Exercise 13.4.6

As in the proof of Theorem 13.4.5, suppose that we have s u ( y ) [ u 1 , . . . , u n , y ] and h [ u 1 , . . . , u n , y ] is an irreducible factor of s u ( y ) when s u ( y ) is regarded as an element of [ u 1 , . . . , u n , y ] . In this exercise we will study how close h is to being an irreducible factor of s u ( y ) in [ x 1 , . . . , x n , y ] .

(a)
We know that the rings [ x 1 , . . . , x n , y ] and [ x 1 , . . . , x n , y ] are both UFDs. Prove that if f [ x 1 , . . . , x n , y ] is irreducible and nonconstant, then it is also irreducible when regarded as an element of [ x 1 , . . . , x n , y ] .
(b)
Prove that s u ( y ) and h are as above, then h is a -multiple of an irreducible factor of s u ( y ) in [ x 1 , . . . , x n , y ] .

Proof.

(a)
Assume that f [ x 1 , . . . , x n , y ] is irreducible and nonconstant.

Suppose that f = gh , where g , h are in [ x 1 , , x n , y ] . There are positive integers r , s such that rg , sh [ x 1 , , x n , y ] (it is not needed to take r , s as small as possible). Then rsf = ( rg ) ( sh ) = g 1 h 1 , where g 1 = rg , h 1 = sh are in [ x 1 , , x n , y ] .

Write rs = p 1 a 1 p r a r the decomposition of rs in prime integers. Note that every prime in is irreducible in [ x 1 , , x n ] , so that

rsf = p 1 a 1 p r a r f

is a decomposition in irreducible factors in [ x 1 , , x n ] . The only units of this UFD are ± 1 , so that the unicity of the decomposition in irreducible factors in [ x 1 , , x n ] shows that the decompositions of g 1 = rg , h 1 = sh are in the form

g 1 = ± p 1 b 1 p r b r f , h 1 = ± p 1 c 1 p r c r ,

or

g 1 = ± p 1 b 1 p r b r , h 1 = ± p 1 c 1 p r c r f .

Therefore h 1 or g 1 is in , thus g or h is in , so is a unit in [ x 1 , , x n , y ] . This proves that f is irreducible in [ x 1 , , x n , y ] .

(b)
Since [ x 1 , . . . , x n , y ] is UFD, s u ( y ) can be uniquely factorized s u ( y ) = h 1 h n , where h 1 , . . . , h n [ x 1 , . . . , x n , y ] and are irreducible. By the part (a), h 1 , . . . , h n are irreducible when regarded as elements of [ x 1 , . . . , x n , y ] . Hence for any irreducible factor h [ x 1 , . . . , x n , y ] of s u ( y ) , there exists i such that h h i . It follows that h is associate to h i , h = λ h i , λ , is the product of h i by a suitable constant from , where h i is a factor of s u ( y ) irreducible in [ x 1 , . . . , x n , y ] .

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2022-07-19 00:00
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