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Let $A \subset B$(integral domain), $B$ is finitely generated over $A$. Let $y_1, \cdots, y_n \in B$ algebraically independent over $A$. Then homomorphism $f:A \to \Omega$(algebraically closed field) can be extended to $A[y_1,\cdots,y_n] \to \Omega$ by $y_i \mapsto 0$ for all $i$.

How can I prove it other than by using direct method? Is there any theorem or generalization? (I skimmed serge lang but failed to find similar one, all theorems looks slightly different.)

How about the case of $y_1, \cdots, y_n$ algebraically dependent?

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up vote 2 down vote accepted

I can answer the last bit. If, say, $y_1=y_2+1$, and your extension takes both $y_1$ and $y_2$ to zero, then it has to take $1$ to zero, which your original $f$ maybe didn't, contradiction. I'm sure you can see from that how to treat other instances of algebraic dependence.

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Thanks. I see that it doesn't have the universal property in the case of algebraic dependence. – Gobi May 9 '11 at 1:48

HINT $\ $ Employ the universal mapping property of polynomial rings.

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Thanks, I got it. Let $\phi: R \to S$ be any ring homomorphism and let $s \in S$ be any element of $S$. Then there is a unique ring homomorphism extension $R[x] \to S$ such that $x \mapsto s$ – Gobi May 9 '11 at 1:50

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