map of a finite extension

Question

The following seems be well-known but I am not able to prove it. Please give me a help.

Let $A \hookrightarrow R$ be an extension of Noetherian domains such that $R$ is a finitely generated $A$-module. Then there exists an $A$-linear map $\varphi : R \to A$ such that $\varphi (1) \neq 0$.

p/s: I also expect a more general result.

Try to imitate the field extension case and clear denominators. — Mohan, Dec 18 '16 at 17:04
Could you give a detail answer? Can we replace 1 by any non-zero element $c \in R$? — Pham Hung Quy, Dec 19 '16 at 06:03

score 2 · Accepted Answer · answered Dec 19 '16 at 18:39

Let $S=A-\{0\}$ and let $K=S^{-1}A\subset S^{-1}R$. Then, we can choose a splitting $\phi:S^{-1}R\to K$ as $K$-vector spaces with $\phi(1)=1$. Let $x_1,\ldots, x_n$ generate $R$ as an $A$-module. Then we can find an $s\in S$ such that $s\phi(x_i)\in A$. Let $g=s\phi$. Easy to see that $g$ maps $R$ to $A$ an it is an $A$-module homomorphism. Also $g(1)=s\neq 0$. For your question in the comment, in general no. For example, if $A=k[x^2,x^3]\subset k[x]=R$, then you can not find such a $g$ with $g(1)=1$.

thanks, in my comment i ask for every non-zero element $c \in G$ whether a map $g: R \to A$ such that $g(c) \neq 0$. It is true as your solution. — Pham Hung Quy, Dec 20 '16 at 01:03

map of a finite extension

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