$R = \mathbb{Z}[\sqrt{-41}]$, show that 3 is irreducible but not prime in $R$

Question

I'm asked to show that 3 is irreducible but not prime in $R = \mathbb{Z}[\sqrt{-41}]$. And if $R$ is a Euclidean domain.

To show that it's not prime I have $(1 + \sqrt{-41})(1 - \sqrt{-41}) = 42 = (3)(14)$. I get that 3 divides 14 but how do I show that 3 does not divide either $(1 + \sqrt{-41})$, $(1-\sqrt{-41})$ to show that it's not prime? $x \mid 42$ but $x$ does not divide 3?

To show that it's irreducible how am I showing that either $(1 + \sqrt{-41})$, $(1 - \sqrt{-41})$ is a unit since clearly $(1 + \sqrt{-41})(1 - \sqrt{-41}) = 42$?

Answer 1

You can generalize quite a few things here. If $p$ is a prime number in $\mathbb{Z}$, $d \not\equiv 1 \pmod 4$ is negative and squarefree, and $|p| < |d|$, then $p$ is irreducible in $\mathbb{Z}[\sqrt{d}]$. Since $p$ is not a square (one of the most important reasons why 1 is not prime), it is irreducible.

Thus, in $\mathbb{Z}[\sqrt{-41}]$, we see that $|3| < |-41|$. The possible norms in this domain less than 41 are all squares: 1, 4, 9, 16, 25, 36, and clearly 3 is not one of these.

If $p$ is irreducible but not prime, then the congruence $x^2 \equiv d \pmod p$ can be solved. Indeed $1^2 \equiv -41 \pmod 3$, leading you to find that $(1 - \sqrt{-41})(1 + \sqrt{-41}) = 42$.

If either $1 - \sqrt{-41}$ or $1 + \sqrt{-41}$ was a unit, its norm would be equal to 1 (we don't have to worry about a norm of $-1$ here). But you've already seen that both of those numbers have norms of 42 (remember that $N(a + b \sqrt{d}) = (a - b \sqrt{d})(a + b \sqrt{d})$).

Saying "$a$ divides $b$" is a convenient shorthand for "$b$ divided by $a$ is a number that's also in this domain." But $$\frac{1 - \sqrt{-41}}{3} \not\in \mathbb{Z}[\sqrt{-41}].$$ Likewise for $1 + \sqrt{-41}$.

You have thus proven that 42 has two distinct factorizations in this domain: $$2 \times 3 \times 7 = (1 - \sqrt{-41})(1 + \sqrt{-41}) = 42.$$ (2 and 7 are also irreducible but not prime.)

If a domain is Euclidean, then it's also a principal ideal domain and a unique factorization domain. This domain is clearly not a Euclidean domain because it's not a unique factorization domain (e.g., 42).

Answer 2

You are close to a proof. You observed that $42=(1+\sqrt{-41})((1-\sqrt{-41})$. So $3$ divides $(1+\sqrt{-41})(1-\sqrt{-41})$. But $3$ divides neither $1+\sqrt{-41}$ nor $1-\sqrt{-41}$.

Next we show that $3$ is irreducible. For suppose that $3=(a+b\sqrt{-41})(c+d\sqrt{-41})$, where $a,b,c,d$ are integers. Then the norm of $a+b\sqrt{-41}$ divides the norm of $3$. Thus $a^2+41b^2$ divides $9$. It is clear that we must have $b=0$. So $a^2$ divides $9$. It follows that $a=\pm 1$, in which case $a+b\sqrt{-41}$ is a unit, or $a=\pm 3$, in which case $c+d\sqrt{-41}=\pm 1$, a unit.

Finally, in a Euclidean domain, every irreducible is prime.

Answer 3

Numbers in this domain are of the form $a + b \sqrt{-41}$, where $a, b \in \mathbb{Z}$. Notice then that $$\frac{1 + \sqrt{-41}}{3} = \frac{1}{3} + \frac{\sqrt{-41}}{3},$$ and $$\frac{3}{1 + \sqrt{-41}} = \frac{1}{14} + \frac{\sqrt{-41}}{14}.$$ Neither of these numbers is of the form $a + b \sqrt{-41}$ with $a, b \in \mathbb{Z}$. This means that $3$ is not a divisor of $1 + \sqrt{-41}$ nor the other way around, but obviously $3$ is a divisor of $42$, proving that $3$ is not prime.

On the last paragraph of your question you've gotten a little confused. What you need to show is that if $xy = 3$, with $x, y \in \mathbb{Z}[\sqrt{-41}]$, either $x$ is a unit or $y$ is. The only solutions are $x = -1$ or $1$, $y = -3$ or $3$. $-1$ and $1$ are units.

Answer 4

The answers given already are all very complete and good. I wanted to give a slightly different perspective on the fact that $3$ is not prime; this perspective is alluded to in Robert Soupe's very nice answer.

If $3 \in R:= \mathbb{Z}[\sqrt{-41}]$ is prime, then $3R$ is a prime ideal of $R$. In particular, this means that $R/3R$ is an integral domain. Since $R$ can be expressed as the quotient $\mathbb{Z}[X]/\langle X^{2}+41 \rangle$ and we can interchange the order in which we take quotients, we see that

$$R/3R \cong \mathbb{Z}[X]/\langle X^{2}+41, 3 \rangle \cong (\mathbb{Z}/3\mathbb{Z})[X]/\langle X^{2}+2 \rangle$$

Of course, $X^{2}+2$ is reducible over $\mathbb{Z}/3\mathbb{Z}$, since it factors as $(X+1)(X+2)$. The images of $X+1$ and $X+2$ in the quotient ring $(\mathbb{Z}/3\mathbb{Z})[X]/\langle X^{2}+2 \rangle$ are thus zero divisors, so $R/3R$ is not a domain, and therefore $3$ is not a prime element of $R$.