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Math Help - 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  \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?

    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  \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
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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  \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  \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
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