Is it true that a morphism of affine algebraic varieties is continuous in Zariski topology? How should I proceed? thank you
Morphism of Affine Algebraic Variety
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1Hint: 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. – 2012-04-05
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0thank you for the reply.But need an explicit answer.it would be nice if you elaborate by an example. – 2012-04-05
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0That 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? – 2012-04-05
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0@MimMim: have you tried thinking of one for yourself? – 2012-04-05
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2What's your definition of a morphism? – 2012-04-06
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0Dear Sir, My Definition of morphism is a polynomial map: $$\mathbb{A}^n\rightarrow \mathbb{A}^m$$ $$x\mapsto (F_1(x),\dots,F_m(x))$$ – 2012-04-06
1 Answers
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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1Dear 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. – 2012-08-12
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0@GeorgesElencwajg Thanks. I edited my answer. – 2012-08-12
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0How do we know S is closed in X? Why are there not other functions which are 0 on S? – 2018-10-04