TITLE: $R = \mathbb{Z}[\sqrt{-41}]$, show that 3 is irreducible but not prime in $R$
QUESTION [7 upvotes]: 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 $(1 + \sqrt{-41})(1 - \sqrt{-41}) = 42$?

REPLY [4 votes]: 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 $R$.