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I'm going through the Rings section of Abstract Algebra by Dummit and Foote, and I have a question about an early Corollary's proof.

Proof: Let R be a finite integral domain and let a be a nonzero element of R. By the cancellation law the map x->ax is an injective function. Since R is finite this map is also surjective...

How does R being finite make the map surjective?

Jonathan Hebert
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  • Note that on the right side of this page under Related you will find that the question has been answered on MSE several times. – André Nicolas Dec 30 '14 at 04:12
  • @AndréNicolas maybe the list changes slightly, but I do not actually see any related questions addressing the main question. The title question has of course been answered several times, but in all cases I'm aware of, the solution to the question here was taken as granted. So we have a surprise case where the real related question is not well-detected by the software :) – rschwieb Dec 30 '14 at 11:35
  • Related: http://math.stackexchange.com/questions/62548 – Watson Jan 19 '17 at 13:04

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Since there are only so many elements, and it is injective (one-to-one) and the domain and range have the same cardinality. Hence it must be surjective. There is not 'space' for it to be otherwise.

user141592
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