# Exercise in Springer: Linear Algebraic Groups (1.4.8 (i) )

This is a pretty bad book to learn algebraic geometry from if you've never seen it before. I'm trying to verify the following assertion. Let $k$ be an algebraically closed field, and $X, Y$ be affine algebraic varieties. If $\psi: k[Y] \rightarrow k[X]$ is a $k$-algebra homomorphism, one defines a map $$\phi: X \rightarrow Y$$ as follows. For $x \in X$, let $F_x: k[X] \rightarrow k$ be the homomorphism given by $F_x(f) = f(x)$. Then $F_x \circ \psi$ is a homomorphism $k[Y] \rightarrow k$, so its kernel is a maximal ideal, necessarily equal to (for exactly one $y \in Y$) $\mathfrak M_y = \{ g \in k[Y] : g(y) = 0\}$. We then set $y = \phi(x)$.

The claim is that $\phi$ is a continuous function.

I haven't gotten too far with this. A closed set in $Y$ looks like $\mathcal V_Y(I)$ for $I$ a radical ideal of $k[Y]$, where $\mathcal V_Y(I)$ is the set of $y \in Y$ for which $h(y) = 0$ for all $h \in I$. To say that $\phi(x) = y$ just means that $\psi(g)(x) = g(y)$ for all $g \in k[Y]$.

So the preimage $\phi^{-1} \mathcal V_Y(I)$ (which I want to show is closed) is the set of all $x \in X$ with the following property: there exists a $y \in Y$ with the property that $h(y) = 0$ for all $h \in I$, and that $g(y) = \psi(g)(x)$ for all $g \in k[Y]$. In particular $\psi(g)(x) = 0$ whenever $g \in I$.

Let $S$ be any subset of $k[Y]$ and $$V(S) = \{ y\in Y: f(y) = 0\ \ \forall f\in S\}.$$

Now we show

$$\phi^{-1} V(S) = V(\psi(S))$$

(Thus $\phi^{-1} V(S)$ is closed and so $\phi$ is continuous).

Let $x\in V(\psi (S))$. Then the maximal ideal $m_x$ contains $\psi(S)$. In particular,

$$S \subset \psi^{-1} \psi(S) \subset \psi^{-1} m_x.$$

Now note that for $\phi(x) = y$, $m_y$ must contains $\psi^{-1} m_x$. So $S \subset m_y$ and $x\in \phi^{-1}V(S)$.

On the other hand, let $x\in \phi^{-1} V(S)$. Then $\phi(x) \in V(S)$ and so $m_{\phi(x)}$ contains $S$. So $\psi (m_{\phi(x)})$ contains $\psi(S)$. As $m_x$ contains $\psi m_{\phi(x)}$ by construction of $\phi$, we have $\psi(S) \subset m_x$ and so $s\in V(\psi(S))$.

• Thanks for your answer. Does this do it for us? The closed sets of $X$ are of the form $V(J)$, where $J$ is an ideal of $k[X]$. But if $S$ is an ideal of $k[Y]$, the image $\psi S$ need not be a ideal of $k[X]$, in the case where $\psi$ is not surjective. – D_S Feb 22 '15 at 6:04
• $V(S) = V(I)$ where $I$ is the ideal generated by $S$. @D_S – user99914 Feb 22 '15 at 6:19
• Ah of course! Thanks so much! – D_S Feb 22 '15 at 20:46