# Prove that ideals are prime

Prove that $\mathfrak{a}=\langle 3,4+\sqrt{-5}\rangle$ and $\mathfrak{b}=\langle7,4+\sqrt{-5}\rangle$ are prime ideals in ring $\mathbb{Z}[\sqrt{-5}]$.

Any ideas?

Thanks.

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Well, what do you know about ideals - what context do you have, what algebra do you know - and what are your own thoughts on the problem? – Mark Bennet Jan 10 '13 at 14:20
@MarkBennet Well, I have to prove (for $\mathfrak{a}$)that if $uv \in \mathfrak{a}$ then $u \in \mathfrak{a}$ or $v \in \mathfrak{a}$. But I don't have idea how to do it. – Cortizol Jan 10 '13 at 14:25
Hint: if $I$ is an ideal contained in the ideal $p$ you want to test for being prime, then $p$ is prime in $R$ if and only if $p/I$ is prime in $R/I$ (becuause the coresponding quotient rings are the same). It would seem an idea to apply this for $I$ generated by an appropriate element of $\Bbb Z$ here. – Marc van Leeuwen Jan 10 '13 at 14:32
Another possible approach is to look at the norms of ideals. – Mark Bennet Jan 10 '13 at 18:08
@MarkBennet Can me show how? – Cortizol Jan 10 '13 at 18:17

This is a "norm of ideal" sort of approach, which does not accurately define the norm, or identify the precise mathematical machinery involved, but is intended to illustrate why it might be a useful tool for this kind of problem. We are dealing with a quadratic extension of the integers $\mathbb Z (\sqrt{-5})$, and the norm of an element $a+b\sqrt{-5}$ is computed as $(a+b\sqrt{-5})(a-b\sqrt{-5})=a^2+5b^2$.

An ideal $\mathfrak{I}$ has a norm too, and the norm of every element in the ideal is a multiple of the norm of the ideal. (The norm can, for example, be defined in terms of the "volume" [area] of a cell in the lattice generated by the elements of the ideal). The norm is an integer, and if the norm of an ideal is prime, the ideal itself must be prime. This is because $N(\mathfrak{a}\mathfrak{b})=N(\mathfrak{a})N(\mathfrak{b})$, and any ideal of norm 1 is the whole ring.

Consider a prime number $p$ which generates the ideal $\mathbb Z(p) \subset \mathbb Z$. In a quadratic extension such a prime has norm $p^2$, as does the associated ideal. It is possible for the ideal associated with $p$ to remain prime in the extension. The only other possibility is that the ideal associated with $p$may become the product of two larger ideals each having norm $p$ - larger ideals generate finer lattices and therefore have smaller norms.

In fact three cases are normally recognised for the behaviour of primes in quadratic extensions - $\langle p \rangle$ is prime, $\langle p\rangle$ is the square of an ideal, or $\langle p\rangle$ is the product of two different ideals.

In the case we are given we have elements of norm 9, 21 and 49. Taking the case of $\mathfrak{a}=\langle 3,4+\sqrt{-5}\rangle$ we see that the ideal is bigger than $\langle 3 \rangle$, but that the generating elements have norms divisible by 3, so all the elements do, so it can't be the whole lot. We therefore have $\langle 3 \rangle \subset \mathfrak{a} \subset \mathbb Z (\sqrt{-5})$ with both inclusions being proper. So $N(\mathfrak{a})|9$ is a proper divisor not equal to 1, hence is 3. Since the norm of the ideal is a prime integer, the ideal is a prime ideal.

The question here starts to show how the primes 3 and 7 behave in the given extension.

NB the norm of a prime ideal need not be a prime integer - it could be a power of a prime.

Note that this can be done quickly, but there are hazards not encountered in this illustration - for example we have $-5 \equiv 3$ mod $4$. Since we are implicitly working with rings of integers we need to be aware that there are differences when $p\equiv1$ mod $4$. You need to know how to compute norms to use the method.

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Thanks a lot for the explanation. – Cortizol Jan 10 '13 at 21:39

A perhaps easier strategy for these kind of problems is to think of $\mathbb{Z}[\sqrt{-5}]$ as $\mathbb{Z}[x]/(x^2+5)$, so that $\sqrt{-5}$ is represented by the image of $x$ in this quotient ring.

Now the first question is simply to decide if the ideal $\mathfrak{a} = (3,4+x)$ is prime in $\mathbb{Z}[x]/(x^2+5)$. But this is easy: We first mod out by $(3)$, and see what we get:

($\mathbb{Z}[x]/(x^2+5))/(3,4+x)= \mathbb{Z}/(3)[x]/(4+x,x^2+5)= (\mathbb{Z}/(3))/(21)=\mathbb{Z}/(3)$, which is an integral domain, hence the ideal must have been prime.

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+1 Pretty nice. – DonAntonio Jan 10 '13 at 14:48
How is $(4+x,x^2+5)=(21)$ in $\mathbb Z_3[x]$? – JSchlather Jan 10 '13 at 14:49
@JacobSchlather: Since we divide by $4+x = 0$ in the quotient, we can replace the $x^2$ in the other generator with $-4$. – Fredrik Meyer Jan 10 '13 at 14:52
@FredrikMeyer You've shown that the ideal generated by $(1+x)$ and $(1+x)(2+x)$ in $\mathbb Z_3[x]$ is equal to 0? – JSchlather Jan 10 '13 at 14:56
@Andrew: I'm not sure if I understand what you mean: your ideal would correspond to $(3,1+2x)$ in $\mathbb{Z}[x]/(x^2+5)$. We can still do the same reduction: We need to calculate $(1+2x,x^2+5)$ modulo three, which is $(1-x,x^2+2)=(3)=(0)$. So the quotient is again $\mathbb{Z}/(3)$. – Fredrik Meyer Mar 8 '13 at 22:15