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 $\displaystyle \mapsto $ ax is injective.

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

$\displaystyle f(x)=f(y)\Rightarrow ax=ay\Rightarrow a(x-y)=0\Rightarrow x-y=0\Rightarrow x=y$

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.

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 $\displaystyle \mapsto $ ax is *surjective.*

Peter

Re: Finite integral domains

Quote:

Originally Posted by

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

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