $V(F)$ is infinite. Prove that $I(V(F))=(F)$.
Here, $V(f)$ is the set of zeroes of the polynomial, and $I(V(F))$ is the ideal $(V(F))$.

Proof: Since $k$ is a field, $k[x,y]$ is a unique factorization domain. So if $F$ is an irreducible polynomial in $k[x,y]$ that's the same as being a prime element. So $F$ generates a prime ideal. Next, we have that $\textbf{rad}(F)=(F)$, since $(F)$ is a prime ideal. So by the Nullstellensatz, $(F)=I(V(F))$.

That is my proof for the problem, but nowhere in my proof did I use the hypothesis that $V(F)$ was infinite. So I was wondering if my proof is valid, or if I made some wrong assumption along the way. Also this proof would work for $k[x_1,...,x_n]$, and the problem only asks for $k[x,y]$, so I'm extra dubious about its correctness.

Answer 1

I’ve just worked through this problem myself, and as posed Perrin doesn’t specify that the field is algebraically closed. In fact it doesn’t need to be, and this is where the condition that $V(F)$ is infinite is used.

There is a lemma (proved in Section 1.1 of Shafarevich’s Basic Algebraic Geometry 1) which states:

For an arbitrary field $k$, let $F,G\in k[X,Y]$ with $F$ irreducible. If $F\nmid G$ then the system of equations $F(x,y)=G(x,y)=0$ has only finitely many solutions.

Now, clearly $(F)\subseteq I(V(F))$. Since $k[X,Y]$ is Noetherian we can write $I(V(F))=(G_1,\ldots, G_n)$ for some $G_i\in k[X,Y]$. We then have $$V(F)=V(I(V(F)))=V(G_1,\ldots,G_n)$$

Then each point $(x,y)\in V(F)$ satisfies $F(x,y)=G_i(x,y)=0$. Since $V(F)$ is infinite, then by the lemma we have $F\mid G_i$ and so $G_i=H_iF$ for some $H_i\in k[X,Y]$. This yields $$I(V(F))=(G_1,\ldots,G_n)=(H_1F,\ldots,H_nF)\subseteq(F)$$

So $I(V(F))=(F)$ and we are done.

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Update:

As noted in the comments, I've been meaning to mention that we don't need to appeal to Shafarevich's book for the lemma, Perrin proves Theorem 5.1 in Section I, which is sufficient for this exercise.

Answer 2

Your proof is correct. In fact, one doesn't need to assume that $V(F)$ is infinite. One can prove it from the other assumptions. Thus you shouldn't need to use it in your proof.

Claim: If $k$ is algebraically closed, and $f\in k[x_1,\ldots,x_n]$, with $n\ge 2$ is a nonunit, then $V(f)$ is infinite.

Proof of claim

First note that since $k$ is algebraically closed, it is infinite. Let $f\in k[x_1,\ldots,x_n]$, where $n\ge 2$. If $f$ is zero, $V(f)$ is clearly infinite, so assume $f\ne 0$. Then $f$ has positive degree in some variable, which we can assume without loss of generality is $x_1$. Let $d$ be the degree of $f$ with respect to $x_1$. Now write $$f=\sum_{i=0}^d g_i(x_2,\ldots,x_n)x_1^i,$$ for some polynomials $g_i\in k[x_2,\ldots,x_n]$, with $g_d\ne 0$. Then for every choice of $a_2,\ldots,a_n$ with $g_d(a_2,\ldots,a_n)\ne 0$, there exists $a_1$ such that $f(a_1,a_2,\ldots,a_n)=0$ since $k$ is algebraically closed.

Thus we just need to show that there are infinitely many choices of $a_2,\ldots,a_n$ with $g_d(a_2,\ldots,a_n)\ne 0$. However, this is much easier. Since $g_d\ne 0$, there is some point $a\in \Bbb{A}_k^{n-1}$ with $g_d(a)\ne 0$. Then choose any line $L$ through $a$. $g_d|_L$ will be a nonzero polynomial on an affine line, and therefore $g_d$ will only have finitely many roots on this line. Hence by infiniteness of $k$, $g_d$ will have infinitely many nonroots on that line.