UIC Master’s Exam- Fall 2007 A3

Problem Statement: Show that a finite integral domain is a field.

Proof: Fix a\in D, a\neq 0 and consider the function \phi(x)=ax for all x\in D. We wish to show that \phi is injective, and thus a has an inverse.

Suppose that \phi(x)=\phi(y), then ax=ay which implies that a(x-y)=0. Since we are in an integral domain we know there are no zero divisors, so either a=0 or x-y=0. We know that a\neq 0 so it must be that x-y=0 which implies that x=y and so \phi is injective. Furthermore, since \phi maps D to D, and D is finite, we know that \phi is onto D. So, there is some x\in D such that \phi(x)=ax=e, thus, x=a^{-1}.

Thus, D is a field.

\Box

Reflection: This is a classic qual problem. It’s very straightforward and there are many ways to approach the proof. The key rests in the fact that D is finite, otherwise we wouldn’t be able to say that \phi is onto even though it maps D to D and is one-to-one.

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This entry was posted in Algebra, Field, Integral Domain, Math. Bookmark the permalink.

One Response to UIC Master’s Exam- Fall 2007 A3

  1. eekelly2388 says:

    As mentioned in the reflection above, there are many ways to approach this proof. Below I have written up the proof using another approach.
    Proof:
    Let D be a finite integral domain and let d\in D. If d=1 then d=d^{-1}, so suppose that d\neq 1. Then we know that all powers of d must also be elements of D. Since D is finite eventually these powers must repeat, so there exist integers m\neq n such that d^m=d^n. Since D is an integral domain we have cancellation and so d^m=d^n implies that d^{m-n}=1 (wlog we may assume that m>n). Since d\neq 1 we know that m-n>1 and it follows that d^{m-n-1}=d^{-1}. Thus, any non-zero element of D is a unit and so D is a field.
    \Box

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