For which $r\in\mathbb{Q}$ is $\mathbb{Q}[x]/\langle x^2 - r\rangle$ a field?

Asked Mar 04 '21 at 12:53

Active Mar 04 '21 at 12:53

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So I know that $\mathbb{Q}[x]$ is a P.I.D. Does this mean that I only have to find where $x^2-r$ is irreducible over $\mathbb{Q}[x]$?

I have tried approaching this by doing the following:

$x^2-r$ can be written as $(x-\sqrt{r})(x+\sqrt{r})$ and thus would be irreducible for all $r$ for which this is not an rational polynomial.

Am I thinking about this correctly?

asked Mar 04 '21 at 12:53

Max

1

Yes, that is correct. – Sofía Marlasca Aparicio Mar 04 '21 at 12:58
@marlasca23 So, this would just be all $r$ that does not have a rational root? Would that be the answer? – Max Mar 04 '21 at 13:01
2

Yes. $\mathbb{Q}[x]/\langle f(x) \rangle$ is a field if and only if $f$ is irreducible. If $f$ has degree $2$, it is irreducible if and only if it has no rational roots. In the case $x^2 - r$, this has no rational roots if and only if $\sqrt{r} \notin \mathbb{Q}$. – Sofía Marlasca Aparicio Mar 04 '21 at 13:06

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