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Thread: Finite integral domains

  1. #1
    Super Member Bernhard's Avatar
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    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?

    Can anyone please help?

    Peter
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Finite integral domains

    Quote Originally Posted by Bernhard View Post
    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 $\displaystyle R$ :

    $\displaystyle f(x)=f(y)\Rightarrow ax=ay\Rightarrow a(x-y)=0\Rightarrow x-y=0\Rightarrow x=y$
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  3. #3
    Member ModusPonens's Avatar
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    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.
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  4. #4
    Super Member Bernhard's Avatar
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    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
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Re: Finite integral domains

    Quote Originally Posted by Bernhard View Post
    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
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