# 5 is Prime But 7 is Not Prime in the Ring $\Z[\sqrt{2}]$

## Problem 224

In the ring
$\Z[\sqrt{2}]=\{a+\sqrt{2}b \mid a, b \in \Z\},$ show that $5$ is a prime element but $7$ is not a prime element.

## Hint.

An element $p$ in a ring $R$ is prime if $p$ is non zero, non unit element and whenever $p$ divide $ab$ for $a, b \in R$, then $p$ divides $a$ or $b$.

Equivalently, an element $p$ in the ring $R$ is prime if the principal ideal $(p)$ generated by $p$ is a nonzero prime ideal of $R$.

## Proof.

### 5 is a prime element in the ring $\Z[\sqrt{2}]$.

We first show that $5$ is prime in the ring $\Z[\sqrt{2}]$.
Suppose that
$5|(a+\sqrt{2}b)(c+\sqrt{2}d)$ for $a+\sqrt{2}b, c+\sqrt{2}d \in \Z[\sqrt{2}]$.
By taking the norm, we obtain
$25| (a^2-2b^2)(c^2-2d^2)$ in $\Z$.
From this, we may assume that $5|a^2-2b^2$.
Now look at the following table.

\begin{array}{ |c|c|c|c| }
\hline
a, b & a^2, b^2 \pmod{5} & 2b^2 \pmod{5} \\
\hline
0 & 0 & 0 \\
1& 1 & 2 \\
2& 4 & 3 \\
3 & 4 & 3\\
4 & 1 & 2\\
\hline
\end{array}

From this table, we see that $a^2-2b^2=0 \pmod{5}$ if and only if $a, b$ are both divisible by $5$.
Therefore $5|a+\sqrt{2}b$, and $5$ is a prime element in $\Z[\sqrt{2}]$.

### 7 is not a prime element in the ring $\Z[\sqrt{2}]$.

Next, we show that $7$ is not a prime element in $\Z[\sqrt{2}]$.
To see this, note that we have
$7=(3+\sqrt{2})(3-\sqrt{2})$ and $7$ does not divide $3+\sqrt{2}$ and $3-\sqrt{2}$.
Hence $7$ is not a prime element in the ring $\Z[\sqrt{2}]$.

## Related Question.

Problem. Prove that the ring $\Z[\sqrt{2}]$ is a Euclidean Domain.

For a proof of this fact, see that post “The Ring $\Z[\sqrt{2}]$ is a Euclidean Domain“.

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### 1 Response

1. 07/08/2017

[…] For a proof of this problem, see the post “5 is prime but 7 is not prime in the ring $Z[sqrt{2}]$“. […]

##### A Prime Ideal in the Ring $\Z[\sqrt{10}]$

Consider the ring $\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}$ and its ideal $P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}.$...

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