Calculating the degree of some extension of $\mathbb{Q}_3$

Question

Let $p=3$ and $\zeta$ be a cube root of unity not equal $1$. Consider the field of $3$-adic numbers $\mathbb{Q}_3$. At the beginning of section 5.4 of Fernando Gouvea's book $p$-adic numbers - An Introduction, he states that the field $\mathbb{Q}_3(\sqrt{2},\zeta)$ is an extension of degree $4$. Furthermore, the fields $\mathbb{Q}_3(\sqrt{2})$ and $\mathbb{Q}_3(\zeta)$ are both extensions of degree $2$. Now I would like to understand why this is true.

• I understand why we have $[\mathbb{Q}_3(\sqrt{2}):\mathbb{Q}_3]=2$. This is true because $f = x^2-2$ is irreducible over $\mathbb{Q}_3$, as 2 is a quadratic nonresidue, so we can apply Eisenstein's criterion. This means that $f$ is the minimal polynomial of $\sqrt{2}$ which has degree $2$. Therefore, the claim holds.
• I am also really sure about the fact that $[\mathbb{Q}_3(\zeta):\mathbb{Q}_3]=2$ is true. As $\zeta$ is a cubic root of unity, the minimal polynomial of $\zeta$ divides $x^3 - 1$ which has $1$ as a root, so it is obviously not irreducible. If we divide $x^3 - 1$ by $x-1$, then we obtain the polynomial $g=x^2+x+1$. This must mean that $g$ is the minimal polynomial of $\zeta$ if the degree of the extension is really $2$. But I do not know how to show why $g$ is irreducible over $\mathbb{Q}_3$.
• To show that $[\mathbb{Q}_3(\zeta,\sqrt{2}):\mathbb{Q}_3] = 4$, we only have to show either $\zeta \not\in \mathbb{Q}_3(\sqrt{2})$ or $\sqrt{2} \not\in \mathbb{Q}_3(\zeta)$, so we can apply the tower law.
• For instance, we assume $\zeta \in \mathbb{Q}_3(\sqrt{2})$. Then there exist coefficients $c_1,c_2 \in \mathbb{Q}_3$ such that $\zeta = c_1 + c_2 \sqrt{2}$ because we know that $\mathbb{Q}_3(\sqrt{2})$ is a vector space of $\mathbb{Q}_3$ of dimension $2$ with basis $\{1,\sqrt{2}\}$. Then $$ 0 = \zeta^2 + \zeta + 1 = (c_1^2 + 2c_2^2+c_1+1) + (2c_1c_2+c_2)\sqrt{2} $$ and therefore $c_1^2 + 2c_2^2+c_1+1=0$ and $2c_1c_2+c_2=0$. But this looks like a really difficult system of equations to solve, as I know nothing about $c_1$ and $c_2$.

Could you please help me with this problem? Thank you in advance!

Answer 1

Here’s an argument that you may find more direct.

First, your $g(X)=X^2+X+1$ has $g(X+1)=X^2+3X+3$, Eisenstein, so irreducible. Not only that, its root $\zeta_3-1$ clearly has (additive) $v_3$-valuation equal to $\frac12$, so that $\Bbb Q_3(\zeta_3)$ is quadratic and (totally) ramified over $\Bbb Q_3$.

Second, since $\Bbb F_3$ has no square root of $2$, you need a quadratic residue field extension to catch $\sqrt2$ as an element algebraic over $\Bbb Q_3$, or in other words, $\Bbb Q_3(\sqrt2\,)$ is quadratic and unramified over $\Bbb Q_3$.

Thus $\Bbb Q_3(\zeta_3)$ and $\Bbb Q_3(\sqrt2\,)$ are different quadratic extensions of $\Bbb Q_3$, so that their compositum is of degree $4$.

Answer 2

A preliminary bit of Kummer theory will certainly do no harm. Let $K$ be a field of characteristic $\neq 2$, such that neither $2$ nor $-3$ are squares in $K$. Then $K(\sqrt 2)$ and $K(\sqrt {-3})=K(\zeta)$ (where $\zeta$ is a primitive 3-rd root of 1) are quadratic extensions, and their compositum $L=K(\sqrt 2,\sqrt {-3})$ is Galois abelian according to Kummer, with group $G$ isomorphic to the subgroup $<[2], [-3]>$ of $K^*/{K^*}^2$ generated by the classes of $2$ and $-3$ mod ${K^*}^2$. Thus $G \cong C_2 \times C_2$ (multiplicative notation) iff $[2]$ and $[-3]$ are linearly independant mod ${K^*}^2$ (here we view $K^*/{K^*}^2$, in additive notation, as a $\mathbf F_2$-vector space), or equivalently $-6\notin {K^*}^2$.

In the case $K=\mathbf Q_3$, writing $v_3$ for the $3$-adic valuation, we have $v_3 (2)=0, v_3(-3)=1$, $v_3(-6)=1$, hence $2,-3,-6$ are not squares and $G \cong C_2 \times C_2$. Note that the same holds for $K=\mathbf Q$.