Tag Archives: irreducible polynomial

A fact about polynomials over a perfect field

Let K be an extension of F, with F a perfect field. Suppose p(x) \in F[x] has no repeated irreducible factors; prove that p(x) has no repeated irreducible factors in K[x].

Suppose p has no repeated irreducible factors. Since F is perfect, the irreducible factors of p are separable (Prop. 37 on 549 in D&F), and so p is separable. (If not, then it would have a repeated irreducible factor.) That is, p has no repeated roots.

Now if p has a repeated irreducible factor over K, then it has repeated roots- a contradiction.

Over an imperfect field of characteristic p, there exist irreducible inseparable polynomials

Let K be an imperfect field of characteristic p. Prove that there exist irreducible inseparable polynomials over K. Conclude that there exist inseparable finite extensions of K.

Let \alpha \in K be an element which is not a pth power in K, and let \zeta be a root of x^p-\alpha. In particular, \zeta \notin K. Now consider q(x) = (x-\zeta)^p = x^p-\alpha. Suppose t(x) is an irreducible factor of q over K. Now t cannot be linear, since \zeta \notin K. But any factor of q of degree at least 2 has a multiple root (since all the roots are \zeta). So t is an irreducible inseparable polynomial over K.

Then K(\zeta) is an inseparable extension of K.

xᵖ-x+a is irreducible and separable over ZZ/(p)

Prove that for any prime p and any nonzero a \in \mathbb{F}_p, q(x) = x^p-x+a is irreducible and separable over \mathbb{F}_p.

Note that D_x(q) = px^{p-1}-1 = -1, so that q and D(q) are relatively prime. So q is separable.

Now let \alpha be a root of q. Using the Frobenius endomorphism, (\alpha+1)^p-(\alpha+1)+a = \alpha^p - \alpha + a = 0, so that \alpha+1 is also a root. By induction, \alpha_k is a root for all k \in \mathbb{F}_p, and since q has degree p, these are all of the roots.

Now \mathbb{F}_p(\alpha+k) = \mathbb{F}_p(\alpha), and in particular the minimal polynomials of \alpha and \alpha+k have the same degree over \mathbb{F}_p – say d. Since q is the product of the minimal polynomials of its roots, we have p = dt for some t. Since p is prime, we have either d=1 (so that \alpha \in \mathbb{F}_p, a contradiction) or d=p, so that q itself is the minimal polynommial of \alpha, hence is irreducible.

Find the irreducible polynomials of degree 1, 2, and 4 over ZZ/(2)

Find all the irreducible polynomials of degree 1, 2, or 4 over \mathbb{F}_2, and verify that their product is x^{16}-x.

Note that there are 2^k (monic) polynomials of degree k, as each non-leading coefficient can be either 0 or 1.

The polynomials of degree 1, p_1(x) = x and p_2(x) = x+1, are both irreducible.

There are 4 polynomials of degree 2, one of which is irreducible.

  1. x^2 = xx is reducible
  2. x^2+1 = (x+1)^2 is reducible
  3. x^2+x = x(x+1) is reducible
  4. p_3(x) = x^2+x+1 has no roots, and so is irreducible.

Before we address the degree 4 polynomials, we prove a lemma.

Lemma 1: If p(x) is a degree 4 polynomial over \mathbb{F}_2 with constant term 1 and p factors as a product of quadratics, then the linear and cubic terms of p are equal. Proof: We have (as we assume) p(x) = (x^2+ax+b)(x^2+cx+d) = x^4 + (a+c)x^3 + (b+d+ac)x^2 + (ad+bc)x+bd. Now bd = 1, so that b = d = 1. So p(x) = x^4 + (a+c)x^3 + acx^2 + (a+c)x + 1, as desired. \square

Lemma 2: \Phi_5(x) = x^4+x^3+x^2+x+1 is irreducible over \mathbb{F}_2. Proof: Clearly this has no roots. By Lemma 1, if \Phi_5(x) factors into two quadratics, then the factors’ linear terms, a and c, satisfy a+c=ac=1, which is impossible over \mathbb{F}_2. \square

There are 16 polynomials of degree 4.

  1. x^4 = xx^3 is reducible
  2. x^4+1 = (x^2+1)^2 is reducible
  3. x^4+x = x(x^3+1) is reducible
  4. x^4+x^2 = x^2(x^2+1) is reducible
  5. x^4+x^3 = x^3(x+1) is reducible
  6. p_4(x) = x^4+x+1 clearly has no roots, and by the lemma, has no quadratic factors. So p_4 is irreducible.
  7. x^4+x^2+1 = (x^2+x+1)^2 is reducible
  8. p_5(x) = x^4+x^3+1 clearly has no roots, and by Lemma 1, has no quadratic factors. So p_5 is irreducible.
  9. x^4+x^2+x = x(x^3+x+1) is reducible
  10. x^4+x^3+x = x(x^3+x^2+1) is reducible
  11. x^4+x^3+x^2 = x^2(x^2+x+1) is reducible
  12. x^4+x^2+x+1 has 1 as a root, so is reducible
  13. x^4+x^3+x^2+1 = (x+1)(x^3+1) is reducible
  14. x^4+x^3+x^2+1 has 1 as a root, so is reducible
  15. x^4+x^3+x^2+x = x(x^3+x^2+x+1) is reducible
  16. p_6(x) = x^4+x^3+x^2+x+1 = \Phi_5(x) is irreducible

So there are six irreducible polynomials of degree 1, 2, or 4 over \mathbb{F}_2:

  1. p_1(x) = x
  2. p_2(x) = x+1
  3. p_3(x) = x^2+x+1
  4. p_4(x) = x^4+x+1
  5. p_5(x) = x^4+x^3+1
  6. p_6(x) = x^4+x^3+x^2+x+1

It is easy (if tedious) to verify that the product of these polynomials is x^{16}-x. (WolframAlpha agrees.)

A finite field extension K of F is a splitting field if and only if every irreducible polynomial over F has no roots in K or splits over K

Let K be a finite extension of a field F. Prove that K is a splitting field over F if and only if every irreducible polynomial over F with a root in K splits over K.

We begin with some lemmas.

Lemma 1: Let K be a finite extension of F, and let \alpha be algebraic over F. If K is a splitting field for a set S \subseteq F[x], then K(\alpha) is a splitting field for S considered as polynomials over F(\alpha). Proof: Certainly the roots of elements of S are in K(\alpha) \supseteq K. Now suppose E is a splitting field for S over F(\alpha); in particular, E contains F and the roots of the polynomials in S, so K \subseteq E. Moreover \alpha \in E, so K(\alpha) \subseteq E. Hence K(\alpha) is a splitting field for S over F(\alpha). \square

Suppose K is a splitting field for some (finite) set S \subseteq F[x]. Let q(x) be irreducible over F, and suppose \alpha and \beta are roots of q with \alpha \in K. By Theorem 8 on page 519 of D&F, \sigma : \alpha \mapsto \beta extends to an isomorphism \sigma : F(\alpha) \rightarrow F(\beta). Now K(\alpha) is the splitting field for S \subseteq F(\alpha)[x], and likewise K(\beta) is the splitting field for S over F(\beta). By Theorem 27 on page 541 of D&F, the isomorphism \sigma extends to an isomorphism \psi : K(\alpha) \rightarrow K(\beta). We can visualize this using the following diagram.

A field diagram

Since \alpha \in K, [K(\alpha):K] = 1, and so [K(\beta):K] = 1, and we have \beta \in K. That is, if an irreducible polynomial over F has any roots in K, then it splits completely over K.

Conversely, suppose K is finite over F and that every irreducible polynomial with a root in K has all roots in K. Since K is a finite extension, it is algebraic (see Theorem 17 on page 526 of D&F). Say K = F(\alpha_1,\ldots,\alpha_n), and let m_i be the minimal polynomial of \alpha_i over F. By our hypothesis, each m_i splits over K. Moreover, any field over which all the m_i split must contain the \alpha_i. So in fact K is the splitting field over F of the set of polynomials m_i.

Prove that a given polynomial is irreducible over QQ

Show that p(x) = x^3 + x^2 - 2x - 1 is irreducible over \mathbb{Q}. Use the fact (to be proven later) that \alpha = 2 \mathsf{cos}(2\pi/7) is a root of p to argue that the regular 7-gon is not constructible by straightedge and compass.

Using the rational root theorem, any rational roots of p(x) must be either 1 or -1. Certainly then p(x) has no rational roots. Since p(x) has degree 3, it is irreducible over \mathbb{Q} if and only if it has no roots (in \mathbb{Q}), so that p(x) is irreducible. In particular, the roots of p(x) lie in degree 3 extensions of \mathbb{Q}.

Suppose now that the regular 7-gon is constructible. The exterior angles of a regular 7-gon have measure 2\pi/7, so in particular, if the regular 7-gon is constructible, then so is the point (\mathsf{cos}(2\pi/7), \mathsf{sin}(2\pi/7)), so that the number 2\mathsf{cos}(2\pi/7) is constructible. But we have seen that any constructible element must have degree a power of 2 over \mathbb{Q}. Given that \alpha is a root of p(x), this yields a contradiction.

Prove that two given polynomials are irreducible over the Gaussian rationals

Show that p(x) = x^3-2 and q(x) = x^3-3 are irreducible over F = \mathbb{Q}(i).

Recall that the Gaussian integers \mathbb{Z}[i] are a Unique Factorization Domain, and that \mathbb{Z}[i] \subseteq \mathbb{Q}(i). We claim that \mathbb{Q}(i) is the field of fractions of \mathbb{Z}[i]. Indeed, if F is a field containing \mathbb{Z}[i], then F also contains any Gaussian rational since we can write \frac{a}{b} + \frac{c}{d}i = \frac{1}{bd}(ad + bci) with a,b,c,d \in \mathbb{Z}. By the universal property of fields of fractions, \mathbb{Q}(i) is the field of fractions of \mathbb{Z}[i].

Recall that 1+i and 3 are irreducible in \mathbb{Z}[i]. So Eisenstein’s criterion applies, showing that p and q are irreducible over \mathbb{Q}(i).

Show that a given family of polynomials is irreducible over ZZ

Let p(x) = x^5 - tx - 1 \in \mathbb{Z}[x]. Show that p(x) is irreducible in \mathbb{Z}[x] unless t \in \{0,2,-1\}.

Note that if p(x) is reducible in \mathbb{Z}[x], then it must have either a linear or a quadratic factor.

Suppose p(x) has a linear factor in \mathbb{Z}[x]; say p(x) = (x+a)(x^4+bx^3+cx^2+dx+e), with these coefficients in \mathbb{Z}. Comparing coefficients, we have the following system of equations: a+b = 0, c+ab = 0, d+ac = 0, e+ad = -t, ae = -1. Since a and e are integers, the last equation yields the two cases (a,e) = (-1,1) and (a,e) = (1,-1). Note also that -a is a root of p(x), so that t = (a^5+1)/a. We can see that in the first case, t = 0, and in the second, t = 2. Indeed, this yields the factorizations x^5 - 1 = (x-1)(x^4+x^3+x^2+x+1) and x^5-2x-1 = (x+1)(x^4-x^3+x^2-x-1). For no other t does p(x) have a linear factor.

Now suppose p(x) has a quadratic factor in \mathbb{Z}[x]; say p(x) = (x^2+ax+b)(x^3+cx^2+dx+e). Again comparing coefficients, we have a+c = 0, b+ac+d = 0, e+ad+bc = 0, ae+bd = -t, and be = -1. Substituting the first two equations into the third, we see that a is a rational root of x^3-2bx+e.

If (b,e) = (-1,1), then a is a rational root of q(x) = x^3+2x-1; by the rational root test, a = \pm 1. Neither of these is a root of q(x), a contradiction.

If (b,e) = (1,-1), then a is a rational root of q(x) = x^3-2x-1; by the rational root theorem, a = -1. Then c = 1 and d = 0, and we have t = -1. Indeed, this corresponds to the factorization x^5+x-1 = (x^2-x+1)(x^3+x^2-1). For no other t does p(x) have a quadratic factor.

Show that a given family of polynomials is irreducible over ZZ

Show that p(x) = x^3 - nx + 2 is irreducible over \mathbb{Q} for n \notin \{-1,3,5\}.

We did this already.

Compute the inverse of a given element in a field extension

Show that p(x) = x^3 + 9x + 6 is irreducible over \mathbb{Q}. Let \theta be a root of p(x). Find the inverse of 1+\theta in \mathbb{Q}(\theta).

p(x) is Eisenstein at 3, and so is irreducible, and thus \mathbb{Q}(\theta) \cong \mathbb{Q}[x]/(p(x)) is a field.

Note that every element of \mathbb{Q}(\theta) has the form a + b\theta + c\theta^2 for some rational numbers a,b,c. Suppose (1+\theta)(a+b\theta+c\theta^2) = 1; comparing coefficients, we see that a-6c = 1, a+b-9c = 0, and b+c = 0. Solving this system, we can see then that (1+\theta)^{-1} = \frac{1}{4}(\theta^2 - \theta + 10).