# Homomorphism of coordinate rings induces a polynomial map

I have a question about the proof of Proposition 2 on page 26 of Fulton's algebraic curves.

Let $\mathbb V\subset A^n$ and $W\subset \mathbb A^m$ be affine varieties, and let $\Gamma(V)$ and $\Gamma(W)$ be their coordinate rings.

Suppose that we have a homomorphism $\alpha\colon \Gamma(V)\longrightarrow\Gamma(W)$. We want to show that there is a polynomial map from $V$ into $W$ which induces $\alpha$.

Choose $T_1,...,T_m\in k[X_1,...,X_n]$ with the property $\alpha(\overline{X_1})=\overline{T_1},\ldots,\alpha(\overline{X_m})=\overline{T_m}$, where bars denote taking residues in $\Gamma(V)$ and $\Gamma(W)$. We thus get a polynomial map $T=(T_1,\ldots,T_m)\colon\mathbb A^n\longrightarrow \mathbb A^m$. This induces a homomorphism $\widetilde{T}\colon k[X_1,\ldots,X_m]\longrightarrow k[X_1,\ldots,X_n]$. Fulton says that it is easy to check that $\widetilde{T}(I(W))\subset I(V)$. Why should this be true?

The first thing to point out is that this correspondence reverses the order. $$\alpha\colon \Gamma(W)\longrightarrow\Gamma(V) \Leftrightarrow T\colon V \longrightarrow W$$

You have the polynomial map $T\colon\mathbb A^n\longrightarrow \mathbb A^m$ and the induced map $\widetilde{T}\colon k[Y_1,\ldots,Y_m]\longrightarrow k[X_1,\ldots,X_n]$ is given by $$g(Y_1,\ldots,Y_m) \mapsto (g\circ T) (X_1,\ldots,X_n) = g(T_1(X_1,\ldots,X_n),\ldots,T_m(X_1,\ldots,X_n))$$ where $\alpha(\overline Y_i) = \overline T_i$. (Here I just changed the notation to distinguish the variables)

If $g \in I(W)$, $g \equiv 0 \pmod{I(W)}$, hence $\mod{I(V)}$ we get $$0 = \alpha(\bar{g}) = \alpha(g(\overline Y_1,\ldots,\overline Y_m)) = g(\alpha(\overline Y_1),\ldots, \alpha(\overline Y_m)) = g(\overline T_1,\ldots,\overline T_m)$$ which means that $\widetilde{T}(g) = g\circ T \in I(V)$.

• why $\alpha$ and $g$ commute?
– lou
Jun 23 at 21:05
• Because $\alpha$ is a morphism of $k$-algebras, hence identity on constants. Jun 23 at 21:10
• Is it correct ? Let $f\in I(W)$ such that $$f=\sum_{i=1}^{d}a_i{Y_1}^{i_1}{Y_2}^{i_2}\dots{Y_m}^{i_m}.$$ Since $\overline{f}=0$ in $\Gamma(W)$ we will have $\alpha(\overline{f})=0$ in $\Gamma(V)$. But $\alpha(\overline{f})| =\alpha(\sum_{i=1}^{d}\overline{a_i{Y_1}^{i_1}{Y_2}^{i_2}\dots{Y_m}^{i_m}}) =\sum_{i=1}^{d}\overline{a_i}\alpha(\overline{{Y_1}^{i_1}}) \alpha(\overline{{Y_2}^{i_2}})\dots\alpha({Y_m}^{i_m}) =\sum_{i=1}^{d}\overline{a_i} \overline{T_1}^{i_1}\overline{T_2}^{i_2}\dots\overline{T_m}^{i_m} =\overline{\tilde{T}\circ f}$, $\overline{\tilde{T}\circ f}=0$ on $\Gamma(V)$. Aug 27 at 13:53

In continuation of the answer given by @Alan I would like to say that it is easier to visualize like this- let $$f\in I(W)$$ such that $$f=\sum_{i=1}^{d}a_i{Y_1}^{i_1}{Y_2}^{i_2}\dots{Y_m}^{i_m}.$$ Since $$\overline{f}=0$$ in $$\Gamma(W)$$ we will have $$\alpha(\overline{f})=0$$ in $$\Gamma(V)$$. But $$\alpha(\overline{f})=\alpha(\sum_{i=1}^{d}\overline{a_i{Y_1}^{i_1}{Y_2}^{i_2}\dots{Y_m}^{i_m}})=\sum_{i=1}^{d}\overline{a_i}\alpha(\overline{{Y_1}^{i_1}}) \alpha(\overline{{Y_2}^{i_2}})\dots\alpha({Y_m}^{i_m})=\sum_{i=1}^{d}\overline{a_i} \overline{T_1}^{i_1}\overline{T_2}^{i_2}\dots\overline{T_m}^{i_m}=\overline{\tilde{T}\circ f}$$ i.e., $$\overline{\tilde{T}\circ f}=0$$ on $$\Gamma(V)$$. Hence, $$\tilde{T}\circ f\in I(V)$$.

Is it correct? It was too long for a comment.