Morphism of Affine Algebraic Variety

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

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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. –  Matt Pressland Apr 5 '12 at 16:57
thank you for the reply.But need an explicit answer.it would be nice if you elaborate by an example. –  Bunuelian Trick Apr 5 '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-Alvarez Apr 5 '12 at 22:40
@MimMim: have you tried thinking of one for yourself? –  Clive Newstead Apr 5 '12 at 22:40
What's your definition of a morphism? –  Dylan Moreland Apr 6 '12 at 1:57
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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.

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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