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Dummit and Foote q9.3.1

Let $R$ be an integral domain with quotient field $F$ and let $p(x)$ be a monic polynomial in $R[x]$. Assume $p(x) = a(x)b(x)$ where $a(x)$ and $b(x)$ are monic polynomials in $F[x]$ of smaller degree than $p(x)$. $\text{Prove that if a(x)} \not \in \text{R[x] then R is not a U.F.D.}$ Deduce that $\mathbb{Z}[2\sqrt2]$ is not a U.F.D.

How to do the last part? 

$\mathbb{Z}[2\sqrt2]$ is integral domain. $F$ be its quotient field. To find $p(x)\in \mathbb{Z}[2\sqrt2][x](=\mathbb{Z}[2\sqrt2, x])$, $a(x) \not \in \mathbb{Z}[2\sqrt2, x]$, $a(x), b(x) \in F[x]$ of smaller degree than $p(x)$ s.t. $p(x) = a(x)b(x)$. $deg_x(p(x)) \geq 2$ because it can't be $0$ or $1$. 

Please give a hint how to find $p(x), a(x), b(x)$. Please do not give solution.

Thanks!

This is a direct solution which is maybe not expected here.

Vinay Deshpande
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  • Easier way: $,\Bbb Z[2\sqrt 2]$ is not a UFD by the Rational Root Test fails: the proper fraction $(2\sqrt 2)/2$ is a root of monic $,x^2 - 2;,$ i.e. UFDs are integrally closed (contain all fractions that are roots of monic polynomials). – Bill Dubuque Oct 08 '23 at 21:33
  • Use $\ x^2-2 = (x-\sqrt 2)(x+\sqrt 2),$ to translate the idea in the prior comment to the D&F hint. $\ \ $ – Bill Dubuque Oct 08 '23 at 21:40

1 Answers1

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Hint: Think about $x^2+2\sqrt 2 x + 2$. Full solution:

Let $p(x)= x^2+2\sqrt 2x +2$
Then, $a(x)=b(x)=x+\sqrt 2 \not \in \mathbb Z[2\sqrt 2,x]$.

razivo
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    Thanks. But it would have been better to get to know how to come up with that $p(x)$. Luckily, I found a simple method which makes it easier to come up with that example:$\$ $deg_x(p(x)) \geq 2$. First try $deg_x(p(x)) =2$. Let $a(x)=x+c$, $b(x)=x+d$, $c,d \in F$. $\$ $x^2+(c+d)x+cd\in \mathbb{Z}[2\sqrt2][x] \implies c+d, cd \in \mathbb{Z}[2\sqrt2]$. Now it's easier to find such $c$ and $d$. – Vinay Deshpande Nov 29 '20 at 11:05
  • What is the quotient field of $\mathbb Z [2\sqrt{2}][x]$? Why the given factors are in the quotient field of $\mathbb Z [2\sqrt{2}][x]$? – Intuition Oct 08 '23 at 20:59
  • Simpler: $ $ use $\ x^2-2 = (x-\sqrt 2)(x+\sqrt 2),,$ whose discovery has simple conceptual motivation - see my comments on the question. – Bill Dubuque Oct 08 '23 at 22:17