# Intermediate field extensions

Let $$\zeta$$ be a primitive 12th-root of unity. How many field extensions are there between $$\mathbb{Q}(\zeta^3)$$ and $$\mathbb{Q}(\zeta)$$?

Here is my solution:

$$\mathbb{Q}(\zeta)$$ is the splitting field of $$X^{12}-1$$, so $$\mathbb{Q}(\zeta):\mathbb{Q}$$ is a Galois extension. The minimal polynomial of $$\zeta$$ is the cyclotomic polynomial $$\Phi_{12}$$, whose degree is 4, therefore $$G(\mathbb{Q}(\zeta):\mathbb{Q})=\mathbb{Z}_{12}^*$$. Then, every non trivial subgroup has order 2, so every intermediate extension of $$\mathbb{Q}(\zeta):\mathbb{Q}$$ is quadratic, which shows that there is no intermediate field extensions between $$\mathbb{Q}(\zeta^3)$$ et $$\mathbb{Q}(\zeta)$$.

I would like to know if my reasoning is right.

• As long as you are confident in that $\Bbb{Q}(\zeta^3)$ is neither of the end points, then, yes, this is one way to conclude. Commented Dec 29, 2020 at 20:22
• What do you mean by "end points"?
– QGM
Commented Dec 29, 2020 at 20:41
• The fields $L=\Bbb{Q}(\zeta)$ and $K=\Bbb{Q}(\zeta^3)$ themselves could be called the end points of $L/K$. If I were a teacher who assigned this as an exercise I would probably insist that a full solution include an argument as to why $L\neq K$, when the "interval" from $K$ to $L$ would have two end points rather than just one :-) Sorry about being unclear. Commented Dec 29, 2020 at 20:45

Since $$[Q(ζ):Q)]=|Z_{12}^*|=4$$ and $$[Q(ζ^3):Q)]=2$$, then we get $$[Q(ζ):Q(ζ^3))]=2$$ and hence your reasons are right.