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Is it true that a morphism of affine algebraic varieties is continuous in Zariski topology? How should I proceed? thank you

Myshkin
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    Hint: a function is continuous if and only if the pre-image of a closed set under the function is closed. So you should try to show that the pre-image of an algebraic variety is itself an algebraic variety, by finding an ideal that it is the variety of. – mdp Apr 05 '12 at 16:57
  • thank you for the reply.But need an explicit answer.it would be nice if you elaborate by an example. – Myshkin Apr 05 '12 at 17:02
  • That might be nice, but it would be best if you tried to use Matt's hint, and then told us where you get stuck. Why do you need an explicit answer? – Mariano Suárez-Álvarez Apr 05 '12 at 22:40
  • @MimMim: have you tried thinking of one for yourself? – Clive Newstead Apr 05 '12 at 22:40
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    What's your definition of a morphism? – Dylan Moreland Apr 06 '12 at 01:57
  • Dear Sir, My Definition of morphism is a polynomial map: $$\mathbb{A}^n\rightarrow \mathbb{A}^m$$ $$x\mapsto (F_1(x),\dots,F_m(x))$$ – Myshkin Apr 06 '12 at 08:16

2 Answers2

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Let $K$ be a field. Let $\mathbb{A}^n$ and $\mathbb{A}^m$ be affine spaces over $K$. Let $X$ be a closed subset of $\mathbb{A}^n$ and $Y$ be a closed subset of $\mathbb{A}^m$.

Let $F_1,\dots,F_m \in K[X_1,\dots,X_n]$. Let $f:X \rightarrow Y$ be a morphism defined by $f(x) = (F_1(x),\dots,F_m(x))$. We prove that $f$ is continuous.

Let $T$ be a closed subset of $Y$. It suffices to prove that $f^{-1}(T)$ is closed in $X$.

Since T is a closed subset of $\mathbb{A}^m$, there exist polynomials $G_1,\dots,G_r \in K[Y_1,\dots,Y_m]$ such that $T$ is the set of common zeros of $G_1,\dots,G_r$.

Let $H_i = G_i(F_1(X_1,\dots,X_n),\dots,F_m(X_1,\dots,X_n))$ for $i = 1,\dots,r$. Let $S$ be the intersection of $X$ and the set of common zeros of $H_1,\dots,H_r$. If $f(x) \in T$, then $H_1(x) = \cdots = H_r(x) = 0$. Hence $x \in S$. Conversely if $x \in S$, then $H_1(x) = \cdots = H_r(x) = 0$. Hence $f(x) \in T$. Hence $f^{-1}(T) = S$. This completes the proof.

Makoto Kato
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    Dear Makoto, +1 for the nice and precise way this is written. Notice that your proof doesn't use that $K$ is algebraically closed nor, come to think of it, that $K$ is a field: every word you write is true for an arbitrary ring $K$. I am not criticizing your excellent answer but emphasizing its (welcome) formal character. – Georges Elencwajg Aug 12 '12 at 11:58
  • @GeorgesElencwajg Thanks. I edited my answer. – Makoto Kato Aug 12 '12 at 18:35
  • How do we know S is closed in X? Why are there not other functions which are 0 on S? – swedishfished Oct 04 '18 at 04:06
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The reverse image of the $\cap$ of an arbitrary family of sets is equal to the $\cap$ of their respective reverse images: article.

Let $R$ be a commutative ring with $1$. Let $\mathfrak{a} \subset R[X_1, \dots, X_n]$ be an ideal.

Then $V(\mathfrak{a}) = \bigcap_{g \in \mathfrak{a}} V(g)$. Thus it sufficies to prove continuity at each $V(g)$ by the link above.

But $f^{-1}(V(g)) = \{x \in R^n : f(x) \in V(g)\}$, but $f(x) \in V(g) \iff g(f(x)) = 0 \iff x \in V(g\circ f) \subset R^{n}$.

Clearly composing with $f$ puts $g \circ f \in R[X_1, \dots, X_n]$, thus taking $V(g\circ f)$ makes sense.

Thus morphisms of these "algebraic sets" over a ring are continuous. Replace $R$ with $k$ and you have your proof. Note that $V(g)$ was taken in $R^m$ (the "codomain") and $V(g\circ f)$ was taken in $R^n$, but notationally we've used "$V$" for both.

Daniel Donnelly
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