Properties of this set $Z_f = \{(b_1, \dots, b_n) \in \mathbb{A}_k^n : f(x_1, \dots, x_m, b_1, \dots, b_n) \in I(V)\}$

Question

I want to solve the following problem:

Let $V \subset \mathbb{A}_k^m$ and $W \subset \mathbb{A}_k^n$ be varieties, with $k$ algebraically closed.

1. Let $f \in k[x_1,\dots,x_m,y_1,\dots,y_n]$. Consider the subset of $\mathbb{A}_k^n$ defined by $$ Z_f = \{(b_1,\dots,b_n) \in \mathbb{A}_k^n : f(x_1,\dots,x_m,b_1,\dots,b_n) \in I(V)\}, $$ where $I(V) \subset k[x_1,\dots,x_m]$ is the ideal of $V$. Show that $Z_f$ is a closed set of $\mathbb{A}_k^n$.
2. Let $f,g \in k[x_1,\dots,x_m,y_1,\dots,y_n]$ such that $f \cdot g \in I(V \times W)$. Show that $$W = (Z_f \cap W) \cup (Z_g \cap W)$$
3. Conclude that the ideal $I(V \times W)$ is prime and, therefore, $V \times W$ is a variety

I list below the my comments about the items above.

• The first one I did do, but I am not sure if it is correct: Take an arbitrary $(a_1,\dots,a_m) \in V$ and lets denote $$\mathfrak{f}(y_1,\dots,y_n) = f(a_1,\dots,a_m,y_1,\dots,y_n) \in k[y_1,\dots,y_n].$$ Consider $Z(\mathfrak{f})$ the set of zeros of $\mathfrak{f}$ and take $(b_1,\dots,b_n) \in Z(\mathfrak{f})$. Then, we get $$\mathfrak{f}(b_1,\dots,b_n) = 0 = f(a_1,\dots,a_m,b_1,\dots,b_n).$$ So $(b_1,\dots,b_n) \in Z_f$. On the other hand, take $(b_1,\dots,b_n) \in Z_f$, then $f(c_1,\dots,c_m,b_1,\dots,b_n) = 0$ for all $(c_1,\dots,c_m) \in V$, in particular, for $(a_1,\dots,a_m) \in V$. So $$f(a_1,\dots,a_m,b_1,\dots,b_n) = 0 = \mathfrak{f}(b_1,\dots,b_n).$$ Then, $(b_1,\dots,b_n) \in Z(\mathfrak{f})$. Thus $Z_f = Z(\mathfrak{f})$ and, consequently, $Z_f$ is closed. $\blacksquare$
• The second one is the tricky part. I tried to brute force into it, but I can't see the path that will get me on the result. I started by taking $fg \in I(V \times W)$, so $f(a_1,\dots,a_m,b_1,\dots,b_m) = 0$ for all $(a_1,\dots,a_m) \in V$ and $(b_1,\dots,b_n) \in W$. Here is where I can't get any further, I tried assuming $f \neq 0$ for all $(a_1,\dots,a_m) \in V$ and I did not see how to show the equality.
• For the third one the conclusion is easy, just take $f \cdot g$ in the ideal and show that $f$ is in there or $g$ is in there. Thus, if $I(V \times W)$ is prime, then $\dfrac{k[x_1,\dots,x_m,y_1,\dots,y_n]}{I(V \times W)}$ is an integral domain and therefore $V \times W$ is an algebraic variety.

Finally, did I prove the first item correctly? The second one, what am I not seeing?

Thank you very much!

Answer 1

Quetion: "Finally, did I prove the first item correctly? The second one, what am I not seeing? Thank you very much!"

Answer: An "algebraic variety" in the sense of Hartshorne, Chapter I assumes $k$ to be algebraically closed. If your field $k$ is not algebraically closed it is not true in general that for integral domains $k \subseteq A,B$ it follows $A\otimes_k B$ is an integral domain. There is an exercise in Hartshorne (Ex II.3.15) saying $V:=Spec(A)$ is geometrically irreducible iff $k_s⊗_k V$ is irreducible where $k_s$ is the separable closure of $k$. But checking if an explicit ideal $I$ has the property that $k_s⊗_kI $ is a prime ideal may not be an easy task. There is a similar property for reduced schemes (HH Ex.II.3.15). If $k$ is algebraically closed it follows $k_s=k$ and then $I$ is prime iff $k_s \otimes_k I =I$ is prime.

Example: As an example:

$$\mathbb{Q}(i)⊗_{\mathbb{Q}} \mathbb{Q}(i)≅\mathbb{Q}[x]/(x^2+1)⊗_{\mathbb{Q}} \mathbb{Q}[x]/(x^2+1)≅\mathbb{Q}(i)⊕\mathbb{Q}(i),$$

which is not an integral domain. Hence $S:=Spec(\mathbb{Q}(i))$ is an integral (reduced and irreducible) scheme of finite type over $T:=Spec(\mathbb{Q})$ but $S\times_T S\cong S \cup S $ ($\cup$ is the disjoint union) is not irreducible