The simple extension $\mathbb Q(i+\sqrt{2})$

Question

I want to

a) show that $i$ and $\sqrt{2}$ are in $\mathbb Q(i+\sqrt{2})$ and that $[\mathbb Q(i+\sqrt{2}):\mathbb Q]=4$

b) find the minimal polynomial of $i+\sqrt{2}$ over $\mathbb Q$.

Now I can solve both parts easily by first doing b) and show that $x^4+6x+1$ is the sought for polynomial of degree 4, whence $[\mathbb Q(i+\sqrt{2}):\mathbb Q]=4$ follows. Independent of this, the first part of a) is of course easy to verify. But how do I go from there to $[\mathbb Q(i+\sqrt{2}):\mathbb Q]=4$ without going via b)?

Answer 1

The standard way is to see (using what you showed) that $\mathbb Q(i + \sqrt{2}) = \mathbb Q(i,\sqrt{2})$ (the inclusion of the right side in the left being obvious, and the opposite inclusion being the first part of (a)).

Now write $\mathbb Q(i,\sqrt{2})$ as an iterated extension, namely $\mathbb Q(\sqrt{2})(i).$ The first of these is quadratic, and the second has degree either $1$ of $2$. But since $i \not\in \mathbb Q(\sqrt{2})$ (there are many ways to check this, but the easiest is perhaps to observe that $\mathbb Q(\sqrt{2}) \subset \mathbb R,$ while $i \not\in \mathbb R$), its degree must be $2$. The second part of (a) follows.

Answer 2

Once you've shown that $i$ and $\sqrt{2}$ are elements of your field, we have

$$\Bbb{Q} \leq \Bbb{Q}[i] \leq \Bbb{Q}[i + \sqrt{2}]$$

These are both degree two extensions, so the extension $\Bbb{Q} \leq \Bbb{Q}[i + \sqrt{2}]$ has degree at most $2 \cdot 2 = 4$. It's easy to see that both extensions are proper, so the degree is at least $4$ as well.

Answer 3

(i+√2)(i+√2) = 1+2√2i, which means √2i is in the extension field. (subtract 1 and divide by 2, which you can do because this is an extension field of Q). (√2i)(i+√2)=-√2+2i

(-√2+2i)+(i+√2)= 3i, so i is in the extension field, as is (i+√2)-i = √2,