## 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 $i$th 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.