Find Polynomial $P$ of a splitting field, such that deg($P$) matches $[L:K]$

Question

I am currently trying to solve the following problem:

Let $K$ be a field. Find a polynomial $P$ with degree $n\geq 2$ and a corresponding splitting field $L$, sucht that i) $[L:K]=n$, ii) $n<[L:K]<n!$ and iii) $[L:K]=n!$.

I think for i) i have $K=\mathbb{Q}$ and $L=\mathbb{Q}(\sqrt{2},\sqrt{3})$. Then the Polynomial $P$ would be $(X-\sqrt{2})(X-\sqrt{3})\in \mathbb{Q}(\sqrt{2},\sqrt{3})$ and deg$(P)=2$. Furthermore $[\mathbb{Q}(\sqrt{2},\sqrt{3}),\mathbb{Q}]=$dim$_{\mathbb{Q}}(\mathbb{Q}(\sqrt{2},\sqrt{3}))=2$. Is this correct or am I getting something wrong?

For ii) and iii) I dont really have an idea and I would be gratefull if anyone could help me.

Answer 1

It is a great idea to take $K=\Bbb{Q}$, as rational polynomials are easy to understand. We of course need that $P$ is rational in this case for the question to even make sense.

Hopefully you have seen that the degree of the splitting field is $n\leq [L:\Bbb{Q}]\leq n!$. Thus, the question is asking for a polynomial satisfying the minimal value, the maximal value, and some intermediate value. To make it as easy as possible, we want our polynomials to have as low a degree as possible, so let's think of quadratic, cubic and quartic polynomials.

If $[L:\Bbb{Q}]=n$, that means the splitting field is the same as adding a root $\alpha$ of $P$ to $\Bbb{Q}$, as it is a fact that $[\Bbb{Q}(\alpha):\Bbb{Q}]=\deg P$. So we are looking for an polynomial where all the roots are generated by any of its roots. Can you find a quadratic polynomial with this property?

For $[L:\Bbb{Q}]=n!$, I find it easiest to think of cubic polynomials. We want a polynomial where one root does not generate the other roots. A great way to look for this is looking at polynomials where some roots are real, and some are imaginary. Can you find a cubic with this property, and how would its splitting field look?

For $n<[L:\Bbb{Q}]<n!$, we want a polynomial with some in-between property. One root does not generate all, but it makes it easier in some sense to generate the others. Note that this is impossible with quadratics, as it is a fact that $[L:\Bbb{Q}]$ divides $n!$, but when $n=2$ we only have the options $1,2,4$. Same problem for cubics, so we will look at quartics. As $4!=24$, we need a splitting field of degree $8$ or $12$. In my mind, I immediately think "what if we can find a splitting field which is the composition of a degree $4$ extension and a degree $2$ extension that intersects trivially?". For example, a totally real extension and an imaginary quadratic extension, maybe something along the lines of the previous example?

I tried not to solve the exercise completely, try it yourself for a bit, and if you get stuck, write me a comment.

Answer 2

Thanks to SomeCallMeTim's advice I think I now have the solutions to my problem:

For i) I have corrected my solution to $P=x^2-2\in\mathbb{Q}$. Then the splitting field would be $\mathbb{Q}(\sqrt{2})$ and the degree of the Polynomial would match $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$.

For ii) we know, as SomeCallMeTim said, that the degree of the polynomial needs to be at least $4$. I chose $P=x^4+2$. Then the roots would be $\sqrt[4]{2}e^{i\frac{\pi}{4}},\sqrt[4]{2}e^{i\frac{3\pi}{4}},\sqrt[4]{2}e^{i\frac{5\pi}{4}}$ and $\sqrt[4]{2}e^{i\frac{7\pi}{4}}$. Then the splitting field would be $\mathbb{Q}(\sqrt[4]{2},i)$ because $e^{i\frac{\pi}{4}}=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$ is in this field. We can use the multiplicative rule of filed extencions to get $[\mathbb{Q}(\sqrt[4]{2},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{2},i):\mathbb{Q}(\sqrt[4]{2})][\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]=8$ which, by looking at the degree of their minimal polynomials, would be $2$ and $4$ respectively.

For i) I chose a similar polynomial of lesser degree $P=x^3+2$. The roots are $\sqrt[3]{2},\sqrt[3]{2}\omega$ and $\sqrt[3]{2}\omega^2$, where $\omega=e^{\frac{2\pi i}{3}}$. With the exact same arguments as in ii) we have $[\mathbb{Q}(\sqrt[3]{2}:\omega)]=6$ whis is the factorial of the degree of the polynomial.