## Tag Archives: computation

### Compute the splitting field of x⁶-4 over QQ

Compute the splitting field of $p(x) = x^6-4$ over $\mathbb{Q}$ and its degree.

Note that $x^6-1$ factors as $x^6-1 = (x+1)(x-1)(x^2+x+1)(x^2-x+1)$. (Using only the difference of squares and sum of cubes formulas familiar to middle schoolers.) Using the quadratic formula (again familiar to middle schoolers) we see that the roots of $p(x)$ are $\pm 1$ and $\pm \dfrac{1}{2} \pm \dfrac{i\sqrt{3}}{2}$.

Now if $\zeta$ is a 6th root of 1, then $\sqrt[6]{4}\zeta$ is a root of $p(x)$, where $\sqrt[6]{4} = \sqrt[3]{2}$ denotes the positive real 6th root of 4 (aka the positive cube root of 2). (Middle schoolers could verify that.) Evidently, then, the splitting field of $p(x)$ is $\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2}(1+i\sqrt{3}, \sqrt[3]{2}(1-i\sqrt{3}) = \mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})$.

Now $\mathbb{Q}(\sqrt[3]{2})$ has degree 3 over $\mathbb{Q}$, and $\mathbb{Q}(i\sqrt{3})$ has degree 2 over $\mathbb{Q}$. (Use Eisenstein for both.) By Corollary 22 on page 529 of D&F, then, $\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})$ has degree (as a middle schooler could compute) $3 \cdot 2 = 6$ over $\mathbb{Q}$.

So we have proved not only that the splitting field of $p(x)$ has degree 6 over $\mathbb{Q}$, but, in a metamathematical twist, also that a middle schooler could prove this as well.

### Compute the splitting field of x⁴+x²+1 over QQ

Compute the splitting field of $p(x) = x^4 + x^2 + 1$ over $\mathbb{Q}$, and its degree.

Note that $p = g \circ h$, where $g(y) = y^2+y+1$ and $h(x) = x^2$. Evidently, the roots of $g(y)$ are $\eta = \dfrac{-1 \pm i\sqrt{3}}{2}$, and $\zeta$ is a root of $p$ if $h(\zeta) = \eta$. This yields the (distinct) roots $\zeta = \pm \sqrt{\dfrac{-1 \pm i\sqrt{3}}{2}}$. There are four of these, and so we have completely factored $p$.

The splitting field of $p(x)$ over $\mathbb{Q}$ is thus $\mathbb{Q}(\pm \sqrt{\dfrac{-1 \pm i\sqrt{3}}{2}}) = \mathbb{Q}(\sqrt{-1/2 \pm \sqrt{-3/4}})$.

Using this previous exercise, we have $\sqrt{-1/2 + \sqrt{-3/4}} = 1/2 + i\sqrt{3}/2$ and $\sqrt{-1/2 - \sqrt{-3/4}} = 1/2 - i\sqrt{3}/2$.

Evidently, then, the splitting field of $p(x)$ is $\mathbb{Q}(i\sqrt{3})$. Since $i\sqrt{3}$ is a root of the irreducible $x^2+3$ (Eisenstein), this extension of $\mathbb{Q}$ has degree 2.

### Compute the splitting field of x⁴+2 over QQ

Compute the splitting field of $p(x) = x^4+2$ over $\mathbb{Q}$ and its degree.

The roots of $p(x)$ exist in $\mathbb{C}$ (if we don’t know this yet, just assume some roots are in $\mathbb{C}$). Let $\zeta = a+bi$ be such a root; then $\zeta^4 = -2$.

Evidently, $\zeta^4 = [a^4-6a^2b^2 + b^4] + [4ab(a^2-b^2]i$. Comparing coefficients, we see that either $a = 0$, $b = 0$, or $a^2 = b^2$. If $a = 0$, then $b^4 = -2$, a contradiction in $\mathbb{R}$. Likewise, if $b = 0$ we get a contradiction. Thus $a^2 = b^2$. Substituting, we have $b^4 - 6b^4 + b^4 = -2$, so $-4b^4 = -2$, and so $b^4 = 1/2$. There is a unique positive 4th root of $1/2$, which we denote by $2^{-1/4}$; so $b = \pm 2^{-1/4}$ and $a = \pm 2^{-1/4}$, and hence $\zeta = \pm 2^{-1/4} \pm 2^{-1/4}i$. There are 4 such roots, and so we have completely factored $p(x)$. (WolframAlpha agrees.)

So the splitting field of $p(x)$ is $\mathbb{Q}(\pm 2^{-1/4} \pm 2^{-1/4}i)$. Note that if $\zeta_1 = 2^{-1/4} + 2^{-1/4}i$ and $\zeta_2 = \pm 2^{-1/4} - 2^{-1/4}i$, then $(\zeta_1 + \zeta_2)/2 = 2^{-1/4}$, and $(\zeta_1 - \zeta_2)/22^{-1/4} = i$. Thus $\mathbb{Q}(\pm 2^{-1/4} \pm 2^{-1/4}i) = \mathbb{Q}(i,2^{-1/4})$.

Note that $2^{-1/4}$ is a root of $q(x) = 2x^4 - 1$, and that the reverse of $q(x)$ is $t(x) = -x^4+2$. Now $t(x)$ is irreducible over the UFD $\mathbb{Z}[i]$, since it is Eisenstein at the irreducible element $1+i$. So $t(x)$ is irreducible over $\mathbb{Q}(i)$, the field of fractions of $\mathbb{Z}[i]$ as a consequence of Gauss’ Lemma. As we showed previously, the reverse of $t$, namely $q$, is also irreducible over $\mathbb{Q}(i)$. So $\mathbb{Q}(i,2^{-1/4})$ has degree 4 over $\mathbb{Q}(i)$, and thus has degree 8 over $\mathbb{Q}$.

### Compute the splitting field of x⁴-2 over QQ

Compute the splitting field of $p(x) = x^4-2$ over $\mathbb{Q}$, as well as its degree.

Let $\sqrt[4]{2}$ denote the positive real fourth root of 2. Evidently, $p(x) = (x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-i\sqrt[4]{2})(x+i\sqrt[4]{2})$. So the splitting field of $p(x)$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[4]{2}, i\sqrt[4]{2}) = \mathbb{Q}(i,\sqrt[4]{2})$.

Note that $\mathbb{Q}(i)$ has degree 2 over $\mathbb{Q}$. Over the UFD $\mathbb{Z}[i]$, $p(x)$ is Eisenstein at $1+i$, hence irreducible, and so by Gauss’ Lemma, is irreducible over $\mathbb{Q}(i)$ (the field of fractions of $\mathbb{Z}[i]$). So $\mathbb{Q}(i,\sqrt[4]{2})$ has degree 4 over $\mathbb{Q}(i)$. We can visualize this scenario in the following diagram.

A field diagram

So $\mathbb{Q}(i,\sqrt[4]{2})$ has degree 8 over $\mathbb{Q}$.

### A procedure for finding a polynomial satisfied by an element of a given algebraic field extension

Let $F$ be a field, $K$ an extension of $F$ of finite degree, and let $\alpha \in K$. Show that if $A$ is the matrix of the linear transformation $\varphi_\alpha$ corresponding to ‘multiplication by $\alpha$‘ (described here) then $\alpha$ is a root of the characteristic polynomial of $A$. Use this result to obtain monic polynomials of degree 3 satisfied by $\alpha = \sqrt[3]{2}$ and $\beta = 1 + \sqrt[3]{2} + \sqrt[3]{4}$.

Let $\Psi : K \rightarrow \mathsf{Mat}_n(F)$ be the $F$-linear transformation described here. If $c(x)$ is the characteristic polynomial of $A$, then we have $c(A) = 0$. On the other hand, $0 = c(A)$ $= c(\Psi(\alpha))$ $= \Psi(c(\alpha))$, and so $c(\alpha) = 0$ since $\Psi$ is injective. So $\alpha$ is a root of $c(\alpha)$.

Consider the basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ of $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$. Evidently, with respect to this basis, the matrix of $\varphi_\alpha$ is $A = \begin{bmatrix} 0 & 0 & 2 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$. As we showed previously, the characteristic polynomial of $A$ is $x^3-2$. So $\sqrt[3]{2}$ satisfies $x^3-2$. (Surprise!)

Similarly, $\varphi_\beta$ has the matrix $B = \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 2 \\ 1 & 1 & 1 \end{bmatrix}$. Evidently, the characteristic polynomial of $B$ is $x^3-3x^2-3x-1$. We can verify that $\beta$ actually satisfies this polynomial (WolframAlpha agrees.)

### Compute the degree of a given extension of the rationals

Compute the degrees of $\mathbb{Q}(\sqrt{3+4i} + \sqrt{3-4i})$ and $\mathbb{Q}(\sqrt{1+\sqrt{-3}} + \sqrt{1-\sqrt{-3}})$ over $\mathbb{Q}$.

Let $\zeta = \sqrt{3+4i} + \sqrt{3-4i}$. Evidently, $\zeta^2 = 16$. (WolframAlpha agrees.) That is, $\zeta$ is a root of $p(x) = x^2 - 16 = (x+4)(x-4)$. Thus the minimal polynomial of $\zeta$ over $\mathbb{Q}$ has degree 1, and the extension $\mathbb{Q}(\zeta)$ has degree 1 over $\mathbb{Q}$.

Similarly, let $\eta = \sqrt{1+\sqrt{-3}} + \sqrt{1-\sqrt{-3}}$. Evidently, $\eta^2 = 6$ (WolframAlpha agrees), so that $\eta$ is a root of $q(x) = x^2-6$. $q$ is irreducible by Eisenstein’s criterion, and so is the minimal polynomial of $\eta$ over $\mathbb{Q}$. The degree of $\mathbb{Q}(\eta)$ over $\mathbb{Q}$ is thus 2.

### Compute the degree of a given extension of the rationals

Find the degree of $\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ over $\mathbb{Q}$.

We have $\sqrt{3+2\sqrt{2}} = \sqrt{3 + \sqrt{8}}$, and $3^2 - 8 = 1^2$ is square over $\mathbb{Q}$. By this previous exercise, $\sqrt{3+2\sqrt{2}} = 1 + \sqrt{2}$. So $\mathbb{Q}(\sqrt{3 + 2\sqrt{2}}) = \mathbb{Q}(1+\sqrt{2}) = \mathbb{Q}(\sqrt{2})$ has degree 2 over $\mathbb{Q}$.

### Compute the degree of a given extension of the rationals

Prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Conclude that $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ has degree 4 over $\mathbb{Q}$. Find an irreducible polynomial satisfied by $\sqrt{2} + \sqrt{3}$.

The $(\supseteq)$ inclusion is clear. To see the $(\subseteq)$ inclusion, let $\zeta = \sqrt{2} + \sqrt{3}$. Then evidently $\frac{1}{2}(\zeta^3 - 9\zeta) = \sqrt{2}$ and $-\frac{1}{2}(\zeta^3 - 11\zeta) = \sqrt{3}$. (WolframAlpha agrees; see here and here.) Thus the extensions are equal.

Recall that $\mathbb{Q}(\sqrt{2})$ has degree 2 over $\mathbb{Q}$. Now we claim that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree 2 over $\mathbb{Q}(\sqrt{2})$. To see this, suppose to the contrary that $\sqrt{3} \in \mathbb{Q}(\sqrt{2})$. Say $(a+b\sqrt{2})^2 = 3$ for some rationals $a$ and $b$. Coparing coefficients, we have $a^2 + 2b^2 = 3$ and $2ab = 0$; if $a = 0$, then $b^2 = \frac{3}{2}$, and if $b = 0$, then $a^2 = 3$. Either case yields a contradiction.

So $\mathbb{Q}(\sqrt{2} + \sqrt{3})$ has degree 4 over $\mathbb{Q}$.

The minimal polynomial of $\zeta$ has degree 4 over $\mathbb{Q}$. We can solve the system $\zeta^4 + a\zeta^3 + b\zeta^2 + c\zeta + d = 0$ over $\mathbb{Q}$, and in so doing we see that $p(x) = x^4-10x^2+1$ is the minimal polynomial of $\sqrt{2} + \sqrt{3}$.

### 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 $\mathbb{Q}$.