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Here is the question I am trying to understand the solution of the last part in it:

Let $R$ be an integral domain with quotient field $F$ and let $p(x)$ be a monic polynomial in $R[x].$ Assume that $p(x) = a(x) b(x)$ where $a(x)$ and $b(x)$ are monic polynomials in $F[x]$ of smaller degree than $p(x).$ Prove that if $a(x) \notin R[x]$ then $R$ is not a $UFD$. Deduce that $\mathbb Z[2 \sqrt{2}]$ is not a $UFD$.

Here is a solution I found here:

$\mathbb{Z}[2\sqrt2]$ is not a UFD.

My question is:

What is the quotient field of $\mathbb Z[2 \sqrt{2}][x]$? Why the given factors in the solution are in the quotient field of $\mathbb Z[2 \sqrt{2}][x]$?

Can someone clarify this to me please?

Anne Bauval
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Intuition
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  • The quotient field of $R:=\mathbb Z[2 \sqrt{2}]$ is $F:=\Bbb Q(\sqrt2)$.

    The quotient field of $R[x]$ is therefore $F(x)$ but this is off topic.

    The "given factors" $a(x)=b(x)=x+\sqrt 2$ are clearly in $F[x]$ (but not in $R[x]$).

     – Anne Bauval Oct 08 '23 at 21:32
  • You don't want the quotient field of $\mathbb{Z}[\sqrt{2}][x]$. You want the polynomials with coefficients in the quotient field of $\mathbb{Z}[\sqrt{2}]$ (which is $\mathbb{Q}(\sqrt{2})$). – Arturo Magidin Oct 08 '23 at 21:33
  • @ArturoMagidin why you are saying this ? Why $\mathbb Z[ \sqrt{2} ]$ and not $\mathbb Z[ 2\sqrt{2} ],$ I do not understand you, could you please explain this? – Intuition Oct 08 '23 at 22:30
  • @AnneBauval why you are saying this ? Why $\mathbb Q[ \sqrt{2} ]$ and not $\mathbb Q[ 2\sqrt{2} ]$ is the quotient field? I do not understand you, could you please explain this? – Intuition Oct 08 '23 at 22:32
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    Sorry, I missed a couple of $2$s; but the point is that you are asking for the quotient field of the polynomial ring $\mathbb{Z}[2\sqrt{2}][x]$; but what you want is the polynomial ring over the quotient field of $\mathbb{Z}[2\sqrt{2}]$; that one is exactly the same as the quotient field of $\mathbb{Z}[\sqrt{2}]$, namely $\mathbb{Q}(\sqrt{2}\mathbb{Q}[\sqrt{2}]=\mathbb{Q}[2\sqrt{2}]=\mathbb{Q}(2\sqrt{2})$. That is, you want the polynomial ring over the quotient field, not the quotient field of the polynomial ring. – Arturo Magidin Oct 08 '23 at 23:04
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    $\mathbb{Q}[2\sqrt{2}]=\mathbb{Q}[\sqrt{2}]$. You clearly have $\subseteq$, and $\sqrt{2}=\frac{1}{2}(2\sqrt{2})\in\mathbb{Q}[2\sqrt{2}]$, giving $\supseteq$. – Arturo Magidin Oct 08 '23 at 23:05

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