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].