Minimal polynomial problem

Question

Show that $\mathbb Q(\sqrt 2 + i)=\mathbb Q(\sqrt 2, i)$ and find minimal polynomial.

My question: Assume that they are equal, then the minimal polynomial of both sides must be the same. To prove and to find the minimal polynomial, I can do both at the same time.

LHS: let $u = \sqrt 2 +i\\u^2=1-2\sqrt2 i\\u^2-1=-2\sqrt2 i\\u^4-2u^2+1=-8 \\u^4-2u^2-9=0$.

Hence $t^4-2t^2-9$ is minimal polynomial of $\sqrt2 +i$

RHS: $\sqrt2 , i$ are roots of a polynomial, $(t^2-2)(t^2+1)=t^4-t^2-2$

It is not the same with LHS. What is wrong? Please, help me

Answer 1

The 'minimal polynomial' (in the sense you imply) need not be the same. In other words, a field extension can very easily be the splitting field for two different polynomials.

A simpler example is that $\mathbb{Q}(\sqrt{2}) = \mathbb{Q}(\sqrt{2}+1)$, but the minimal polynomial of $\sqrt{2}$ is $t^2-2$, while the minimal polynomial of $\sqrt{2}+1$ is $t^2-2t-1$.

Answer 2

Hints:

1) $\mathbb Q(\sqrt 2 + i) \subset \mathbb Q(\sqrt 2, i)$ is obvious.

2) Show that the degree of $\mathbb Q(\sqrt 2, i)$ over $\mathbb Q$ is 4.

3) You have already shown that the degree of $\mathbb Q(\sqrt 2 + i)$ over $\mathbb Q$ is $\leq 4$ (why?)

4) Use the multiplicativity of degrees of field extensions.