Prove that if $w \in \mathbb{Z}[\sqrt{3}]$ and $N(w)$ is a prime, then $w$ is prime also

Question

Prove that if $w$ is an extended integer, $w \in\mathbb{Z}[\sqrt{3}]$, and $N(w)$ is a prime in the ordinary integers, then $w$ is a prime. From this, conclude that $7+2\sqrt{3}$ is a prime in $\mathbb{Z}[\sqrt{3}]$.

I'm kinda having a tough time dealing with primes so I'm wondering if this proof makes sense:

We say that $w = a+b\sqrt{3}$ and $N(w)$ is prime. Let $w = a+b\sqrt{3} = k\cdot z$ where $k$ and $z$ are integers. We now have $N(w) = N(k)\cdot N(z)$, but we know that $N(w)$ is prime which implies that either $N(k)$ is unit or $N(z)$ is unit. If $N(k)$ is unit, then $k$ is also unit. Similarly, if $N(z)$ is unit, then $z$ is also unit. Since $w = k\cdot z$ and either $k$ or $z$ is unit, we can conclude that $w$ is prime hence the proof is complete.

We have $w = a+b\sqrt{3} = 7+2\sqrt{3}$, so $N(w) = a^2-3b^2 = (7)^2-3(2)^2 = 49-3(4) = 37$. Since $37$ is a positive integer prime, this implies that $N(w)$ is prime. From the proof above, we can conclude that if $N(w)$ is prime then $w$ is also prime, hence $7+2\sqrt{3}$ is a prime in $\mathbb{Z}[\sqrt{3}]$.

Thank you for any help you may be able to offer.

Answer 1

Your proof is good, but as Starfall points out, you have proven that $w$ is irreducible but not necessarily prime. I'll get back to that later.

When you say

Let $w = a + b \sqrt 3 = k \cdot z$ where $k$ and $z$ are integers.

it feels a little redundant because you have already defined what $w$ is. More importantly, perhaps, you might want to clarify that $k$ and $z$ are integers from $\mathbb{Z}[\sqrt 3]$, not necessarily rational integers.

Suppose for example $k = 74 - 3 \sqrt 3$ and $z = 2 + \sqrt 3$. Then $kz = 37$. But $N(z) = 1$, so $z$ is a unit, just as expected.

So $7 + 2 \sqrt 3$ is indeed irreducible, but that alone does not guarantee that every number in this domain with a norm nontrivially divisible by $37$ is itself divisible by $7 + 2 \sqrt 3$. After all, in $\mathbb{Z}[\sqrt{10}]$ we see that $37$ is irredubiclbe yet $3 \times 37 = (11 - \sqrt{10})(11 + \sqrt{10})$.

The most straightforward thing then, oops, sorry, gotta go

Also love your choice of $w$ for a variable. Wonderfully confusing for those familiar with Eisenstein integers and $\omega$. Mwahahahahahahahaha!

Answer 2

I think that your proof is adequate for the level that you are in. Generalizing it to other rings, it would look the same (aside from stylistic differences of notation and terminology) as the proof of Theorem $3.11$ in Keith Conrad's Factoring in Quadratic Fields handout for his University of Connecticut students (I was able to pull up the PDF, but for some reason I'm unable to pull up the URL without a bunch of Googlecruft).

If a number $n$ in a ring $R$ has a norm that is prime in $\mathbb Z$, then $n$ is irreducible in $R$.

To prove that $n$ is indeed irreducible and prime, you might need to jump ahead in your textbook. Maybe your professor is planning to get there, maybe not, I don't know. I would use Theorem $7.4$ from the Conrad handout.

An ideal whose norm is prime in $\mathbb Z$ is a prime ideal.

Also $7.1$:

If an ideal is prime, then its conjugate ideal is also prime.

Have you studied ideals yet? So the number $7 + 2 \sqrt 3$ has a prime norm, and its conjugate $7 - 2 \sqrt 3$ also has that same prime norm. The number $7 + 2 \sqrt 3$ generates the principal ideal $\langle 7 + 2 \sqrt 3 \rangle$, and likewise $7 - 2 \sqrt 3$ generates $\langle 7 - 2 \sqrt 3 \rangle$. By Conrad's Theorem $6.3$,

The norm of a principal ideal is the same as the absolute value of the norm of the generating number.

$\langle 7 + 2 \sqrt 3 \rangle$ and $\langle 7 - 2 \sqrt 3 \rangle$ are prime ideals. Somewhere in there I'm sure it says that if a principal ideal is prime, then the generating number is prime. Therefore $7 + 2 \sqrt 3$ and $7 - 2 \sqrt 3$ are both prime.

Then it's not strictly necessary to know whether or not $\mathbb Z[\sqrt 3]$ is or is not a unique factorization domain (it is, though) in order to show that these numbers with a norm of $37$ are indeed prime. We see for example that numbers with a norm of $19$ are prime in $\mathbb Z[\sqrt 30]$, which is not a unique factorization domain at all.

Answer 3

Your proof not only makes sense, it is very standard. If you look ahead a little bit in your textbook, I think you will find pretty much the same proof, though probably with $a$ and $b$ or $\alpha$ and $\beta$ instead of $k$ and $z$. I think it's silly to get hung up on your choice of variables, but it needs to be mentioned so we can move on to more important things.

You seem to have some doubts, and I am wondering if it has to do with your awareness of conjugates. Given $a + b \sqrt{3}$, the conjugate is $a - b \sqrt{3}$, and it has the same norm. Furthermore, if the norm is an odd number coprime to 3, this number is not divisible by its conjugate nor a multiple thereof.

Notice that $$\frac{7 + 2\sqrt{3}}{7 - 2\sqrt{3}} = \frac{61 + 28\sqrt{3}}{37}$$ and $$\frac{7 - 2\sqrt{3}}{7 + 2\sqrt{3}} = \frac{61 - 28\sqrt{3}}{37}.$$

This does nothing to challenge the primality of $7 + 2\sqrt{3}$. For example, $N((7 - \sqrt{3})^2) = 37^2$, but since $7 + 2\sqrt{3}$ doesn't divide $7 - 2\sqrt{3}$, it doesn't actually matter that much that it doesn't divide $(7 - 2\sqrt{3})^2$ either.

But if a number in this domain is divisible by 37, it is also divisible by both $7 - 2\sqrt{3}$ and $7 + 2\sqrt{3}$. Assuming certain things have been proven in your class already, this fact follows from the standard proof you've presented. It would help if we knew whether you've covered irreducibles and primes and already, and ideals, principal or otherwise.

Answer 4

The most important fact here is that the norm is multiplicative: $N(pq) = N(p) N(q)$. For example, $N(1 + 2 \sqrt 3) = -11$, $N(4 + \sqrt 3) = 13$, $(1 + 2 \sqrt 3)(4 + \sqrt 3) = 10 + 9 \sqrt 3$ and $N(10 + 9 \sqrt 3) = -143$. This proves that $10 + 9 \sqrt 3$ is not a prime number, since we've seen it is the product of two numbers, neither of them units.

The norm takes one of these "extended" numbers, and gives you a number in good ol' $\mathbb Z$ (the norm may be negative, but if you want, you can adjust it so that it only gives positive integers for nonzero numbers).

So if the norm of a number $x$ is a number that is prime in $\mathbb Z$, and the norm is multiplicative, then the only way $ab = x$ is either $a$ or $b$ is a unit, which means that $N(a)$ or $N(b)$ must be 1 or $-1$. Because for a prime prime in $\mathbb Z$, its only divisors are 1, $-1$, the prime itself, and itself times $-1$.

Maybe your intuition is telling you that if $N(x)$ is a prime in $\mathbb Z$, then $x$ must itself also be prime in the given domain we're looking at, whether that domain is a unique factorization domain or not. Your intuition is correct, but that might be getting too far ahead in your course outline.

Answer 5

I will first prove

(1) An element $p \in R$ is prime id and only if the residue class ring $R/pR$ of the residue classes modulo $\pi$ is a domain.

In fact, the residue class ring oes not have a zero divisor if $ab \equiv 0 \bmod p$ implies that $a \equiv 0 \bmod p$ or $b \equiv 0 \bmod p$. But this is just a version of the definition of a prime element, which states that an element ist prime if $p \mid ab$ implies that $p \mid a$ or $p \mid b$.

Now we claim

(2) If $k$ is a quadratic number field with ring of integers $R ? {\mathcal O}_k$, then each $\pi \in {\mathcal O}_k$ with prime norm is prime.

We will show that the residue class ring $R/\pi R$ is a domain by showing that it is isomorphic to the field with $p$
elements.

To this end let $\{1, \omega\}$ be an integral basis of ${\mathcal O}_k$; then $\pi = a+b\omega$ for integers $a, b \in {\mathbb Z}$. We claim that $b$ is not divisible by $\pi$ (and thus not divisible by $p = |\pi \pi'|$). In fact, $\pi \mid b$ implies $\pi \mid a$ since $a = \pi - b\omega$, and taking norms we find $p \mid a^2$ and $p \mid b^2$. Since $p$ is prime, this implies that $p \mid a$ and $p \mid b$. But then $\pi = a+b\omega$ would be divisible by $p$, hence $\pi'$ would be a unit: contradiction.

Thus there exists an integer $c \in {\mathbb Z}$ with $bc \equiv 1 \bmod p$, and in particular we have $bc \equiv 1 \bmod \pi {\mathcal O}_k$. We find $b\omega \equiv -a \bmod \pi$, after multiplying through by $c$ thus $\omega \equiv -ac \bmod \pi {\mathcal O}_k$. If any $\gamma = r+s\omega \in {\mathcal O}_k$ is given, then we find $\gamma \equiv r - sac \bmod \pi {\mathcal O}_k$, thus modulo $\pi$ every element is congruent to an ordinary integer. Reducing this number modulo $p$ (and $p$ is a multiple of $\pi$) we find that $\gamma$ is congruent to one of the numbers $0, 1, 2, \ldots, p-1$ modulo $\pi$.

Now it is easy to show that there are no zero divisors in the ring of residue classes: If we had $\alpha \beta \equiv 0 \bmod \pi$ and if $A, B \in \{0, 1, \ldots, p-1\}$ are integers with $\alpha \equiv A \bmod \pi {\mathcal O}_k$ and $\beta \equiv B \bmod \pi {\mathcal O}_k$, then $\pi \mid AB$; taking norms yields $p \mid A^2B^2$, hence $p \mid A$ or $p \mid B$. Thus $A = 0$ or $B = 0$, and therefore $\alpha \equiv A = 0 \bmod \pi$ or $\beta \equiv B = 0 \bmod \pi$.

Answer 6

What you have proven is that if $ N(w) $ is prime, then $ w $ is irreducible. To prove that it is prime, you need to show, additionally, that $ \mathbf Z[\sqrt{3}] $ is a principal ideal domain, which would imply that every irreducible element is prime.

This is not the only way forward, however - indeed, if $ K $ is any number field and $ z \in \mathcal O_K $ is such that $ N(z) $ is prime, then $ (z) $ is always a prime ideal of $ \mathcal O_K $. This general result is obvious if we extend the norm map to ideals, and recall basic facts about Dedekind domains. Alternatively, it follows if one knows that the norm of an algebraic integer is the submodule index of the principal ideal it generates in $ \mathcal O_K $.