Is $\mathbb{Z}[\sqrt{15}]$ a UFD?

Question

Let $R=\mathbb{Z}[\sqrt{15}]=\{a+b\sqrt{15}:a,b\in\mathbb{Z}\}$.

1. How do I show that $(3,\sqrt{15})$ is a maximal ideal but not a principal ideal?
2. How do I show that $(3,\sqrt{15})^2$ is a principal ideal?
3. How do I show that $R$ is (not) a UFD?

What I have done:

1. If $(3,\sqrt{15})$ is a maximal ideal, then I must show that $R/(3,\sqrt{15})$ is a field. I thought that this holds:$R/(3,\sqrt{15})=\mathbb{Z}/3\mathbb{Z}$. Is this correct? How do I proceed from here?
2. I know that $(3,\sqrt{15})^2=(9,3\sqrt{15},15)$. How can I use this?
3. I'm afraid that I don't know where to start with this one. Maybe one of the statements above can help?

Thanks for taking the time!

Answer 1

How to prove that $(3,\sqrt{15})$ is not principal.

Assume $(3,\sqrt{15})$ is principal - that is $(3,\sqrt{15})=(a+b\sqrt{15})$ for some $a,b\in\mathbb Z$.

Write $\alpha=a+b\sqrt{15}$.

What is $N(\alpha)=a^2-15b^2$? It must be a divisor of $N(3)=9$ and $N(\sqrt{15})=15$. We know $N(\alpha)$ cannot be $\pm 1$, or $\alpha$ would be a unit, and $(3,\sqrt{15})$ is not all of $R$.

So you need $N(\alpha)$ to be a divisor of $9$ and $15$ but not $\pm 1$. That means:

$$N(\alpha)=a^2-15b^2=\pm 3$$

Prove this isn't possible. Look modulo $5$.

How to prove $(3,\sqrt{15})^2$ is principal

As you've noted, $I=(3,\sqrt{15})^2=(9,3\sqrt{15},15)$. But if an ideal contains $9$ and $15$, it contains $2\cdot 9-15=3$.

Also, if an ideal of $R$ contains $3$, then it contains $3\sqrt{15}$.

So what is the ideal $I$?

The UFD question

We've shown that $3$ must be irreducible, by showing it has no non-unit divisors.

So $15=3\cdot 5=\sqrt{15}\sqrt{15}$. Can $3$ divide $\sqrt{15}$?

Answer 2

Maybe I've got it all backwards, but I think these questions are best answered in the reverse of the order you have posed them.

3. How do I show that $R$ is (not) a UFD?

Given primes $p$ and $q$, if $\mathbb{Z}[\sqrt{pq}]$ is a unique factorization domain, then both $p$ and $q$ split in this domain. That's what happens to 2 and 3 in $\mathbb{Z}[\sqrt{6}]$, for example. If $p = (a - b \sqrt{pq})(a + b \sqrt{pq})$, $q = (c - d \sqrt{pq})(c + d \sqrt{pq})$ and $(a + b \sqrt{pq})(c + d \sqrt{pq}) = \sqrt{pq}$, then $pq = (\sqrt{pq})^2$ represents two incomplete factorizations.

But 3 is irreducible in $\mathbb{Z}[\sqrt{15}]$, as there is no solution to $x^2 \equiv \pm 3 \pmod 5$. Takes just a little more effort to show 5 is likewise irreducible. Then $15 = 3 \times 5 = (\sqrt{15})$ constitutes the presentation of two distinct, complete factorizations, and therefore $\mathbb{Z}[\sqrt{15}]$ is not a unique factorization domain.

2. How do I show that $(3, \sqrt{15})^2$ is a principal ideal?

You have already determined that $\langle 3, \sqrt{15} \rangle^2 = \langle 9, 3 \sqrt{15}, 15 \rangle$. Two elements are enough to generate an ideal, which means that one of these three elements is kind of redundant. Since 9 and 15 are both integers of $\mathbb{Z}$ and $\mathbb{Z}$ is a Euclidean domain, we can boil $\langle 9, 15 \rangle$ to $\langle 3 \rangle$, leaving us with $\langle 3, 3 \sqrt{15} \rangle$. But since $\langle 3 \rangle$ absorbs multiplication of 3 by $\sqrt{15}$, then $\langle 3 \sqrt{15} \rangle$ turns out to be entirely redundant.

Congratulations, you have have factored the principal ideal $\langle 3 \rangle$ as the square of the ideal $\langle 3, \sqrt{15} \rangle$. The number 3 is irreducible, but the ideal $\langle 3 \rangle$ is "composite". Which brings us to your first question:

1. How do I show that $(3, \sqrt{15})$ is a maximal ideal but not a principal ideal?

If $\langle 3, \sqrt{15} \rangle$ was not maximal, it could either be properly contained in a maximal ideal or it could be all of $\mathbb{Z}[\sqrt{15}]$. The latter is quite easy to disprove: show that $1 \not\in \langle 3, \sqrt{15} \rangle$. The former takes more work, though there probably is some shortcut I'm unaware of.

If it was the case that $\langle 3, \sqrt{15} \rangle$ is properly contained in a different ideal, that ideal would have to properly contain $\langle 3 \rangle$ and $\langle \sqrt{15} \rangle$ without actually being $\langle 3, \sqrt{15} \rangle$. The only ideal that satisfies this criterion is the whole ring.

So $\langle 3, \sqrt{15} \rangle$ is a maximal ideal. If it was a principal ideal, we could find a single number $a \in \mathbb{Z}[\sqrt{15}]$ such that $\langle a \rangle = \langle 3, \sqrt{15} \rangle$. This $\langle a \rangle$ must be such that it contains numbers with norms divisible by 9 and numbers with norms divisible by 15. Which would mean that $a$ itself has a norm of 3 and therefore 3 splits in this ring. But we already saw that the number 3 is in fact irreducible.

Answer 3

It's nice to see people studying non-UFDs besides $\textbf{Z}[\sqrt{-5}]$.

If $\langle 3, \sqrt{15} \rangle$ was a principal ideal, we could find a number $n \in \textbf{Z}[\sqrt{15}]$ such that $\langle 3, \sqrt{15} \rangle = \langle n \rangle$. Then $\langle n \rangle$ contains numbers with a norm of $9$ and numbers with a norm of $15$. Since $\gcd(9, 15) = 3$, that means $N(n) = -3$ or $3$. But $x^2 - 15y^2 = \pm 3$ has no solutions. Therefore $\langle 3, \sqrt{15} \rangle$ is not principal. But since it contains every number in this ring with a norm divisible by $3$ (such as $3 + \sqrt{15}$), it is maximal.

As for $\langle 3, \sqrt{15} \rangle^2$ being principal, you're halfway there with $\langle 9, 3 \sqrt{15}, 15 \rangle$. This ideal contains numbers with norms of $81, 135, 45$. These are all divisible by $9$ so in fact $\langle 3, \sqrt{15} \rangle^2 = \langle 3 \rangle$.

The third question follows from the preceding.

Answer 4

Your question can easily be generalized to $\mathbb{Z}[\sqrt{15m}]$ for positive, squarefree $m$ such that $\gcd(15, m) = 1$ and $m \equiv 1 \pmod 4$.

Then $15my^2 \equiv 0$ or $5 \pmod{10}$ and thus neither $3$ nor $-3$ can be the result of $x^2 - 15my^2$ (because no square ends in $2, 3, 7, 8$ in base $10$), which means that $3$ is irreducible. But it's not prime, because $3 | (\sqrt{15m})^2$ but $3 \not| \sqrt{15m}$. The existence of irreducibles that are not primes is a dead giveaway that we're not in a UFD.

And that means that neither the principal ideal generated by $3$ nor the principal ideal generated by $\sqrt{15m}$ are maximal, and therefore they must both be contained within a maximal ideal jointly generated by $3$ and $\sqrt{15m}$.

Answer 5

$$\mathbb Z[\sqrt{15}]\simeq\mathbb Z[X]/(X^2-15),$$ so $$\mathbb Z[\sqrt{15}]/(3,\sqrt{15})\simeq\frac{\mathbb Z[X]/(X^2-15)}{(3,X,X^2-15)/(X^2-15)}\simeq\mathbb Z[X]/(3,X)\simeq\mathbb Z/3\mathbb Z.$$ Thus $(3,\sqrt{15})$ is a maximal ideal.