Show that $\phi^{-1}((y_1-a_1,\ldots,y_n-a_n))=(x_1-b_1,\ldots,x_m-b_m)$ for points in affine varieties $\psi((a_1,\ldots,a_n))=(b_1,\ldots,b_m)$

Question

Let $A=k[x_1,\ldots,x_m]$ and $B=k[y_1,\ldots,y_n]$ over some algebraically closed field $k$. Let $\psi : \mathbb{A}^n\rightarrow \mathbb{A}^m$ be the morphism corresponding to a $k$-algebra homomorphism $\phi: A\rightarrow B$. Let $a=(a_1,\ldots,a_n)\in \mathbb{A}^n$ and $\psi(a)=(b_1,\ldots,b_m)\in\mathbb{A}^m$. I want to show that $\phi^{-1}((y_1-a_1,\ldots,y_n-a_n))=(x_1-b_1,\ldots,x_m-b_m)$.

I know that since $\phi$ is a $k$-algebra homomorphism and $A$ and $B$ are finitely generated $k$-algebra, then the preimage of a maximal ideal is again a maximal ideal. I also know that by Hilberts Nullstellensatz, then a point $(a_1,\ldots,a_n)$ in a affine variety corresponds to the maximal ideal $(a_1-p_1,\ldots,x_n-a_n)$. And by how $\psi$ is defined on the point $a$, then this seems to be natural that $\phi^{-1}((y_1-a_1,\ldots,y_n-a_n))=(x_1-b_1,\ldots,x_m-b_m)$. However I was wondering if there were a more rigorously way of doing it or seeing it. My argument seems a little loose.

Answer 1

Your argument is the right idea. Let me try to give you some help.

A $k$-point in $\Bbb A^n$ is the same as a map from $\Bbb A^0\to \Bbb A^n$ picking out the point we care about. On the coordinate algebra side, this is represented by the map $k[y_1,\cdots,y_n]\to k$ given by $y_i\mapsto a_i$, and the preimage of the maximal ideal $(0)\subset k$ is exactly the maximal ideal $(y_1-a_1,\cdots,y_n-a_n)$.

Composing this with our map $\Bbb A^n\to \Bbb A^m$, we get a map $\Bbb A^0\to \Bbb A^m$ which corresponds to $k[x_1,\cdots,x_m]\to k$. The map $\Bbb A^0\to \Bbb A^m$ picks out $(b_1,\cdots,b_m)$, so we have that our map on coordinate rings $k[x_1,\cdots,x_m]\to k$ is given by $x_j\mapsto b_j$, and the preimage of the maximal ideal $(0)\subset k$ is exactly the maximal $(x_1-b_1,\cdots,x_m-b_m)$.

Now look at the sequence of maps $k[x_1,\cdots,x_m]\to k[y_1,\cdots,y_n]\to k$. The preimage of $(0)$ along the first map is $(y_1-a_1,\cdots,y_n-a_n)$, and the preimage of $(0)$ along the composite is $(x_1-b_1,\cdots,x_m-b_m)$. But this means that the preimage of $(y_1-a_1,\cdots,y_n-a_n)$ is $(x_1-b_1,\cdots,x_m-b_m)$ by, for instance, the second-to-last line of this answer.