Thread: Finite integral domains

I am reading Dummit and Foote Chapter 10 - Introduction to Rings

On page 228 Corollary 3 reads:

Any finite integral domain is a field

Proof:

Let R be a finite integral domain and let a be a non-zero element of R.

By the cancellation law the map x ax is an injective function. Since R is finite, this map is surjective. In particular, there is some b such that ab = 1, i.e. a is a unit in R. Since a was an arbitrary element, R is a field.


My problem is as follows: how do we explicitly demonstrate that since R is finite, the map x ax is injective?

Can anyone please help?

Peter

Originally Posted by Bernhard

My problem is as follows: how do we explicitly demonstrate that since R is finite, the map x ax is injective.

It is not necessary for to be finite. If , the map , is injective for any integral domain :

You probably mean surjective in your question. Just make a venn diagram of an injective function from a 4 element set and another 4 element set. You'll get the principle by doing that.

Sorry Fernando - my question was wrong!

I meant: How do we explicitly show that since R is finite, the map x ax is surjective.

Peter

Originally Posted by Bernhard

I meant: How do we explicitly show that since R is finite, the map x ax is surjective.

One way: the map , is bijective i.e. . Suppose is not surjective, then and . As is finite, which is absurd