# Finite integral domains

• Sep 3rd 2011, 07:21 AM
Bernhard
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 $\mapsto$ ax is an injective function. Since R is finite, this map is surjective. In particular, there is some b $\in$ 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 $\mapsto$ ax is injective?

Peter
• Sep 3rd 2011, 10:01 AM
FernandoRevilla
Re: Finite integral domains
Quote:

Originally Posted by Bernhard
My problem is as follows: how do we explicitly demonstrate that since R is finite, the map x $\mapsto$ ax is injective.

It is not necessary for $R$ to be finite. If $0\neq a\in R$, the map $f:R\to R$ , $f(x)=ax$ is injective for any integral domain $R$ :

$f(x)=f(y)\Rightarrow ax=ay\Rightarrow a(x-y)=0\Rightarrow x-y=0\Rightarrow x=y$
• Sep 3rd 2011, 10:06 AM
ModusPonens
Re: Finite integral domains
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.
• Sep 3rd 2011, 05:48 PM
Bernhard
Re: Finite integral domains
Sorry Fernando - my question was wrong!

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

Peter
• Sep 4th 2011, 12:15 AM
FernandoRevilla
Re: Finite integral domains
Quote:

Originally Posted by Bernhard
I meant: How do we explicitly show that since R is finite, the map x $\mapsto$ ax is surjective.

One way: the map $f:R\to f(R)\subset R$ , $f(x)=ax$ is bijective i.e. $\textrm{card}(R)=\textrm{card}(f(R))$ . Suppose $f:R\to R$ is not surjective, then $f(R)\subset R$ and $f(R)\neq R$ . As $R$ is finite, $\textrm{card}(R)=\textrm{card}(f(R))<\textrm{card} (R)$ which is absurd