I think this must have been questioned before, but after searching, I couldn't find it.

I thought of considering a set of $\{x_1, ..., x_n\}$ such that $\lt x_1, ... x_n\gt = A$. The hypotesis shows easily that $\lt f(x_1), ... f(x_n)\gt = A$

Also thought about having an ideal $I_1=Nu(f)$ and define $I_n$ such that I take an $x_n \in A -I_{n-1}$ and $I_n=I_{n-1} \cdot \lt x_n \gt$. Since A is Noetherian, for some m, $I_m=A$, but couldn't find a way with this to show that $f^{-1}(0)=0$ or anything else that shows that f is inyective.

One last thing I thought was the inverse of the first thought: I take a set that generates A, then the inverse of that set must generate A.

All of this trouble is because I'm assumming $A$ is infinite. If $A$ is finite it's easy because they have the same amount of elements


marked as duplicate by rschwieb abstract-algebra Oct 19 '18 at 10:58

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  • 1
    $\begingroup$ You were right about this already being on the site. I found this by searching [modules] surjective noetherian. Also related. $\endgroup$ – rschwieb Oct 19 '18 at 10:58

Hint Consider the ideals $I_n = \ker f^n$.


Then $I_n \subseteq I_{n+1}$, so this sequence stabilizes at some point, say at $I_k$. Pick $x$ such that $f(x)=0$. Since $f$ is onto, we can always find $y$ such that $f^k(y) = x$. Then $y\in I_{k+1}$ since $f(x)=0$, but this ideal is $I_k$, so $f^k(y) =x =0$. Whence $f$ is injective.

  • $\begingroup$ Thank you! I didn't fully understand the end at first, but now I see that $I_k = I_{k+1}$ so, since $y \in I_{k+1}$ then $y \in I_k$. $\endgroup$ – Dani Seidler Oct 19 '18 at 9:35

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