why if I proved it for $\mathbb Q[\sqrt{-2}],$ it will be correct for $\mathbb Q(\sqrt{-2}),$?

Question

Here is the problem I am trying to solve:

Prove that the polynomial $p(x) = x^4 - 4x^2 + 8x + 2$ is irreducible over the quadratic field $F = \mathbb Q(\sqrt{-2}) = \{ a + b\sqrt{-2}\,|\, a,b \in \mathbb Q\}.$

I have seen many links here to its solution as you can see below:

Irreducibility of $p(x)=x^4-4x^2+8x+2$ over $\mathbb{Q}(\sqrt{-2})$- Dummit Foote Abstract algebra $9.4.10$

Clarification regarding hint given for showing $p(x)$ is irreducible over $\mathbb{Q}(\sqrt{-2})$.

But no-one of these links answered the following question:

why if I proved it for $\mathbb Q[\sqrt{-2}],$ it will be correct for $\mathbb Q(\sqrt{-2}),$?

Could anyone clarify the answer of this question to me please?

Answer 1

Because $\mathbb Q[\sqrt{-2}]$ and $\mathbb Q(\sqrt{-2})$ are the exact same thing. $\mathbb Q[\sqrt{-2}]$ is already a field, as can be seen from

$$\frac{1}{a + b\sqrt{-2}} = \frac{a}{a^2+2b^2} - \frac{b}{a^2+2b^2}\sqrt{-2}$$

so there are no additional elements in $\mathbb Q(\sqrt 2)$ that aren't already present in $\mathbb Q[\sqrt{-2}]$.