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This is an excerpt (paraphrased) from Dummit and Foote (I believe 3rd edition):

Let $V \subset \mathbb{A}^n$ be an algebraic set and let $f \in k[V]$. If $J$ is the ideal generated by $I(V)$ and $x_{n+1}f-1$ in $k[x_1,\ldots,x_{n+1}]$, show that $J = I(Z(J))$.

The $J \subset I(Z(J))$ always holds, it is the $I(Z(J)) \subset J$ direction I am having trouble with.

I think have shown that $Z(J) = \{(v,\frac{1}{f(v)}):v \in V_f\}$ is but I don’t know what conclusions to draw from this.

Here, $V_f = \{v \in V: f(v) \neq 0\}$.

Possibly, one can observe that $g = g+x_{n+1}f-1$ on $Z(J)$. Also, if we fix $\frac{1}{f}$ in the $x_{n+1}$ coordinate we will get a rational fractions of polynomials in $k(x_1,\ldots,x_n)$, $g(x_1,\ldots,x_n,\frac{1}{f})$ that vanishes on $V_f$. Still not sure how to proceed.

Clarification: $k$ is not neccessarily algebraically closed.

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  • $\begingroup$ Since $V$ is an affine algebraic set, $J=I(V)$. Did you use the fact that $\sqrt{J}=I(Z(J))$? So all you need is to show that $J$ is a radical ideal. $\endgroup$ Jan 3 at 4:23
  • $\begingroup$ k is not neccessarily algebraically closed so you can’t use strong nullstellensatz. Unless I am missing something, there is nothing in the problem-formulation about k being algebraically closed. $\endgroup$
    – Ben123
    Jan 3 at 4:27
  • $\begingroup$ I see. Then maybe you should include it in the statement of your post to avoid any future confusion. $\endgroup$ Jan 3 at 4:29
  • $\begingroup$ I’ve added it @ShyamalSayak. $\endgroup$
    – Ben123
    Jan 3 at 4:31
  • $\begingroup$ Unless I am mistaken, $Z(J)$ is not $V_f \times \{1/f\}$. For example, set $V = \{0\}$ in $\mathbf{R}$, and let $f = x$. Then $J = (x,xy-1)$, whose zero locus in $\mathbf{R}^2$ is empty. $\endgroup$
    – While I Am
    Jan 3 at 4:42

1 Answer 1

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Here's a proof without using that $k$ is algebraically closed (cf. https://feryll.blogspot.com/2015/11/generic-and-explicit-algebraic.html which is not quite complete if V is reducible).

Choose a representative $f\in k[x_1,x_2,\dots,x_n]$ of the original $f\in k[V]$. Let $W=Z(J)\subset \mathbb{A^{n+1}}$. We need to prove that $I(W) = J$. We have that $J\subseteq I(W)$ by definition.

Suppose that $g\in I(W) \subset k[x_1,x_2,\dots,x_{n+1}]$. For some integer $d\geq 0$, we can write:

$$g = g_0(x_1,\dots,x_n)+x_{n+1}g_1(x_1,\dots,x_n)+\dots+(x_{n+1})^dg_d(x_1,\dots,x_n).$$

We want to show that $g\in J$, that is, $g = 0 \pmod{J}$. Since $fx_{n+1} = 1 \pmod{J}$, i.e., $f$ is invertible mod $J$, it is enough to show that $f^Ng = 0 \pmod{J}$ for some integer $N\geq 0$. We have that

$$f^dg = f^dg_0(x_1,\dots,x_n)+f^{d-1}g_1(x_1,\dots,x_n)+\dots+g_d(x_1,\dots,x_n) \pmod{J}$$

also vanishes on W. The RHS is an element of $k[x_1,x_2,\dots,x_n]$ which vanishes on $\pi(W) = V_f$, where $\pi\colon A^{n+1}\to \mathbb{A}^n$ is the projection. Thus $f^{d+1}g$ is an element of $k[x_1,x_2,\dots,x_n]$ which vanishes on $V$. Thus $f^{d+1}g \in I(V)\subset J$ so $f^{d+1}g = 0 \pmod{J}$ and hence $g=0 \pmod{J}$ as observed above.

If $V$ is irreducible, then $f^dg$ already vanishes on $V$ as well (D&F Exc. 15.2.11), but for $V$ reducible we need $f^{d+1}g$.

Here's a proof using that $k$ is algebraically closed.

By the Nullstellensatz, it is enough to show that $J$ is radical. Equivalently, we have to prove that $k[x_1,x_2,\dots,x_{n+1}]/J$ has trivial nilradical. But this ring is the following localization (see Example 2 on p. 708 in D&F):

$$k[V][x_{n+1}]/(fx_{n+1}-1) \cong k[V]_f$$

and $k[V]=k[x_1,x_2,\dots,x_n]/I(V)$ has trivial nilradical. It follows that $k[V]_f$ has trivial nilradical (cf., Prop 42 (2) in D&F).

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