113 Class Problems: Irreducible Elements and Unique

Factorization Domains

1. Is the polynomial 2x2 − 4 irreducible in C[x]? How about in R[x], Q[x] or Z[x]?Solutions:

2. If a polynomial f(x) ∈ Q[x] has no roots in Q must it be irreducible in Q[x]?

Solutions:

EC c Cx t

UT 6 ReducibleCx 2 2 4 2 x Tz x fr6 RE e IRC

in Ecsc

ReducibleIRCx 2 2 4 Z x 152 Cx T2 in RexCx 2 2 4 axe b Coxed a b c d C

I 9 3 Vz Vz Vz C GutradictionIrreducible in Cx

26 3 zc.cz z Reducible in ZCxZC y Eca

No x2 Cx2 is reducible ai Cx but has

no roots in

3. Consider the subring Z[√−5] ⊂ C

(a) Prove that Z[√−5] = {a+ b

√−5|a, b ∈ Z}

(b) If a + b√−5 ∈ Z[

√−5] is non-zero, what is the minimum possible value of |a +

b√−5|2, the square of the absolute value?

(c) Using part (b) determine Z[√−5]∗, the units in Z[

√−5].

(d) Prove that 2, 3, 1 +√−5, 1−

√−5 are non-associated elements of Z[

√−5].

(e) Prove that 2, 3, 1 +√−5, 1−

√−5 are irreducible elements of Z[

√−5].

(f) Prove that Z[√−5] is not a UFD.

Solutions:

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a

Fs S E Z Ects atb Fs a bee

I la tbFsl2 art 5b2 a b C Z

lax bfs 12 if a b C Z Min value is when aand b O

I L D E Z Cfs late 113T I

2 3 1 121211312 1 1 12 11312 1 I

Cfs t

d 2 3 I f s l f s are pairwise non associated as

2 Cfs tft and non is a negative of another

2 BE Ectse

2 2 3 22 12121 312 I E 1,2 4

a Sbl 2 has no integer solutions la I _I n 4

1 12 1 c Effs I 12 4 11312 1 BE ECFSI

z irreducible

Same logic shows 3 It F s l Fs are all irreducible

G Z 3 I Fsk yestwo non associatedirreducible Factorization

Cfs not a VFD