Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$

Question

I would like to know how to solve part $ii)$ of the following problem:

Let $K /\mathbb{Q}$ be a splitting field for $f(X) =X^4-3X^2+5$.

i) Prove that $f(X)$ is irreducible in $\mathbb{Q}[X]$

ii) Prove that $K$ has degree $8$ over $\mathbb{Q}$.

iii) Determine the Galois group of the extension $K/\mathbb{Q}$ and show how it acts on the roots of $f$.

I've done part i), and have found the roots of $f$ explicitly as:

$$\pm\bigg(\frac{3\pm\sqrt{-11}}{2}\bigg)^{1/2}$$

but am not sure how to show that the extension has degree $8$. If $x_1$ is the root where both of the $\pm$ signs above are $+$ and $x_2$ is the root where only the outer sign is a $+$, then $K = \mathbb{Q}(x_1,x_2)$. By part $i)$, $x_1$ has degree $4$ over $\mathbb{Q}$ and then $x_2$ has degree $1$ or $2$ over $\mathbb{Q}(x_1)$, but I'm not sure how to show that this degree is $2$, or prove the result by other means.

Due to the ordering of the parts, I would expect there to be an answer for ii) that doesn't require computing the entire Galois group of the extension, so would appreciate something along these lines.

Answer 1

After some thought, I've found a very short answer that uses a minimal amount of computation:

With notation as in the question, we have: $$x_1x_2 = \sqrt{5}, x_1^2 = \frac{3+\sqrt{-11}}{2}.$$ Thus $K$ contains the subfield $F = \mathbb{Q}(\sqrt{5},\sqrt{-11})$ which is Galois and degree $4$ over $\mathbb{Q}$, with Galois group $G'\cong V_4$, generated by $\sigma$ and $\tau$, where $\sigma$ fixes $\sqrt{5}$ and permutes $\pm\sqrt{-11}$ and $\tau$ fixes $\sqrt{-11}$ and permutes $\pm\sqrt{5}$.

If $F=K$, then $x_i \in F$ and then the relations above immediately give that $\sigma\tau(x_1) = \pm x_2$ and $\sigma\tau(x_2)=\mp x_1$. But then $\sigma\tau \in G'$ has order $4$, a contradiction, so we must have that $K$ is strictly larger than $F$, so must be of degree $8$ over $\mathbb{Q}$ as desired.

Answer 2

Over $\mathbb{Q}(\sqrt{-11})$ you have two polynomials $$x^2-3-\sqrt{-11}=0$$ and $$y^2-3+\sqrt{-11}=0$$

Now verify that $$(xy)^2=20$$ so that if $y\in\mathbb{Q}(\sqrt{-11},x)$ then $\sqrt{5} \in \mathbb{Q}(\sqrt{-11},x)$. So it only remains to verify that this cannot happen.

For this one approach is to assume $$\sqrt{5} =a+b\sqrt{-11}+(c+d\sqrt{-11})x$$ and solving this for $x$ we are reduced to $x\in \mathbb{Q}(\sqrt{-11},\sqrt{5})$.

Which would mean $$\sqrt{3+\sqrt{-11}}=a+b\sqrt{-11}+c\sqrt{5}+d\sqrt{-11}\sqrt{5}$$

lets write this as $$\sqrt{3+\sqrt{-11}}=p+q\sqrt{5}$$ where $p=a+b\sqrt{-11}$ and $q=c+d\sqrt{-11}$ elements of $\mathbb{Q}(\sqrt{-11})$.

Squaring we have $$3+\sqrt{-11}=p^2+5q^2+2qp\sqrt{5}$$ if $pq\neq 0$ then $\sqrt{5}\in \mathbb{Q}(\sqrt{-11})$. So $pq=0$. If $q=0$ then $\sqrt{3+\sqrt{-11}}\in \mathbb{Q}(\sqrt{-11})$. And if $p=0$ then $\sqrt{3+\sqrt{-11}}=(c+d\sqrt{-11})\sqrt{5}$ which is also impossible.

Answer 3

Rightly $$ x_1=\sqrt{\frac{3+\sqrt{-11}}{2}}= \frac{1}{2}(\sqrt{\sqrt{20}+3}+i\sqrt{\sqrt{20}-3}) $$ and $$ x_2=\sqrt{\frac{3-\sqrt{-11}}{2}}= \frac{1}{2}(\sqrt{\sqrt{20}+3}-i\sqrt{\sqrt{20}-3}) $$ (any determination thereof) suffice to generate the splitting field.

If $x_2\in\mathbb{Q}(x_1)$, also $x_1+x_2=\sqrt{\sqrt{20}+3}$ belongs to $\mathbb{Q}(x_1)$; however, $\alpha=\sqrt{\sqrt{20}+3}$ has degree $4$ over $\mathbb{Q}$ and therefore $\mathbb{Q}(x_1)=\mathbb{Q}(\alpha)$ would be a subset of the reals, which it is not.

---

Why does $\alpha$ have degree $4$? We clearly have $\alpha^2\in\mathbb{Q}(\sqrt{5})$, so we just need to show that $\alpha$ cannot be written as $$ \alpha=a+b\sqrt{5} $$ for rational $a$ and $b$. This means $$ 2\sqrt{5}+3=a^2+5b^2+2ab\sqrt{5} $$ and so $b=a^{-1}$, hence $$ a^4-3a^2+5=0 $$ which is exactly the equation we started with and that has no rational root.

Answer 4

Consider the polynomial $Z^2 - 3Z + 5$. It has two roots, and they both belong to the quadratic extension $L = \mathbb{Q}(\sqrt{-11})$. The roots of your original polynomial are the square roots of these guys, so they each belong to two quadratic extensionf of $L$, and all four of them belong to one extension $K$ of $L$ of degree at most $4$ (over $L$), so degree at most $8$ over $\mathbb{Q}$.

This explicitly constructs a field that includes all of the roots of your polynomial, so it contains the splitting field $K$, but it is not immediately obvious that it is that splitting field. Perhaps a smaller field suffices?

To complete the proof, start with the splitting field $K$ which contains the four roots of the polynomial. Being a field, it must also contain the squares of these roots, and those are none other than the roots of the quadratic. So $K$ does indeed contain $L$. Now, we need to show that it is indeed a degree $4$ extension of it, and not $L$ itself nor a quadratic extension of it. I leave the details to you.

Update. Over $K = \mathbb{Q}(\sqrt{-11})$, i.e. in the field extension of degree $2$, the polynomial $f$ factors as$$\left(X^2 - {{3 + \sqrt{-11}}\over2}\right)\left(X^2 - {{3 - \sqrt{-11}}\over2}\right).$$Obviously, it does not split further over $K$ (check that ${{3 \pm \sqrt{-11}}\over2}$ are not squares in $K$), so we need to adjoin some roots of $f$. Let$$\alpha\text{ be a root of }X^2 - {{3 + \sqrt{-11}}\over2},\text{ }\beta \text{ be a root of }X^2 - {{3 - \sqrt{-11}}\over2}.$$We need to have both of them in the splitting field. Then we will obviously get $-\alpha$ and $-\beta$ there and thus $f$ will split in $K(\alpha, \beta)$. So the only remaining question is as follows: if we adjoin $ \alpha$, could we not automatically drag $\beta$ along, or in other words, is $K(\alpha, \beta) = K(\alpha)$, or equivalently in this case, is $K(\alpha) = K(\beta)$? When $\alpha$ and $\beta$ are roots of polynomials, say $X^2 - 2$ and $X^3 - 3$, we would not even ask this sort of question, since it is pretty clear that adjoining $\sqrt{2}$ would not allow us to get $\sqrt{3}$ as a $\mathbb{Q}$-linear combination of $\sqrt{2}$ and $1$, but this case is slightly less obvious. So is $\beta \in K(\alpha)$, or equivalently, are $\alpha$ and $\beta$ linearly dependent over $K$? Well, assume$$\beta = c\alpha + d,\quad c, \,d \in K.$$Squaring it will give us$$\beta^2 = c^2\alpha^2 + 2cd\alpha + d^2.$$In other words,$$\beta^2 - c^2\alpha^2 - d^2 = 2cd\alpha.$$The left-hand side of the above is an element of $K$ but $\alpha$ is not there, hence the only possibilities are $c = 0$ or $d = 0$. Well, $c = 0$ would imply that $\beta \in K$, which is not true. And $d = 0$ would imply that$$c = {\beta\over\alpha} \in K$$or$$c^2 = {{\beta^2}\over{\alpha^2}} = {{3 - \sqrt{-11}}\over{3 + \sqrt{-11}}} = {{(3 - \sqrt{-11})^2}\over{20}}$$is a square in $K$. That would imply that $20$ is a square in $K$, which is not the case (check this). Thus, we conclude that $K(\alpha, \beta) \supsetneq K(\alpha)$. So we have the following tower of quadratic field extensions$$\mathbb{Q} \subsetneq K \subsetneq K(\alpha) \subsetneq K(\alpha, \beta).$$Since the degree of a field extension is multiplicative, we get$$[K(\alpha, \beta) : \mathbb{Q}] = 2^3 = 8,$$as desired.