Is $1+\sqrt{5}$ a prime under the $\mathbb{Z}[{\sqrt{5}}]$ domain?

Definitions are indeed very important. Two answerers on this one question have been led astray thinking you're asking about $\mathbb{Z}[\sqrt{-5}]$, which is a big but understandable mistake, one that many askers on this site have made.

There are at least two different things you might mean by $\mathbb{Z}[\sqrt{5}]$. One is the ring of algebraic integers of the form $a + b \sqrt{5}$, with $a, b \in \mathbb{Z}$; this includes numbers like $1 + \sqrt{5}$ and $-3 + 2 \sqrt{5}$. Or you could mean the ring of all algebraic integers in $\mathbb{Q}(\sqrt{5})$, which could potentially include numbers like $$\frac{3}{4} + \frac{7 \sqrt{5}}{3},$$ but I'm getting ahead of myself.

If you mean the former, the answer is clear: $1 + \sqrt{5}$ is not prime. Since $(1 + \sqrt{5})(1 - \sqrt{5}) = -4$ but $1 + \sqrt{5}$ is not a divisor of $-2$ nor $2$, it follows that $1 + \sqrt{5}$ is irreducible but not prime (I'm taking your word for it that the number is irreducible).

If you mean the latter, then you have overlooked a couple of things, because $1 + \sqrt{5}$ is actually composite. Verify that $$\frac{1 + \sqrt{5}}{2}$$ is an algebraic integer with minimal polynomial $x^2 - x - 1$ (ever hear of a thing called the golden ratio?) Then $$2 \left(\frac{1 + \sqrt{5}}{2}\right) = 1 + \sqrt{5}.$$


In a ring $R$, we say a non-unit, nonzero $p \in R$ is prime if for any $a, b \in R \setminus \{ 0 \}$ such that $p | ab$, we have either $p | a$ or $p | b$. Mathmo123 showed $1 + \sqrt{5}$ wasn't prime.


No, it's not prime. Compare $6$. Is $6$ prime in $\textbf Z$? If it was, we'd see that whenever $6 \mid ab$, either $6 \mid a$ or $6 \mid b$. Yet $6 \mid 3 \times 4$ but $6 \nmid 3$ nor $4$.

Likewise, in $\textbf Z[\sqrt{-5}$, $(1 + \sqrt{-5}) \mid 2 \times 3$ but...

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Algebraic Number Theory

Integer Rings

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