# $A$ Noetherian and $f:A \rightarrow A$ suryective then $f$ inyective [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 19 '18 at 10:58

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

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

Solution.

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.

• 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 $y \in I_k$. – Dani Seidler Oct 19 '18 at 9:35