6.7.3. We know that ZŒ
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6.7.1. Let R be a commutative ring with identity element. Let J be an ideal in RŒx. Show that for each k 0, the set Ak D fa 2 R W a D 0 or there exists f 2 J with leading term axkg is an ideal in R, and that Ak AkC1 for all k 0.
6.7.2. Suppose that R is Noetherian and ' W R ! S is a surjective ring homomorphism. Show that S is Noetherian.
p
6.7.3. We know that ZΠ5 is not a UFD and, therefore, not a PID.
p
Show that ZΠ5 is Noetherian. Hint: Find a Noetherian ring R and a
p
surjective homomorphism ' W R ! ZŒ
5.
6.7.4. Show that ZŒx C xQŒx is not Noetherian.
6.7.5. Show that a Noetherian domain in which every irreducible element is prime is a unique factorization domain.
6.8. Irreducibility Criteria
In this section, we will consider some elementary techniques for determining whether a polynomial is irreducible.
We restrict ourselves to the problem of determining whether a polyno mial in ZŒx is irreducible. Recall that an integer polynomial factors over the integers if, and only if, it factors over the rational numbers, according to Lemma 6.6.9 and Corollary 6.6.10.
A basic technique in testing for irreducibility is to reduce the polyno mial modulo a prime. For any prime p, the natural homomorphism of Z onto Zp extends to a homomorphism of ZŒx onto ZpŒx, p W P ai xi 7!
PŒai pxi .
Proposition 6.8.1. Fix a prime p. Suppose that a polynomial f .x/ D P ai xi 2 ZŒx has positive degree and that its leading coefficientis not divisible by the prime p. If p.f .x// is irreducible in ZpŒx, then f .x/ is irreducible in QŒx.
Proof. The assumption that the leading coefficient of f is not divisible by p means that deg.p.f // D deg.f /. Suppose that f .x/ has a factorization f .x/ D g.x/h.x/ in ZŒx, with the degree of g.x/ and of h.x/ positive. Then p does not divide the leading coefficients of g.x/ and h.x/, so p.g.x// and p.h.x// have positive degree. Moreover, p.f .x// D p.g.x//p.h.x//.
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Efficient algorithms are known for factorization in ZpŒx. The com mon computer algebra packages such as Mathematica and Maple have these algorithms built in.
The
Mathematica
command
Factor[f, Modulus ! p] can be used for reducing a polynomial f modulo a prime p and factoring the reduction. Unfortunately, the condition of the proposition is merely a sufficient condition. It is quite possible (but rare) for a polynomial to be irreducible over Q but nevertheless for its reductions modulo every prime to be reducible.
Example 6.8.2. Let f .x/ D 83 C 82 x 99 x2
87 x3 17 x4. The reduction of f modulo 3 is f2 C x C x4g, which is irreducible over Z3.
Hence, f is irreducible over Q.
Example 6.8.3. Let f .x/ D
91
63 x 73 x2 C 22 x3 C 50 x4. The reduction of f modulo 17 is 16 6 C 12 x C 5 x2 C 12 x3 C x4, which is irreducible over Z17. Therefore, f is irreducible over Q.
A related sufficient condition for irreducibility is Eisenstein’s criterion: Proposition 6.8.4. (Eisenstein’s criterion). Consider a monic polynomial f .x/ D xnCan 1xn 1C Ca1xCa0 with integer coefficients. Suppose that p is a prime that divides all the coefficients ai and such that p2 does not divide a0. Then f .x/ is irreducible over Q.
Proof. If f has a proper factorization over Q, then it also has a proper factorization over Z, with both factors monic polynomials. Write f .x/ D a.x/b.x/, where a.x/ D PriD0 ˛ixi and b.x/ D PsjD0 ˇixj . Since a0 D ˛0ˇ0, exactly one of ˛0 and ˇ0 is divisible by p; suppose without loss of generality that p divides ˇ0, and p does not divide ˛0. Considering the equations a1 D ˇ1˛0 C ˇ0˛1 : : : as 1 D ˇs 1˛0 C C ˇ0˛s 1 as D ˛0 C ˇs 1˛1 C C ˇ0˛s; we obtain by induction that ˇj is divisible by p for all j (0 j s 1).
Finally, the last equation yields that ˛0 is divisible by p, a contradiction.
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Example 6.8.5. x3 C 14x C 7 is irreducible by the Eisenstein criterion.
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Example 6.8.6. Sometimes the Eisenstein criterion can be applied after a linear change of variables. For example, for the so–called cyclotomic
polynomial
f .x/ D xp 1 C xp 2 C C x2 C x C 1; where p is a prime, we have p 1
X
p
f .x C 1/ D xs:
s C 1
sD0
This is irreducible by Eisenstein’s criterion, so f is irreducible as well.
You are asked to provide the details for this example in Exercise 6.8.2.
There is a simple criterion for a polynomial in ZŒx to have (or not to have) a linear factor, the so–called rational root test.
Proposition 6.8.7. (Rational root test) Let f .x/ D anxn C an 1xn 1 C a1x C a0 2 ZŒx. If r=s is a rational root of f , where r and s are relatively prime, then s divides an and r divides a0.
Proof. Exercise 6.6.3.
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A quadratic or cubic polynomial is irreducible if, and only if, it has no linear factors, so the rational root test is a definitive test for irreducibility for such polynomials. The rational root test can sometimes be used as an adjunct to prove irreducibility of higher degree polynomials: If an integer polynomial of degree n has no rational root, but for some prime p its reduction mod p has irreducible factors of degrees 1 and n 1, then then f is irreducible (Exercise 6.8.1).
Exercises 6.8 6.8.1. Show that if a polynomial f .x/ 2 ZŒx of degree n has no rational root, but for some prime p its reduction mod p has irreducible factors of degrees 1 and n 1, then then f is irreducible.
6.8.2. Provide the details for Example 6.8.6.
6.8.3. Show that for each natural number n
.x
1/.x
2/.x
3/ .x n/
1
is irreducible over the rationals.
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6.8.4. Show that for each natural number n ¤ 4
.x
1/.x
2/.x
3/ .x
n/ C 1 is irreducible over the rationals.
6.8.5. Determine whether the following polynomials are irreducible over the rationals. You may wish to do computer computations of factorizations modulo primes.
(a)
8
60 x 54 x2 C 89 x3 55 x4
(b)
42
55 x 66 x2 C 44 x3 (c)
42
55 x 66 x2 C 44 x3 C x4 (d) 5 C 49 x C 15 x2 27 x3
(e) 96 C 53 x 26 x2 C 21 x3 75 x4
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