Determine whether $\sqrt{-5}$ is irreducible and/or prime in $\mathbb Z[\sqrt{-5}]$.

What is a prime $p>5$ which is prime in $\mathbb Z [\sqrt{-5}]$ not prime in $\mathbb Z [\sqrt{5}]$?

For the first one, I believe $\sqrt{−5}$ is irreducible as $N(\sqrt{−5}) = 5$ and the only integers that divide $5$ are $1$ (where all elements with norm $1$ are units) and $5$ (where only $±\sqrt{−5}$ have norm $5$). Is this a good explanation? I'm guessing that $\sqrt{-5}$ is prime but I'm not sure how to justify why.

For the second one, I can't think of any primes over 5.

Help would be great!

  • $\begingroup$ Your idea doesn't really work because $\sqrt{-5}$ is not even an element of the ring $\Bbb{Z}[\sqrt5]$ - the elements of that ring are all real numbers. Try something different. Can you convince yourself that $11$ cannot be written in the form $a^2+5b^2$, but can be written in the form $a^2-5b^2$? $\endgroup$ – Jyrki Lahtonen Mar 17 '18 at 12:27
  • $\begingroup$ @JyrkiLahtonen OP is considering $\Bbb Z [\sqrt {-5}] $ $\endgroup$ – krirkrirk Mar 17 '18 at 12:38
  • $\begingroup$ @krirkrirk Oops, I was largely discussing the question in the title. $\endgroup$ – Jyrki Lahtonen Mar 21 '18 at 21:55

The condition for $p$ to split in a quadratic field of discriminant $D$ is that


The discriminant of $\mathbb{Z}(\sqrt{-5})$ is $D=-20$, and of $\mathbb{Z}(\sqrt{5})$ is $D=5$, thus you seek a $p$ such that

$$\left(\frac{-20}{p}\right)=-1, \left(\frac{5}{p}\right)=+1$$

Now $$\left(\frac{5}{p}\right)=\left(\frac{p}{5}\right)$$ and this is $+1$ if and only $p\equiv \pm 1 (\mod 5)$.

In this case

$$\left(\frac{-20}{p}\right)=\left(\frac{-1}{p}\right)$$ so $p$ must also satisfy $$\left(\frac{-1}{p}\right)=-1$$ or that

$p\equiv 3 (\mod 4)$.

The smallest such prime is $p=11$.

And indeed,


and on the other had it is easy to see that

$$a^2+5b^2=11$$ is impossible.

One can also see that $19$ is another such prime. Thus in fact the set of such primes are those of the forms $20n+11$ and $20n-1$.

  • $\begingroup$ Thank you for explaining it so clearly. Am I correct in guessing $\sqrt{-5}$ is prime in $\mathbb Z[\sqrt{-5}]$? How would I go about justifying that? $\endgroup$ – user529584 Mar 17 '18 at 18:04
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    $\begingroup$ The norm of $\sqrt{-5}$ is $5$, thus it is prime. $\endgroup$ – Rene Schipperus Mar 17 '18 at 19:47

Yes, $\sqrt{-5}$ is irreducible and prime in $\mathbb{Z}[\sqrt{-5}]$. If it wasn't, it would be possible to find numbers in the domain such that $ab = c \sqrt{-5}$ yet $\sqrt{-5}$ divides neither $a$ nor $b$. Since the norm is multiplicative, that would require $N(ab) = 5$ and so the only possibilities are $N(a) = 1$, $N(b) = 5$ or vice-versa.

For the second part of your question, by $\mathbb{Z}[\sqrt{5}]$ do you mean $\mathbb{Z}[\phi]$ where $$\phi = \frac{1 + \sqrt{5}}{2}?$$ I'll assume that you do. Since $x^2 \equiv \pm 2 \pmod 5$ is insoluble in integers, we can be assured that primes ending in $3$ or $7$ are at least irreducible in both domains (and of course prime in $\mathbb{Z}[\phi]$).

So what we're looking for is that $x^2 \equiv -5 \pmod p$ has no solutions but $x^2 \equiv 5 \pmod p$ does. I'd work out the Legendre symbol for you but I'm running late to dinner.


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