Let $X$ and $Y$ be affine variety, i.e. irreducible closed subsets of some $A_K^n$ and $A_K^m$. Let $f\colon X\to Y$ be a morphism of variety, i.e. $f$ is continuous and for every open subset $V$ of $Y$ and for every regular function $g\in O_Y(V)$ (i.e. $g$ is locally a quotient of polynomial functions on $Y$: if $a$ is a point of $V$, then exits an opens subset $a\in W\subseteq V$ and exist $h,k\in A(Y)$, with $A(Y)$ be the ring of polynomial functions in $Y$, such that $g_{|W}=\frac{h}{k}_{|W}$) , then $g\circ f\colon f^{-1}(V)\to K$ is regular, i.e. $g\circ f\in O_X(f^{-1}(V))$.

Let $Z$ be a closed and irreducible subset of $X$. Let $f_{Z}\colon Z \to \overline{f(Z)}$ with $\overline{f(Z)}$ the closure of $f(Z)$ in $Y$. Then $Z$ and $\overline{f(Z)}$ are affine varieties again. I want to prove that $f_{|Z}$ is a morphism of variety in the sense of the above definition.

Now i easily see that $f_{|Z}$ is contunuous. For the other point:

Step 1: Let $h\in A(\overline{f(Z)})$. Then exist a polynomial function $H$ in $A(Y)$ such that $H_{|\overline{f(Z)}}=h.$ But $H\circ f \in O_X(X)$ since $f$ is a morphism and i know that $O_X(X)\simeq A(X)$. Then $(H\circ f)_{|Z}\in A(Z)$ but $(H\circ f)_{|Z}=h\circ f_{|Z}$.

So $h\circ f_{|Z}\in A(Z)$.

Step 2: Let $V$ be an open subset of $\overline{f(Z)}$ and let $g\colon V \to K$ be an element in $O_{\overline{f(Z)}}(V)$. I want to prove that $g\circ f_{|Z}\in O_Z(f_{|Z}^{-1}(V))$. Let $a$ be in $f_{|Z}^{-1}(V)$, then $f(a)\in V$, then $g$ is regular in $f(a)$. Then exist an open subset $f(a)\in W\subseteq V$ and $h,k \in A(\overline{f(Z)})$ such that $g_{|W}=\frac{h}{k}_{|W}$. Then $(g\circ f_{|Z})_{|{f^{-1}(W)}}=(\frac{h\circ f_{|Z}}{k\circ f_{|Z}})_{|f^{-1}(W)}$. And by step 1 $h\circ f_{|Z}$ and $k\circ f_{|Z}$ are in $A(Z)$.

1) Is my proof correct?

2) Is there a shorter way to prove that? Or, better, how can i simplify my proof?

3) The key-points that implies $f_{|Z}$ again a morphism are (a) that every polynomial function in $\overline{f(Z)}$ is the restriction of a polynomial function in $Y$, and (b) that $O_X(X)\simeq A(X)$? In other words, in my step 1 all the passages are necessary?

  • $\begingroup$ If you are not sure start from $X= V(J)$ the zero set of some prime ideal $J \subset K[x_1,\ldots,x_n]$. Closed irreducible subset means $X = V(J)=V(I(X))$ and $J$ is a prime ideal. $K[X] = K[x_1,\ldots,x_n]/J$, $K(X) = \text{Frac}(K[X])$ and $\frac{u}{v} \in K(X)$ is regular on $U$ if the denominator has no zeros $\in U$. $\endgroup$
    – reuns
    Nov 7, 2017 at 13:28

1 Answer 1


Another characterization of morphism makes what you want easier to see: a function $f: X \rightarrow Y$ is a morphism if and only if there exist polynomials $h_1, ... , h_m \in K[T_1, ... , T_n]$ such that

$$f(x_1, ... , x_n) = (h_1(x_1, ... , x_n), ... , h_m(x_1, ... , x_n))$$

for all $x = (x_1, ... , x_n) \in X$. This follows from the equivalence of categories between finitely generated $K$-algebras which are integral domains, and affine varieties.


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