Tag Archives: rational numbers

Compute the degree of a given extension of QQ

Compute the degree of \mathbb{Q}(\alpha) over \mathbb{Q}, where \alpha is 2+\sqrt{3} or 1 + \sqrt[3]{2} + \sqrt[3]{4}.


Since 2+\sqrt{3} \in \mathbb{Q}(\sqrt{3}), the degree of \mathbb{Q}(2+\sqrt{3}) over \mathbb{Q} is at most 2. We can solve the linear system \alpha^2 + a\alpha + b = 0 in \mathbb{Q}(\sqrt{3}) (as a vector space over \mathbb{Q}) to find a polynomial satisfied by \alpha; evidently 2+\sqrt{3} is a root of p(x) = x^2 - 4x + 1. (WolframAlpha agrees.) Evidently, p(x+1) = x^2-2x-2 (WolframAlpha agrees), which is irreducible by Eisenstein’s criterion; so p(x) is irreducible. Thus p(x) is the minimal polynomial of 2+\sqrt{3} over \mathbb{Q}, and so the degree of \mathbb{Q}(2+\sqrt{3}) over \mathbb{Q} is 2.

In a similar fashion, \beta = 1+\sqrt[3]{2} + \sqrt[3]{4}, as an element of \mathbb{Q}(\sqrt[3]{2}), has degree at most 3 over \mathbb{Q}. Evidently, \beta is a root of q(x) = x^3 - 3x^2 - 3x - 1 (WolframAlpha agrees). Evidently, q(x+1) = x^3-6x-6, which is irreducible by Eisenstein. So q(x) is the minimal polynomial of \beta, and thus \mathbb{Q}(\beta) has degree 3 over \mathbb{Q}.

Compute the minimal polynomial of an algebraic number over QQ

Find the minimal polynomial of 1+i over \mathbb{Q}.


Note that 1+i \in \mathbb{Q}(i), and that \mathbb{Q}(i) has degree 2 over \mathbb{Q} (since i is a root of the irreducible x^2+1). That is, the degree of \mathbb{Q}(1+i) over \mathbb{Q} is at most 2.

Knowing an upper bound on the degree of the minimal polynomial of \alpha = 1+i, we can compute the powers of \alpha and solve the linear system \alpha^2 + a\alpha + b = 0.

Evidently, 1+i is a root of p(x) = x^2-2x+2. (WolframAlpha agrees.) Moreover, p(x) is irreducible over \mathbb{Q} by Eisenstein’s criterion; so it is the minimal polynomial of 1+i over \mathbb{Q}.

As vector spaces over QQ, RR is isomorphic to any finite direct power of itself

Prove that as vector spaces over \mathbb{Q}, \mathbb{R} and \mathbb{R}^n are isomorphic for any positive natural number n.


Note that \mathbb{Q} is countable. By this previous exercise, any basis of \mathbb{R} over \mathbb{Q} must have cardinality \mathsf{card}\ \mathbb{R}. Likewise, any basis of \mathbb{R}^n over \mathbb{Q} must have cardinality \mathsf{card}\ \mathbb{R}^n = \mathsf{card}\ \mathbb{R}.

Thus, as vector spaces over the rationals, \mathbb{R} \cong_\mathbb{Q} \bigoplus_\mathbb{R} \mathbb{Q} \cong_\mathbb{Q} \mathbb{R}^n. In particular, \mathbb{R} and \mathbb{R}^n are isomorphic as abelian groups.

Show that two tensor products are isomorphic as modules

Show that \mathbb{Q} \otimes_\mathbb{Q} \mathbb{Q} and \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q} are isomorphic as left \mathbb{Q}-modules.


Let \frac{a}{b} \otimes \frac{c}{d} be a simple tensor in \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q}. We have \frac{a}{b} \otimes \frac{c}{d} = \frac{ad}{bd} \otimes \frac{c}{d} = \frac{a}{bd} \cdot d \otimes \frac{c}{d} = \frac{a}{bd} \otimes d \cdot \frac{c}{d} = \frac{a}{bd} \otimes \frac{c}{1} = \frac{ac}{bd} \otimes \frac{1}{1}. In particular, every simple tensor (hence every element of \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q}) can be written as q \otimes 1 = q \cdot (1 \otimes 1) where q \in \mathbb{Q}. This gives (by the universal property of free modules) a unique \mathbb{Q}-module homomorphism \Phi : \mathbb{Q} \rightarrow \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q} such that q \mapsto q \otimes 1. Certainly \Phi is surjective.

Certainly every simple tensor (and every element) in \mathbb{Q} \otimes_\mathbb{Q} \mathbb{Q} can be written in the form q \cdot (1 \otimes 1). By the universal property of free modules, this yields a surjective \mathbb{Q}-module homomorphism \Theta : \mathbb{Q} \rightarrow \mathbb{Q} \otimes_\mathbb{Q} \mathbb{Q}.

Now define \psi : \mathbb{Q} \times \mathbb{Q} \rightarrow \mathbb{Q} by (a,b) \mapsto ab. Certainly \psi is \mathbb{Q}-bilinear (and also \mathbb{Z}-bilinear), and so induces a \mathbb{Q}-module homomorphism \Psi : \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q} \rightarrow Q such that \Psi(a \otimes b) = ab. Certainly \Psi is surjective, and so \Phi and \Theta are isomorphisms. Thus \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q} \cong_\mathbb{Q} \mathbb{Q} \cong_\mathbb{Q} \mathbb{Q} \otimes_\mathbb{Q} \mathbb{Q}.

If the product of two rational polynomials is an integer polynomial, then the pairwise product of any two coefficients is an integer

Let f(x), g(x) \in \mathbb{Q}[x] such that f(x)g(x) \in \mathbb{Z}[x]. Prove that the product of any coefficient of f(x) with any coefficient of g(x) is an integer.


Note that f(x)g(x) is in \mathbb{Z}[x] and factors in \mathbb{Q}[x]. By Gauss’ Lemma, there exist r,s \in \mathbb{Q} such that rf, sg \in \mathbb{Z}[x] and (rf)(sg) = fg. Since \mathbb{Q} is an integral domain, in fact rs = 1. Let f_i and g_i denote the coefficients of f and g, respectively; we have rf_i \in \mathbb{Z} and r^{-1}g_i \in \mathbb{Z}, so that f_ig_j \in \mathbb{Z} for all i and j.

Note that the proof still works if we replace \mathbb{Z} by an arbitrary unique factorization domain and \mathbb{Q} by its field of fractions.

Characterize the solutions of a linear Diophantine equation over QQ[x]

Suppose f(x) and g(x) are nonzero polynomials in \mathbb{Q}[x] with greatest common divisor d(x).

  1. Given h(x) \in \mathbb{Q}[x], show that there exist a(x), b(x) \in \mathbb{Q}[x] such that h = af + bg if and only if h is divisible by d.
  2. Given h, if a_0 and b_0 are a particular solution to the equation a_0f + b_0g = h, prove that every solution is of the form a = a_0 + m\frac{g}{d} and b = b_0 - m \frac{f}{d} for some m \in \mathbb{Q}[x].

We will approach this problem from a more abstract point of view.

Lemma: Let R be a Bezout domain and let a,b \in R be nonzero. Suppose (a,b) = (d) and a = dt. If bc \in (a), then c \in (t). Proof: Write b = du. Since (a,b) = (d), we have ax + by = d for some x,y \in R. Since bc \in (a), we have bc = ak for some k. Now bcx = akx, so that bcx = k(d-by), and thus b(cx + yk) = kd. Then u(cx + yk) = k. Now dtu(cx + yk) = au(cx + yk) = ak = bc = duc, so that t(cx + yk) = c, and we have c \in (t). \square

This lemma generalizes part (a) of this previous exercise.

Lemma: Let R be a Bezout domain and let a,b,h \in R with (a,b) = (d). There exist x,y \in R such that ax + by = h if and only if d|h. Proof: If d|h, then h \in (d) = (a,b), so that x and y exist with h = ax + by. Conversely, if x and y exist with ax + by = h, then h \in (a,b) = (d), so that d|h. \square

Part (a) of this problem follows because \mathbb{Q}[x] is a Euclidean domain, hence a principal ideal domain, and thus a Bezout domain.

Lemma: Let R be a Bezout domain and let a,b,x_0,y_0,h \in R such that a,b \neq 0, (a,b) = (d), d|h, and ax_0 + by_0 = h. If x,y \in R such that ax + by = h, then x = x_0 + m\frac{b}{d} and y = y_0 - m\frac{a}{d} for some m \in R. Proof: Note that (ax + by) - (ax_0 + by_0) = h - h = 0. Then a(x - x_0) + b(y - y_0) = 0, and thus a(x - x_0) = b(y_0 - y). Now a(x - x_0) is in (b), so that by the first lemma, x - x_0 \in (\frac{b}{d}). Say m\frac{b}{d} = x - x_0; then x = x_0 + m\frac{b}{d}. Plugging this back into a(x - x_0) = b(y_0 - y) and solving for y (using the fact that d divides a), we see that y = y_0 - m\frac{a}{d}. \square

Again, this solves the stated problem because \mathbb{Q}[x] is a Bezout domain. But more generally, this lemma allows us to solve linear Diophantine equations in two variables over any Bezout domain.

Compute the greatest common divisor of two polynomials over QQ

Compute the greatest common divisor of a(x) = x^3 + 4x^2 + x - 6 and b(x) = x^5 - 6x + 5 in \mathbb{Q}[x] and write it as a linear combination of a(x) and b(x).


Using the long division algorithm for polynomials, we have b(x) = a(x)(x^2 - 4x + 15) + (-50x^2 - 45x + 95), a(x) = (-50x^2 - 45x + 95)(\frac{-1}{50}x - \frac{31}{500}) + (\frac{11}{100}x - \frac{11}{100}), and -50x^2 - 45x + 95 = (\frac{11}{100}x - \frac{11}{100})(\frac{-5000}{11}x - \frac{9500}{11}). Thus the greatest common divisor of a(x) and b(x) is \frac{11}{100}x + \frac{11}{100}.

Backtracking, we have \frac{11}{100}x - \frac{11}{100} = a(x)(\frac{-1}{50}x^3 + \frac{9}{500}x^2  - \frac{13}{250}x + \frac{7}{100}) + b(x)(\frac{1}{50}x + \frac{31}{500}).

Compute greatest common divisors in QQ[x]

Compute the greatest common divisor of a(x) = x^5 + 2x^3 + x^2 + x + 1 and b(x) = x^5 + x^4 + 2x^3 + 2x^2 + 2x + 1 in \mathbb{Q}[x] and write it as a linear combination of a(x) and b(x).


Using the long division algorithm for polynomials, we have a(x) = b(x) + (x^4 + x^2 + x), b(x) = x(x^4 + x^2 + x) + (x^3 + x + 1), and x^4 + x^2 + x = x(x^3 + x + 1). So the greatest common divisor of a(x) and b(x) is x^3 + x + 1. Backtracking, we see that x^3 + x + 1 = a(x)(x+1) - b(x)(x). Indeed, a(x) = (x^3 + x + 1)(x^2+1) and b(x) = (x^3 + x + 1)(x^2 + x + 1).

Compute the greatest common divisor of x³-2 and x+1 in QQ[x]

Compute the greatest common divisor in a(x) = x^3 - 2 and b(x) = x+1 in \mathbb{Q}[x] and write it as a linear combination of a(x) and b(x).


Using polynomial long division, we find that a(x) = b(x)(x^2-x+1) + (-3). Thus, 1 = \frac{-1}{3}a(x) + \frac{1}{3}(x^2-x+1)b(x), and in particular \mathsf{gcd}(a(x),b(x)) = 1.

Exhibit infinitely many group automorphisms on the nonzero rationals

Exhibit infinitely many group automorphisms on the multiplicative group of nonzero rationals. Show that as a ring, \mathbb{Q} has only the trivial automorphism.


Note that every element of \mathbb{Q} can be written uniquely as \prod p_i^{k_i}, where p_i are primes and k_i are integers. In fact, we can define p : \mathbb{N} \rightarrow \mathbb{Z} so that p_i is the ith prime, since for every element of \mathbb{Q} only finitely many of the k_i are not 1.

Let \sigma be an automorphism of \mathbb{N}, and define \varphi_\sigma : \mathbb{Q}^\times \rightarrow \mathbb{Q}^\times by \varphi_\sigma(\prod p_i^{k_i}) = \prod p_i^{k_{\sigma(i)}}. Note that \varphi_{\sigma} is bijective since \varphi_{\sigma^{-1}} is a two-sided inverse. Moreover, \varphi_\sigma is a multiplicative group homomorphism since \varphi_\sigma((\prod p_i^{k_i})(\prod p_i^{\ell_i})) = \varphi_\sigma(\prod p_i^{k_i+\ell_i}) = \varphi_\sigma(\prod p_i^{(k+\ell)_i}) = \prod p_i^{(k+\ell)_{\sigma(i)}} = \prod p_i^{k_{\sigma(i)} + \ell_{\sigma(i)}} = (\prod p_i^{k_{\sigma(i)}})(\prod p_i^{\ell_{\sigma(i)}}) = \varphi_{\sigma}(\prod p_i^{k_i}) \varphi_\sigma(\prod p_i^{\ell_i}).

We can see that the automorphisms \varphi_\sigma are pairwise distinct as follows. Suppose \sigma,\tau \in \mathsf{Sym}(\mathbb{N}) are distinct; then there exists t \in \mathbb{N} such that \sigma^{-1}(t) \neq \tau^{-1}(t). Now \varphi_\sigma(p_t) = p_{\sigma^{-1}(t)} and \varphi_\tau(p_t) = p_{\tau^{-1}(t)} are distinct. Thus we have infinitely many group automorphisms on \mathbb{Q}^\times.

Now let \varphi : \mathbb{Q} \rightarrow \mathbb{Q} be a ring isomorphism; since \varphi is surjective, we have \varphi(1) = 1. Then \varphi(n) = n for all integers n, and we have \varphi(a/b) = \varphi(a)\varphi(b)^{-1} = a/b for all rational numbers a/b. Thus \varphi is the identity on \mathbb{Q}. So \mathbb{Q} has only one ring automorphism, namely the identity.

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