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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?

Brain
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  • $\mathbb{Q}[\sqrt{-2}]\cong \mathbb{Q}[x]/\langle x^2+2 \rangle$ and you want to show that $1/x \in \mathbb{Q}[x]/\langle x^2+2 \rangle .$ So invert it. We have $x^2+2=0$ so $1=-x^2-1$ etc. – Marius S.L. Oct 10 '23 at 01:47
  • $\Bbb Q[\sqrt{-2}]$ is already a field (hint: compute an inverse by rationalizing the denominator - see here for this and more general viewpoints). – Bill Dubuque Oct 10 '23 at 02:02

1 Answers1

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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 $\mathbb Q[\sqrt{-2}]$.

mweiss
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  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Oct 10 '23 at 02:03
  • @BillDubuque Unless I am mistaken, at the time I posted this solution the problem had not yet been marked as a duplicate. – mweiss Oct 11 '23 at 19:10
  • But it is a dupe, as any experienced user should be well aware. Please read the linked site policy. Please refrain from answering questions that are obviously dupes since having hundreds of dupes of common exercises makes it difficult if not impossible to search the site for good answers (its primary purpose). – Bill Dubuque Oct 11 '23 at 19:17