Find a primitive element in the splitting field of $x^4-8x^2+15$.

Question

I'm trying to find a primitive element in the splitting field of $x^4-8x^2+15$. I don't know in general what should I do with this kind of questions. Should I solve the roots and get the Galois group?

Answer 1

Let $E$ be the splitting field of this polynomial $P(X) = X^4-8X^2+15$ over $\mathbb{Q}$. Since this is a biquadratic polynomial, it is easy to solve. In fact: $$P(X) = (X^2-3)(X^2-5)$$ So we see that $E = \mathbb{Q}(\sqrt{3},\sqrt{5})$, right? This is already the first big step.

You could proceed as follows: Let $G = \text{Gal}(E/\mathbb{Q})$. Can you write down the elements of $G$ explicitly? Now let $\alpha = \sqrt{3}+\sqrt{5} \in E$. Then the following equivalence holds: $$ \mathbb{Q}(\alpha) = E \Leftrightarrow \sigma(\alpha) \not= \alpha \quad \forall \sigma \in G $$

This is a consequence of the fundamental theorem of galois theory (I can elaborate on this if necessary).

Now by computing $\sigma(\alpha)$ for every element of the group and simply noting that it is different from $\alpha$, you have your primitive element.

Note: This is a very general method and maybe unnecessary in this specific case. If you have a field extension $\mathbb{Q}(\alpha_1,\alpha_2,\dots,\alpha_n)$ then trying $\lambda_1\alpha_1+\dots+\lambda_n\alpha_n$ as your primitive element (with $\lambda_i \in \mathbb{Q}$ and $\lambda_i \not= 0$) will most likely give you something that works (there are only finitely many such tuples $(\lambda_1,\dots,\lambda_n)$ that will not work). Actually proving that such an element works can be harder. I hope I didn't make things overly complicated.