Show that $\textrm{Hom}(V,W)\overset{*}{\rightarrow}\textrm{Hom}(k[W],k[V])$ is bijective.

Question

Let $V\subseteq\mathbb A^n(K),W\subseteq\mathbb A^m(K)$ be affine algebraic varieties and let $k[V],k[W]$ be their coordinate rings. Let $\textrm{Hom}(V,W)$ be the set of morphisms (i.e. regular mappings) $V\rightarrow W$ and let $\textrm{Hom}(k[W],k[V])$ be the set of ring homomorphisms $k[W]\rightarrow k[V].$ Each morphism $f:V\rightarrow W$ induces a ring homomorphism $f^*$ acting like $f^*(\varphi)=\varphi\circ f$ where $\varphi\in k[W]$.

Show that $\textrm{Hom}(V,W)\overset{*}{\rightarrow}\textrm{Hom}(k[W],k[V])$ is bijective.

First I want to show $*$ is injective. Let $f\in\textrm{Hom}(V,W)$ be a non-zero morphism. Then there exists $x_0$ such that $f(x_0)\ne0.$

Since $\textrm{I}(f(x_0))$ is proper ideal, there exists $g\notin\textrm{I}(f(x_0))$, hence, $g\circ f$ is non-zero.

Is my proof of injectivity correct? And how can I show that $*$ is surjective?

Answer 1

First, I need to point out a common pitfall. The coordinate rings $k[W]$ and $k[V]$ actually have the structure of $k$-algebras. We definitely have a bijection between regular maps $f : V \to W$ and $k$-algebra homs from $k[W] \to k[V]$. But that does not mean we have a bijection between regular maps and ring homs. In fact, we don't. For a concrete example of a ring hom that isn't a $k$-algebra hom, see here. Ok, on with the show!

As is mentioned in the comments, we need a bit more of an argument to check injectivity. However your broad idea can still be salvaged:

First, let $f \neq g \in \text{Hom}(V,W)$. Since $f \neq g$, we can find an $x_0$ with $fx_0 \neq gx_0$. Then the maximal ideals corresponding to $fx_0$ and $gx_0$ in $K[W]$ are distinct, so we can find a function $\varphi \in K[W]$ which is in one ideal (say, $\mathfrak{m}_{fx_0}$) but not the other ($\mathfrak{m}_{gx_0}$). But now we see $\varphi f x_0 = 0$ and $\varphi g x_0$ is not $0$. So $f^* \varphi \neq g^* \varphi$, and $f^* \neq g^*$.

As for surjectivity, it's a bit harder to figure out without having seen the argument before. The idea is to use the coordinate functions on $W$ to tell us where a point in $V$ must be sent.

Let $\varphi : k[W] \to k[V]$, and write $k[W] = k[w_1, \ldots, w_n] / I(W)$. Recall in this context $w_i$ is the function sending a point in $W$ to its $i$th coordinate.

Now we define $v_i = \varphi w_i \in k[V]$. These assemble to give us a map $f : V \to \mathbb{A}^n$ defined by $f(p) = (v_1 p, v_2 p, \ldots, v_n p)$. You should meditate some on the construction of this map, and why it should be an obvious thing to consider. We're taking a point in $V$, and seeing where it must go if we want our map to be compatible with $\varphi$.

It's now routine to check that $f$ actually sends $V$ to $W$. Moreover, we can check (and you should) that $f^* = \varphi$. Though again, we cooked up $f$ to make this work.

---

I hope this helps ^_^

Answer 2

Injectivity: If $f, g: V\to W$ are such that $f^*=g^*$. Let $p\in V$. $p$ will correspond to an ideal $I_p:=(x_1-a_1,\dots, x_n-a_n)$.

$$f^*=g^*: k[y_1\dots, y_m]/I(W)\to k[x_1\dots, x_n]/I(V)$$

Thus, ${f^*}^{-1}(I_p/I(V))={g^*}^{-1}(I_p/I(V))$ in $k[W].$

$f(p)$ is given by the maximal ideal ${f^*}^{-1}(I_p/I(V))$ in $k[W]$, which will be by the Hilbert Nullstellensatz to $(y_1-b_1,\cdots, y_m-b_m)$ for some $b_i$'s. Similarly $g(p)$ is given by the maximal ideal ${g^*}^{-1}(I_p/I(V))$ in $k[W]$.

Thus, $f(p)=g(p)$.

Surjectivity: Let $$\psi: k[y_1\dots, y_m]/I(W)\to k[x_1\dots, x_n]/I(V).$$

Let $f_i:=\psi(y_i)$ be the polynomials in $x_i$'s.

Now $f_i$ 's define a morphism $f:V\to \mathbb{A}^m$ sending $$(b_1,\dots, b_n)\mapsto (f_1(\bar{b}),\dots, f_m(\bar{b})).$$

Now check that $f$ maps $V\subset \mathbb{A}^n$ into $W\subset \mathbb{A}^m$.