A Course on Mathematical Logic by Shashi Mohan Srivastava

Author:Shashi Mohan Srivastava
Language: eng
Format: epub, pdf
Publisher: Springer New York, New York, NY

We close this section by giving an application in algebra.

Proposition 5.5.9.

Let p > 1 be prime and

If the map

is injective, then it is surjective.

Proof.

Suppose f is not surjective. Let {a 1,…,a k } be the set of all coefficients of f 1,…,f n . Let b 1,…,b n ∈F p alg be such that (b 1,…,b n ) is not in the range of f. Let be the subfield of F p alg generated by {a 1,…,a k ,b 1,…,b n }. By Proposition 5.5.2, is finite. But now we have an injective map that is not surjective. This is impossible since is a finite set.

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