## Tag Archives: integers

### 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.

### Exhibit a nontrivial alternating bilinear map from a tensor square

Let $R = \mathbb{Z}[x,y]$ and $I = (x,y)$.

1. Prove that if $ax + by = a^\prime x + b^\prime y$ in $R$, then there exists $h \in R$ such that $a^\prime = a + yh$ and $b^\prime = b - xh$.
2. Prove that the map $\varphi(ax+by,cx+dy) = ad-bc \mod (x,y)$ is a well-defined, alternating, bilinear map $I \times I \rightarrow R/I$.

Note that $(a-a^\prime)x = (b^\prime - b)y$. Since $x$ and $y$ are irreducible (hence prime) in the UFD $R$, we have that $y$ divides $a - a^\prime$ and $x$ divides $b^\prime - b$. Say $a - a^\prime = yh$ and $b^\prime - b = xh^\prime$. Since $R$ is a domain, we see that $h = h^\prime$, and thus $a^\prime = a - yh$ and $b^\prime = b + xh$.

Suppose now that $a_1x + b_1y = a_2x + b_2y$ and $c_1x + d_1y = c_2x + d_2y$. Then there exist $h,g \in R$ such that $a_2 = a_1 - yh$, $b_2 = b_1 + xh$, $c_2 = c_1 - yg$, and $d_2 = d_1 + xg$. Evidently, $a_2d_2 - b_2c_2 \equiv a_1d_1 - b_1c_1 \mod (x,y)$. Thus $\varphi$ is well defined. Note that $\varphi((a_1+a_2)x + (b_1+b_2)y, cx+dy) = (a_1+a_2)d - (b_1+b_2)c$ $= (a_1d - b_1c) + (a_2d - b_2c)$ $= \varphi(a_1x+b_1y,cx+dy) + \varphi(a_2x+b_2y,c_x+d_y)$, so that $\varphi$ is linear in the first argument; similarly it is linear in the second argument. Moreover, $\varphi((ax+by)r,cx+dy) = \varphi(arx+bry,cx+dy)$ $= ard-brc$ $= \varphi(ax+by,r(cx+dy))$. Thus $\varphi$ is $R$-bilinear. Since $\varphi(ax+by,ax+by) = ab-ab = 0$, $\varphi$ is alternating. Finally, note that $\varphi(1,0) = 1$. Thus there exists a nontrivial alternating $R$-bilinear map on $I \times I$.

### Factor some ideals in ZZ

Find the prime factorizations of the ideals $A = (70)$ and $B = (150)$ in $\mathbb{Z}$. Compute $((70),(150))$.

Evidently $70 = 2 \cdot 5 \cdot 7$ and $150 = 2 \cdot 3 \cdot 5^2$, so that $(70) = (2)(5)(7)$ and $(150) = (2)(3)(5)^2$. Since $\mathbb{Z}$ is a unique factorization domain, $(2)$, $(3)$, $(5)$, and $(7)$ are prime since 2, 3, 5, and 7 are prime.

Now $((70),(150)) = (70,150) = (10)$, since $70 = 7 \cdot 10$, $150 = 15 \cdot 10$, and $10 = 150 - 2 \cdot 70$.

### Over ZZ, if (a,b) is maximal then gcd(a,b) is prime

Let $a,b \in \mathbb{Z}$ such that $(a,b)$ is maximal as an ideal in $\mathbb{Z}$. What can be said about $a$ and $b$?

Since $\mathbb{Z}$ is a principal ideal domain, $(a,b) = (d)$ for some $d$. If $(d)$ is maximal, then it is prime, so that $d$ is a prime element. Moreover, $d$ is a greatest common divisor of $a$ and $b$. So $\mathsf{gcd}(a,b)$ is prime.

### Divisibility among rational integers in an algebraic number field

Let $E = \mathbb{Q}(\theta)$ be an algebraic extension of $\mathbb{Q}$ and let $a,b \in \mathbb{Z}$. Recall that if $\alpha,\beta \in E$ are algebraic integers, we say that $\beta|\alpha$ if the quotient $\alpha/\beta$ is an algebraic integer. Prove that $b|a$ in $E$ if and only if $b|a$ in $\mathbb{Q}$.

If $b|a$ in $\mathbb{Q}$, then $a/b$ is a rational integer. Certainly $a/b \in E$ is an algebraic integer; so $b|a$ in $E$.

Now suppose $b|a$ in $E$. Since $a/b$ is an algebraic integer and a rational number, $a/b$ is a rational integer. So $b|a$ in $\mathbb{Q}$.

### The injective hull of the ZZ-module ZZ is QQ

Prove that the injective hull of the $\mathbb{Z}$-module $\mathbb{Z}$ is $\mathbb{Q}$.

Recall that an injective hull of a module $M$ is an injective module $Q$ such that there exists an injective homomorphism $M \rightarrow Q$ and such that every injective module homomorphism from $M$ to an injective module $A$ lifts to an injective homomorphism from $Q$ to $A$. We will take it as given that injective hulls exist and are unique up to isomorphism.

Let $H$ be the injective hull of $\mathbb{Z}$ as a $\mathbb{Z}$-module. Since $\mathbb{Q}$ is injective as a $\mathbb{Z}$-module and the inclusion $\mathbb{Z} \rightarrow \mathbb{Q}$ is injective, there exists an injective abelian group homomorphism $H \rightarrow \mathbb{Q}$. We will identify $H$ with its image in $\mathbb{Q}$.

Note that $H$ is divisible as an abelian group, and since $\mathbb{Z} \subseteq H$, for all integers $n \neq 0$ there exists $q \in H$ such that $nq = 1$. In particular, $q = \frac{1}{n} \in H$ for all $n \neq 0$. Since $H \subseteq \mathbb{Q}$ is a $\mathbb{Z}$-submodule, we have $\frac{a}{b} \in H$ for all $a,b \in \mathbb{Z}$ with $b \neq 0$. Thus $H = \mathbb{Q}$.

### Nontrivial divisible ZZ-modules are not projective

Prove that no nontrivial, divisible $\mathbb{Z}$-module is projective. Deduce that $\mathbb{Q}$ is not projective.

We begin with a lemma.

Lemma: Let $F$ be a free $\mathbb{Z}$-module. Then $\bigcap_{\mathbb{N}^+} nF = 0$. Proof: Suppose $B = \{e_i\}_I$ is a basis for $F$ indexed by a set $I$, and let $x \in \bigcap_{\mathbb{N}^+} nF$. For all $n$, there exist $k_{i,n} \in \mathbb{Z}$ such that $x = \sum nk_{i,n}e_i$ and all but finitely many $k_{i,n}$ are zero. Since $F$ is free on $B$, if $x = \sum \ell_i e_i$, we have $n|\ell_i$ for all $n \in \mathbb{N}^+$ so that $\ell_i = 0$. Thus $x = 0$. Hence $\bigcap_{\mathbb{N}^+} nF = 0$. $\square$

Now suppose the nontrivial, divisible $\mathbb{Z}$-module $Q$ is projective. Then there exists a module $N$ such that $Q \oplus N$ is free. By the lemma, $0 = \bigcap_{\mathbb{N}^+} n(Q \oplus N)$ $= \bigcap_{\mathbb{N}^+} nQ \oplus nN$. Since $Q$ is injective and $\mathbb{Z}$ is a principal ideal domain, $nQ = Q$ for all $n \in \mathbb{Z}$. So we have $0 = \bigcap_{\mathbb{N}^+} Q \oplus nN = Q \oplus \bigcap_{\mathbb{N}^+} nN$. Note, however, that this module contains $Q \oplus 0$, and since $Q$ is nontrivial, we have a contradiction. So $\mathbb{Q}$ is not projective.

In particular, since $\mathbb{Q}$ is nontrivial and divisible (as we showed here), it is not projective as a $\mathbb{Z}$-module.

### Finite abelian groups are neither injective nor projective as ZZ-modules

Let $A$ be a nonzero finite abelian group. Prove that, as a $\mathbb{Z}$-module, $A$ is neither injective nor projective.

Recall that every free $\mathbb{Z}$-module is torsion free, since $\mathbb{Z}$ is an integral domain. Now if $A$ is a nontrivial finite abelian group, $A \oplus B$ contains nonzero torsion elements for any choice of $\mathbb{Z}$-module $B$. Thus $A$ is not a direct summand of a free module, and so cannot be projective.

Now recall that every nontrivial finite abelian group is a direct sum of nontrivial cyclic groups. Consider the cyclic $\mathbb{Z}$-module $\mathbb{Z}/(n)$. Note that the inclusion map $\mathbb{Z} \rightarrow \mathbb{Q}$ is injective, and we have the natural projection $\pi : \mathbb{Z} \rightarrow \mathbb{Z}/(n)$. Suppose $\mathbb{Z}/(n)$ is injective. Then $\pi$ lifts to a surjective homomorphism $\Pi : \mathbb{Q} \rightarrow \mathbb{Z}/(n)$. Via the first isomorphism theorem, $\mathbb{Q}/\mathsf{ker}\ \Pi \cong \mathbb{Z}/(n)$. However, we showed in this previous exercise that $\mathbb{Q}$ has no proper subgroups of finite index, which yields a contradiction. In particular, $\mathbb{Z}/(n)$ is not injective. By this previous exercise, $A$ is not injective.

### Construct a polynomial over ZZ having a given root

Construct a degree 4 polynomial in $\mathbb{Z}[x]$ having $\sqrt{2}+i$ as a root. Construct a degree 6 polynomial having $\sqrt{2}\sqrt[3]{3}$ as a root.

For no particular reason whatsoever, let $x_1 = \sqrt{2}+i$, $x_2 = \sqrt{2} - i$, $x_3 = - \sqrt{2} +i$, and $x_4 = -\sqrt{2} - i$. Letting $\sigma_i$ denote the elementary symmetric polynomials in four variables, and making these substitutions, we see that $\sigma_1 = 0$, $\sigma_2 = -2$, $\sigma_3 = 0$, and $\sigma_4 = 9$. Thus $p(x) = x^4 - 2x^2 + 9$ has $\sqrt{2}+i$ as a root.

Note that $\alpha_1 = \sqrt{2}$ is a root of $a(x) = x^2-2$ and $\beta_1 = \sqrt[3]{3}$ is a root of $b(x) = x^3-3$, and that the other roots of these polynomials are $\alpha_2 = -\sqrt{2}$, $\beta_2 = \frac{-1}{2} + \frac{\sqrt{3}}{2}i$, and $\beta_3 = \frac{-1}{2} - \frac{\sqrt{3}}{2}i$. By Corollary 3.12 in TAN, $\prod_i \prod_j (x-\alpha_i\beta_j)$ is a polynomial in $\mathbb{Z}[x]$. A rather tedious calculation reveals that this polynomial is $x^6-72$.

### Decide whether a given cubic polynomial is irreducible over the rationals

Decide whether the following two polynomials are irreducible over $\mathbb{Q}$: $p(x) = x^3 - x + 2$ and $q(x) = x^3 - 12x^2 + 44x - 52$.

Note that if a cubic polynomial in $\mathbb{Z}[x]$ is reducible over $\mathbb{Q}$, then it must have a linear factor- that is, it has a root in $\mathbb{Q}$. By the rational root theorem, there are finitely many candidates for this root. If none of these is a root, then the polynomial is irreducible in $\mathbb{Q}[x]$.

By the rational root theorem, if $p(x)$ has a root in $\mathbb{Q}$ it is in the set $\{ \pm 2/\pm 1, \pm 1/\pm 1 \} = \{ 2, -2, 1, -1\}$. Since $p(1) = p(-1) = 2$, $p(2) = 8$, and $p(-2) = -4$, $p(x)$ is irreducible in $\mathbb{Q}[x]$.

Again by the rational root theorem, if $q(x)$ has a root in $\mathbb{Q}$ then it is in the set $\{ \pm 1, \pm 2, \pm 4, \pm 13, \pm 26, \pm 52 \}$. It is easy but rather tedious to see that $q(1) = -19$, $q(-1) = -109$, $q(2) = -4$, $q(-2) = -196$, $q(4) = -4$, $q(-4) = -484$, $q(13) = 689$, $q(-13) = -4849$, $q(26) = 10556$, $q(-26) = -26884$, $q(52) = 110396$, and $q(-52) = -175396$. Thus $q(x)$ is irreducible in $\mathbb{Q}[x]$.