# Question about exc. $15.2.21.$b in Dummit and Foote.

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.

• 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. Commented Jan 3 at 4:23
• 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. Commented Jan 3 at 4:27
• I see. Then maybe you should include it in the statement of your post to avoid any future confusion. Commented Jan 3 at 4:29
• I’ve added it @ShyamalSayak. Commented Jan 3 at 4:31
• 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. Commented Jan 3 at 4:42

## 1 Answer

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).