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}.