Give an example of a cubic polynomial in $\mathbb{Q}[x]$ whose splitting field over $\mathbb{Q}$ has degree $8$ or explain why none exists

Question

I understand that the degree of the splitting field is at most $n!$ where $n$ is the degree of the polynomial, so the degree of the splitting field must be $<6$ and hence cannot equal $8$. I don't understand why this is the case, or maybe there is a more intuitive understanding to this problem.

Any help appreciated!

Answer 1

I assume you have heard that the degree of a splitting field for a degree $n$ polynomial is at most $n!$ but don't have good intuition as to why this is so.

Here is an intuitive argument.

You need to know this fact: If $K$ is a subfield of $\mathbb C$, and if $f(x)$ is an irreducible polynomial of degree $n$ in $K[x]$, and if $\alpha \in \mathbb C$ is a root of $f(x)$, then $[K(\alpha):K]=n$. (Actually it is unnecessary to assume that $K$ is a subfield of $\mathbb C$, but whatever...)

Suppose $f_{(n)}(x)$ is your degree $n$ polynomial with $\mathbb Q$ coefficients, which you want to find the splitting field for.

Let $\alpha_1 \in \mathbb C$ be a root of $f_{(n)}(x)$. In the worst case scenario, $f_{(n)}(x)$ is irreducible in $\mathbb Q[x]$. By the fact above, we have $$[\mathbb Q(\alpha_1):\mathbb Q] = n.$$

Since $\alpha_1$ is a root of $f_{(n)}(x)$, we can factorise $f_{(n)}(x)$ in $\mathbb Q(\alpha_1)[x]$ as $$f_{(n)}(x) = (x-\alpha_1)f_{(n-1)}(x)$$ where $f_{(n-1)}(x)$ is some degree $n-1$ polynomial with coefficients in $\mathbb Q(\alpha_1)$.

Let $\alpha_2\in \mathbb C$ be a root of $f_{(n-1)}(x)$. In the worst case scenario, $f_{(n-1)}(x)$ is irreducible in $\mathbb Q(\alpha_1)[x]$. Using the above fact once more, you find that $$[\mathbb Q(\alpha_1,\alpha_2) : \mathbb Q(\alpha_1) ] = n-1.$$ But this means that $$[\mathbb Q(\alpha_1, \alpha_2): \mathbb Q] = [\mathbb Q(\alpha_1,\alpha_2) : \mathbb Q(\alpha_1) ][\mathbb Q(\alpha_1) : \mathbb Q ] = n(n-1).$$

Now factorise $f_{(n)}(x)$ in $\mathbb Q(\alpha_1, \alpha_2)[x]$ as $$f_{(n)}(x) = (x-\alpha_1)(x-\alpha_2)f_{(n-2)}(x)$$ where $f_{(n-2)}(x)$ is a degree $n-2$ polynomial with coefficients in $\mathbb Q(\alpha_1, \alpha_2)$...

I hope you can see where this is going. Eventually, you end up with $f_{(n)}(x)$ fully factorised as $$ f_{(n)}(x) = (x-\alpha_1)(x-\alpha_2) \dots (x-\alpha_n).$$ and in the worst case scenario, $$[\mathbb Q(\alpha_1, \alpha_2 \dots \alpha_n) : \mathbb Q] = n(n-1)\dots 1 = n!.$$

Answer 2

Let be $f(x)=x^3+bx^2+cx+d=(x-\alpha)(x-\beta)(x-\gamma)\Rightarrow x^3+bx^2+cx+d=x^3+(-\alpha-\beta-\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma$

we have $ -\alpha-\beta-\gamma=b\Rightarrow \gamma=-\alpha-\beta-b \Rightarrow \mathbb Q(\gamma)\subseteq \mathbb Q(\alpha,\beta) $. If $\alpha\notin \mathbb Q(\beta)$ then $ [\mathbb Q(\alpha,\beta):\mathbb Q]=6 $ or $\alpha\in \mathbb Q(\beta)$ then $ [\mathbb Q(\alpha,\beta):\mathbb Q]=3$

Answer 3

The degrees of splitting fields behave more rigidly than that, thanks to Lagrange's theorem. Indeed, if $ f $ is a separable polynomial in $ K[X] $ for any field $ K $ with splitting field $ L/K $, then $ L/K $ is Galois. Let $ G $ be its Galois group. It is a basic result that $ |G| = [L : K] $; on the other hand, since $ L $ is generated over $ K $ by the roots of $ f $, $ G $ has a faithful action on the roots of $ f $, which gives rise to an embedding $ G \to S_n $ where $ n $ is the degree of $ f $. By Lagrange, $ |G| = [L : K] $ divides $ |S_n| = n! $.