text: # Finite integral domains
  
  • Sep 3rd 2011, 06:21 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 $\displaystyle \mapsto$ ax is an injective function. Since R is finite, this map is surjective. In particular, there is some b $\displaystyle \in$ 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 $\displaystyle \mapsto$ ax is injective?
  
  Peter
  • Sep 3rd 2011, 09:01 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 $\displaystyle \mapsto$ 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 $\displaystyle R$ :
  
  $\displaystyle f(x)=f(y)\Rightarrow ax=ay\Rightarrow a(x-y)=0\Rightarrow x-y=0\Rightarrow x=y$
  • Sep 3rd 2011, 09:06 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, 04:48 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 $\displaystyle \mapsto$ ax is surjective.
  
  Peter
  • Sep 3rd 2011, 11:15 PM
  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 $\displaystyle \mapsto$ 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