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Exercise 13.4.6
As in the proof of Theorem 13.4.5, suppose that we have and is an irreducible factor of when is regarded as an element of . In this exercise we will study how close is to being an irreducible factor of in .
- (a)
- We know that the rings and are both UFDs. Prove that if is irreducible and nonconstant, then it is also irreducible when regarded as an element of .
- (b)
- Prove that and are as above, then is a -multiple of an irreducible factor of in .
Proof.
- (a)
-
Assume that
is irreducible and nonconstant.
Suppose that , where are in . There are positive integers such that (it is not needed to take as small as possible). Then , where are in .
Write the decomposition of in prime integers. Note that every prime in is irreducible in , so that
is a decomposition in irreducible factors in . The only units of this UFD are , so that the unicity of the decomposition in irreducible factors in shows that the decompositions of are in the form
or
Therefore or is in , thus or is in , so is a unit in . This proves that is irreducible in .
- (b)
- Since is UFD, can be uniquely factorized , where and are irreducible. By the part (a), are irreducible when regarded as elements of . Hence for any irreducible factor of , there exists such that . It follows that is associate to , , is the product of by a suitable constant from , where is a factor of irreducible in .
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