Let $\Bbbk$ be an algebraically closed field of characteristic $0$ (just to keep it simple). Assume we are given three affine sets $X\subseteq\mathbb{A}_{\Bbbk}^n,Y\subseteq\mathbb{A}_{\Bbbk}^m$ and $Z\subseteq\mathbb{A}_{\Bbbk}^l$ and two polynomial maps $f:X\to Z$ and $g:Y\to Z$. I would like to show that $$X\times_Z Y:=\left\{(a,b)\in X\times Y\mid f(a)=g(b)\right\}$$ is an affine set by providing explicitly its vanishing ideal.

Assume that $\Bbbk[X]=\Bbbk[X_1,\ldots,X_n]/\langle p_1,\ldots,p_r\rangle, \Bbbk[Y]=\Bbbk[Y_1,\ldots,Y_m]/\langle q_1,\ldots,q_s\rangle$ and that $\Bbbk[Z]=\Bbbk[Z_1,\ldots,Z_l]/\langle o_1,\ldots,o_t\rangle$. My guess is that $$\mathcal{J}(X\times_Z Y)=\langle P_1,\ldots,P_n,Q_1,\ldots,Q_m,Z_1( F-G),\ldots,Z_l(F-G)\rangle\subseteq \Bbbk[X_1\ldots,X_n,Y_1\ldots,Y_m],$$ where $$ P_i(X_1,\ldots,X_n,Y_1,\ldots,Y_m)=p_i(X_1,\ldots,X_n), \\ Q_j(X_1,\ldots,X_n,Y_1,\ldots,Y_m)=q_j(Y_1,\ldots,Y_m), \\ \big(Z_k(F-G)\big)(X_1,\ldots,X_n,Y_1,\ldots,Y_m) = f_k(X_1,\ldots,X_n)-g_k(Y_1,\ldots,Y_m) $$ (the $k$-th component of the polynomial functions). It's not difficult to show that $$ \mathcal{J}(X\times_Z Y)\supseteq\langle P_1,\ldots,P_n,Q_1,\ldots,Q_m,Z_1( F-G),\ldots,Z_l(F-G)\rangle $$ but I am having some troubles in showing the converse.

Does anybody have an idea?


Thinking of $X\times_Z Y$ embedded in $\mathbb{A}^{n}\times \mathbb{A}^m$ you can write down the fiber product as an intersection $$ \mathbb{A}^n \times Y \cap X\times \mathbb{A}^m \cap \{(a,b)\mid f(a)=g(b) \} $$ and use the Nullstellensatz. You'll get the radical of the ideal you wrote.

However, for this ideal to be radical or not depends on $f$ and $g$ (assuming $X$ and $Y$ reduced).

Example: Consider $X = \mathcal{V}(y-x^2)\subset \mathbb{A}^2$, $Y = \mathcal{V}(z) \subset \mathbb{A}^1$, $Z = \mathbb{A}^1$, $f(x,y)=y$ and $g(z)=z$. Then in $k[x,y,z]$ we have

$$ \left< y-x^2, z, y-z \right> = \left< x^2, z, y \right> $$ which is not radical. Geometrically we are restricting the projection $f$ to the parabola $X$ and $X\times_Z Y$ is the fiber over the point $\mathcal{V}(z) \in \mathbb{A}^1$ that is a double point. Note that in this case $\mathcal{J}(X\times_Z Y) = \left< x, z, y \right> $.

If we choose $g(z)=z+1$, then $$ \left< y-x^2, z, y-z-1 \right> = \left< 1-x^2, z, y-1 \right> $$ which give you two distinct points and the ideal is radical.

  • $\begingroup$ Let me see if I understood your point. You are saying that $\mathcal{J}(X\times_Z Y)=\sqrt{\mathcal{J}(X\times \mathbb{A}^m)+\mathcal{J}(\mathbb{A}^n\times Y)+\mathcal{J}(\{(a,b)\mid f(a)=g(b)\})}$, aren't you? And that now there are few things that I can say in general about the radical of the sum. Thus, how can I prove (in a very elementary way) that $X\times_Z Y$ is an affine set? Can I, in fact? $\endgroup$ – Ender Wiggins Sep 3 '18 at 7:52
  • $\begingroup$ Indeed I can: $X\times_ZY\subseteq \mathcal{V}\mathcal{J}(X\times_ZY)=\mathcal{V}(P_i,Q_j,Z_k(F-G))=\{(a,b)\in\mathbb{A}^{n+m}\mid p_i(a)=P_i(a,b)=0,q_j(b)=Q_j(a,b)=0,f_k(a)-g_k(b)=Z_k(F-G)(a,b)=0\}=X\times_ZY$, isn't it? Whence in fact $$X\times_ZY=\mathcal{V}(P_i,Q_j,Z_k(F-G)).$$ $\endgroup$ – Ender Wiggins Sep 3 '18 at 8:01
  • $\begingroup$ Just included an example. Tell me if it helps. $\endgroup$ – Alan Muniz Sep 3 '18 at 12:41
  • 1
    $\begingroup$ Perfectly clear, thanks. I realized that I asked the question in the wrong way. I originally wanted to show that $X \times_Z Y$ is an affine set. For this I don't need the vanishing ideal, but any set of polynomials whose vanishing set is $X \times_Z Y$ and the one I guessed does the job. In fact, I made the question misleading by involving the vanishing ideal. Nevertheless, your answer is pertinent: +1 and my gratitude for the effort. $\endgroup$ – Ender Wiggins Sep 5 '18 at 12:02

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