Exercise 1.28

Question

Let $f_1,\ldots,f_m$ be elements of $k[t_1,\ldots,t_n]$. They determine a \emph{polynomial mapping} $\varphi \colon k^n 	o k^m$: if $x \in k^n$, the coordinates of $\varphi(x)$ are $f_1(x),\ldots,f_m(x)$.
  \par Let $X,Y$ be affine algebraic varieties in $k^n,k^m$ respectively. A mapping $\varphi\colon X 	o Y$ is said to be \emph{regular} if $\varphi$ is the restriction to $X$ of a polynomial mapping from $k^n$ to $k^m$.
  \par If $\eta$ is a polynomial function on $Y$, then $\eta \circ \varphi$ is a polynomial function on $X$. Hence $\varphi$ induces a $k$-algebra homomorphism $P(Y) 	o P(X)$, namely $\eta \mapsto \eta \circ \varphi$. Show that in this way we obtain a one-to-one correspondence between the regular mappings $X 	o Y$ and the $k$-algebra homomorphisms $P(Y) 	o P(X)$.

Amit Awasthi · Accepted Answer

\begin{proof}
  We have defined a map $\alpha\colon\operatorname{Hom}(X,Y) 	o \operatorname{Hom}(P(Y),P(X))$. This map is injective since $\eta(\varphi_1) = \eta(\varphi_2)$ for all polynomial functions $\eta$ on $Y$ implies $\varphi_1 = \varphi_2$, by considering $\eta$ to be the coordinate functions $\xi_i$.
  \par It remains to show $\alpha$ is surjective. Suppose we have a homomorphism $h \colon P(Y) 	o P(X)$; denote $	ilde{h}\colon k[t_1,\ldots,t_m] 	o P(X)$ as the lift of $h$ to $k[t_1,\ldots,t_m]$. Consider the elements $	ilde{h}(t_i) \in P(X)$. $	ilde{h}(t_i)$ is the image of a polynomial in $k[t_1,\ldots,t_n]$, and so lifting each $	ilde{h}(t_i)$ to some $H_i \in k[t_1,\ldots,t_n]$, we get a regular map $\psi \colon X 	o k^m$ defined by $x \mapsto (H_1(x),\ldots,H_n(x))$.
  \par We need to show $\psi(X) \subseteq Y$, i.e., that for any $x \in X$ and any $f \in I(X)$, $f(\psi(x)) = 0$. But $f(\psi(x)) = f(H_1(x),\ldots,H_n(x))$, and $f$ is a polynomial with $	ilde{h}$ a $k$-algebra homomorphism, hence $f(H_1(x),\ldots,H_n(x)) = 	ilde{h}(f)(x) = 0$ since $f \in I(Y)$. Now $\alpha$ is surjective since we have $\alpha(\psi) = h$:
  \begin{equation*}
    \alpha(\psi)(\eta) = (\eta \circ \psi) = \eta \circ (H_1,\ldots,H_n) = \eta(H_1,\ldots,H_n) = h(\eta).\qedhere
  \end{equation*}
\end{proof}

Exercise 1.28

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Exercise 1.28

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