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.

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.

Irreducible Polynomial Over the Ring of Polynomials Over Integral Domain
Let $R$ be an integral domain and let $S=R[t]$ be the polynomial ring in $t$ over $R$. Let $n$ be a positive integer.
Prove that the polynomial
\[f(x)=x^n-t\]
in the ring $S[x]$ is irreducible in $S[x]$.
Proof.
Consider the principal ideal $(t)$ generated by $t$ […]

Characteristic of an Integral Domain is 0 or a Prime Number
Let $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.
Definition of the characteristic of a ring.
The characteristic of a commutative ring $R$ with $1$ is defined as […]

Nilpotent Element a in a Ring and Unit Element $1-ab$
Let $R$ be a commutative ring with $1 \neq 0$.
An element $a\in R$ is called nilpotent if $a^n=0$ for some positive integer $n$.
Then prove that if $a$ is a nilpotent element of $R$, then $1-ab$ is a unit for all $b \in R$.
We give two proofs.
Proof 1.
Since $a$ […]

In a Principal Ideal Domain (PID), a Prime Ideal is a Maximal Ideal
Let $R$ be a principal ideal domain (PID) and let $P$ be a nonzero prime ideal in $R$.
Show that $P$ is a maximal ideal in $R$.
Definition
A commutative ring $R$ is a principal ideal domain (PID) if $R$ is a domain and any ideal $I$ is generated by a single element […]

Equivalent Conditions For a Prime Ideal in a Commutative Ring
Let $R$ be a commutative ring and let $P$ be an ideal of $R$. Prove that the following statements are equivalent:
(a) The ideal $P$ is a prime ideal.
(b) For any two ideals $I$ and $J$, if $IJ \subset P$ then we have either $I \subset P$ or $J \subset P$.
Proof. […]

If Every Proper Ideal of a Commutative Ring is a Prime Ideal, then It is a Field.
Let $R$ be a commutative ring with $1$.
Prove that if every proper ideal of $R$ is a prime ideal, then $R$ is a field.
Proof.
As the zero ideal $(0)$ of $R$ is a proper ideal, it is a prime ideal by assumption.
Hence $R=R/\{0\}$ is an integral […]

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
\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}.\]
Show that $p$ is a prime ideal of the ring $\Z[\sqrt{10}]$.
Definition of a prime ideal.
An ideal $P$ of a ring $R$ is […]

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