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

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?

• $Hom(V, W)$ is not an abelian group. So 'injectivity' will require more argument. Aug 22 '21 at 17:48

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 ^_^

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 $$W\subset \mathbb{A}^m$$.