## 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 $p$th 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 $n \notin \{-1,3,5\}$.

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 $(1+\theta)^{-1} = \frac{1}{4}(\theta^2 - \theta + 10)$.