# Coordinate ring homomorphisms extend to form local ring

This is problem 2.21 from Fulton's Algebraic Curves:

Let $\varphi:V\to W$ be a polynomial map of affine varieties, $\tilde{\varphi}:\Gamma(W)\to\Gamma(V)$ the induced map on coordinate rings. Suppose $P\in V$, $\varphi(P)=Q$. Show that $\tilde{\varphi}$ extends uniquely to a ring homomorphism (also written $\tilde{\varphi}$) from $\mathcal{O}_Q(W)$ to $\mathcal{O}_P(V)$. Show that $\tilde{\varphi}(\mathfrak{m}_Q(W))\subset\mathfrak{m}_P(V)$.

My approach so far for the first part is that since $k\subset\Gamma(V)\subset\mathcal{O}_P(V)\subset k(V)$, and similarly $k\subset\Gamma(W)\subset\mathcal{O}_Q(W)\subset k(W)$, one can naturally extend $\tilde{\varphi}$ to local rings by considering the inclusion $g\in\Gamma(W)\mapsto\frac{g}{1}\in\mathcal{O}_Q(W)$ (and similarly $f\in\Gamma(V)\mapsto\frac{f}{1}\in\mathcal{O}_P(V)$). At this point I'm stuck, and I'm not even sure if this is a smart approach.

• $\mathcal{O}_P$ is the stalk of the structure sheaf at $P$? Stalks are colimits, and colimits are functors, so you can show you have a morphism between the stalks if you have a morphism between the sheaves. – ziggurism Nov 14 '17 at 3:21
• Elements are fractions with denominators nonvanishing at the point. You can pull back each of the numerator and denominator. Check it is well defined (ie up to rewriting fractions) and a homomorphism. If the numerator vanishes at the point in the target then its pullback vanishes at the point in the domain. – Zach Teitler Nov 14 '17 at 3:31

The map $\tilde{\varphi}$ is given by pre-composition: \begin{align*} \tilde{\varphi}:\Gamma(W) &\to \Gamma(V)\\ f &\mapsto f \circ \varphi \, . \end{align*} As mentioned in the comments, elements of $\DeclareMathOperator{\O}{\mathcal{O}} \O_Q(W)$ are of the form $g = \frac{g_1}{g_2}$ where $g_1, g_2 \in \Gamma(W)$ and $g_2(Q) \neq 0$. We can extend $\tilde{\varphi}$ to $\O_Q(W)$ simply by applying our previous definition of $\tilde{\varphi}$ to the numerator and denominator: $$\DeclareMathOperator{\varphit}{\tilde{\varphi}} \varphit\left(\frac{g_1}{g_2}\right) := \frac{\varphit(g_1)}{\varphit(g_2)} \, .$$
Since $\varphi(P) = Q$, then $$(\tilde{\varphi}(g_2))(P) = (g_2 \circ \varphi)(P) = g_2(Q) \neq 0$$ so $\varphit\left(\frac{g_1}{g_2}\right)$ is indeed an element of $\O_P(V)$. You can check that if $g_1/g_2 = h_1/h_2$, our definition for $\varphit$ gives the same answer on both representatives, hence is well-defined.
You can prove the statement about the maximal ideals in a similar way, again by using the fact that $\varphi(P) = Q$.