[PDF] WORKSHEET &# 8 IRREDUCIBLE POLYNOMIALS



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WORKSHEET &# 8 IRREDUCIBLE POLYNOMIALS

Then f is irreducible if and only if f(a) 6= 0 for all a2k Proposition 0 4 Suppose that a;b2kwith a6= 0 Then f(x) 2k[x] is irreducible if and only if f(ax+b) 2k[x] is irreducible Theorem 0 5 (Reduction mod p) Suppose that f2Z[x] is a monic1 polynomial of degree >0 Set f p 2Z modp[x] to be the reduction mod pof f (ie, take the coe cients



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WORKSHEET # 8

IRREDUCIBLE POLYNOMIALS

We recall several dierent ways we have to prove that a given polynomial is irreducible. As always,kis a eld.

Theorem 0.1(Gauss' Lemma).Suppose thatf2Z[x]is monic of degree>0. Thenfis irreducible inZ[x]if and only if it is irreducible when viewed as an element ofQ[x]. Lemma 0.2.A degree one polynomialf2k[x]is always irreducible.

Proposition 0.3.Suppose thatf2k[x]has degree 2 or 3. Thenfis irreducible if and only iff(a)6= 0for all

a2k. Proposition 0.4.Suppose thata;b2kwitha6= 0. Thenf(x)2k[x]is irreducible if and only iff(ax+b)2k[x] is irreducible. Theorem 0.5(Reduction modp).Suppose thatf2Z[x]is a monic1polynomial of degree>0. Setfp2Zmodp[x]

to be the reduction modpoff(ie, take the coecients modp). Iffp2Zmodp[x]is irreducible for some primep,

thenfis irreducible inZ[x].

WARNING: The converse need not be true.

Theorem 0.6(Eisenstein's Criterion).Suppose thatf=xn+an1xn1++a1x1+a02Z[x]and also that there is a primepsuch thatpjaifor allibut thatp2doesNOTdividea0. Thenfis irreducible.

1.Consider the polynomialf(x) =x3+x2+x+2. In which of the following rings of polynomials isfirreducible?

Justify your answer.

(a)R[x] (b)C[x] (c)Zmod2[x] (d)Zmod3[x] (e)Zmod5[x] (f)Q[x]

Solution:

(a)

It is reducible (= not irredu cible)b ecauseit is a cubic p olynomialand therefore has a ro ot. Thusfcan

be factored asf(x) = (x)g(x). (b) The ro otfrom (a) is also a complex n umber,and so fis reducible inC[x] as well. (c) Mo d2, f2=x3+x2+x, which has a root atx= 0 and so is reducible. (d) Mo d3, f3=x3+x2+x+2. 0 is not a root,f2(1) = 5 = 26= 0, and nallyf2(2) = 8+4+2+2 = 16 = 16= 0.

In particular,f3isirreducible.

(e) Mo d5, f5=x3+x2+x+ 2. Note 1 is a root, and sof5is reducible. (f)fis irreducible sincef3is irreducible by Theorem 0.5.1 The same is true as long as the leading coecient is not divisible byp. 1

WORKSHEET #8 2

2.Show thatx4+ 1 is irreducible inQ[x] but not irreducible inR[x].

Hint:ForQ[x], use Proposition 0.4. ForR[x], try a factorization into two linear terms Solution:First considerf(x) =x4+ 1 so thatf(x+ 1) = (x+ 1)4+ 1 =x4+ 4x3+ 6x2+ 4x2+ 2. Eisenstein's criterion applies and sof(x+ 1) is irreducible inQ[x]. But thus so isf(x) by Proposition 0.4.

For the second part, consider

(x2p2x+ 1)(x2+p2x+ 1) =x4+p2x3+x2p2x32x2p2x+x2+p2x+ 1 =x4+ 1 which proves thatf(x) is reducible.

3.Consider 3x2+4x+32Zmod5[x]. Show it factors both as (3x+2)(x+4) and as (4x+1)(2x+3). Explain why

thisdoes NOTcontradict unique factorization of polynomials.

Solution:First note that

(3x+ 2)(x+ 4) = 3x2+ 12x+ 2x+ 8 = 3x2+ 4x+ 3 and that (4x+ 1)(2x+ 3) = 8x2+ 12x+ 2x+ 3 = 3x2+ 4x+ 3

On the other hand, 23 = 1 inZmod5, and so

(4x+ 1)(2x+ 3) = (4x+ 1)(23)(2x+ 3) = ((4x+ 1)2)(3(2x+ 3)) = (8x+ 2)(6x+ 9) = (3x+ 2)(x+ 4):

This completes the proof.

4.Completely factor all the polynomials in question 1. into irreducible polynomials in each of the rings (c){(f).

Solution:

(c)Zmod2[x],f=x(x2+x+ 1) (d)Zmod3[x],f= (x3+x2+x+ 2) (e)Zmod5[x],f= (x1)(x2+ 2x+ 3) (f)Q[x],f= (x3+x2+x+ 2)quotesdbs_dbs5.pdfusesText_10