Finding whether an extension is normal

Question

I have $\mathbb Q(i,\sqrt5)$ and I think I found the base for the extension on $\mathbb Q$ as ${ a+b\sqrt5 +ci + di\sqrt5 ; a,b,c,d\in \mathbb Q}$

But I don't know at which polynomial to look in order to see if it is normal or not.

Answer 1

A field extension is said to be normal, if the minimal polynomial of every element of the larger field, splits completely within the field itself.

However, there are some other conditions, as it turns out, for a finite field extension to be normal.

In particular, finite field extensions which are splitting fields of separable polynomials (not irreducible, but separable) are normal as well as separable (it goes the other way also). One either defines Galois extensions to have this property, or there is a separate definition, which is shown to be equivalent to normal,separable.

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It turns out that $\mathbb Q(i,\sqrt 5)$ is the splitting field of the polynomial $(x^2+1)(x^2 - 5)$. Why? Because the field contains each of the roots of the polynomial, and it is easy to see that any other field with this property contains $\mathbb Q(i,\sqrt 5)$ by definition of $\mathbb Q(i,\sqrt 5)$ being the smallest field containing $i$ and $\sqrt 5$.

Note that $(x^2 + 1)(x^2 - 5)$ is separable as all its roots are of multiplicity one. Of course, the extension is finite, of degree four.

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Thus, $\mathbb Q(i,\sqrt 5)$ is a normal extension of $\mathbb Q$. It is, infact, a Galois extension, with Galois group $\mathbb Z_2 \times \mathbb Z_2$.

Answer 2

You can probably find the minimal polynomials of $i$ and $\sqrt(5)$. The product of these certainly contains $\mathbb{Q}(i,\sqrt{5})$ inside its decomposition field.