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I'm trying to show that $x^2+3x-1$ is irreducible in $\mathbb{Z}[\sqrt{13}]$. I have that the roots are $\frac{-3+\sqrt{13}}{2}$ and $\frac{-3-\sqrt{13}}{2}$, but I don't believe this is enough to show irreducibility. What else would I need?

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Usually, we write $\mathbb Z[\sqrt{13}]$, with round brackets indicating rational functions of $\sqrt{13}$ – Thomas Andrews Jan 3 '13 at 19:34
Yes, please fix the notation as indicated. $\mathbb Z(\sqrt{13})$ has no standard meaning in this context. – PatrickR Jan 4 '13 at 0:14
up vote 5 down vote accepted

In fact, since $x^2+3x-1$ is a (monic) quadratic polynomial, it is enough; the only way it could factor is into linear factors, which would require that its roots lie in the given ring.

Note that this is not true for quartics and higher: for example, $x^4+4=(x^2+2x+2)(x^2–2x+2)$ factors over $\mathbb{Z}$ even though its roots certainly are not integers.

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What about $6x^2-4x-3=(2x+1)(3x-2)$? It is reducible in $\mathbb{Z}[x]$, but has no roots in $\mathbb{Z}$. – Frank White Jan 3 '13 at 20:02
You also need the condition of monic. – dinoboy Jan 3 '13 at 20:07
@Frank: Sorry, I should have said monic, which avoids that problem. – Zev Chonoles Jan 3 '13 at 20:08

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