Prove that there are infinitely many primes in $\mathbf Q[\sqrt{d}]$. Prove that there are infinitely many primes in $\mathbb Z[\sqrt{d}]$. I don't know how to prove this, but I think that the proof will be similar to proving that there are infinitely many primes in $\mathbb Z$
 A: It seems now from the comments that despite the wording of the problem, we are asked to show there are infinitely many irreducibles. This can be done by a straightforward modification of the standard "Euclid" proof, so for fun we show that there are infinitely many primes among the integers of $\mathbb{Q}[\sqrt{d}]$.
If $d=-1$, $2$, or $-2$, there is no problem, since the irreducibles  in these cases are prime, and it is easy to show there are infinitely many irreducibles. So we can assume that $d$ has an odd prime divisor. Let $q$ be the smallest odd prime divisor of $d$, and let $a$ be the smallest quadratic non-residue of $q$. Then $a$ is prime.
By Dirichlet's Theorem on primes in arithmetic progressions, there are infinitely many odd primes of the form $a+kq$, and therefore infinitely many such odd primes that do not divide $d$. Any such prime $p$ is a quadratic non-residue of $q$.  We will show that $p$ is a prime in the integers of $\mathbb{Q}[\sqrt{d}]$.
So we want to show that if $x$ and $y$ are algebraic integers in $\mathbb{Q}[\sqrt{d}]$, and $p$ divides $xy$, then $p$ divides $x$ or $p$ divides $y$. 
Note that $p$ divides the norm $x\bar{x}y\bar{y}$ of $xy$. So $p$ divides one of $x\bar{x}$ or $y\bar{y}$, say $x\bar{x}$. Now the details depend a little on whether $d$ is congruent to $1$ modulo $4$ or not. The argument is a bit simpler if $d\not\equiv 1\pmod{4}$.
In that case, $x$ has the shape $m+n\sqrt{d}$ where $m$ and $n$ are integers. We have that $p$ divides $m^2-dn^2$. If $p$ does not divide $m$, we have contradicted the fact that $p$ is a quadratic non-residue of $q$. So $p$ divides $m$, and since $p$ does not divide $d$, it follows that $p$ divides $n$, so $p$ divides $x$. 
If $d\equiv 1\pmod{4}$, it could be that $x=\frac{m+n\sqrt{d}}{2}$ where $m$ and $n$ are odd. But again we get that $p$ divides $m^2-dn^2$, and again we conclude that $p$ divides $m$ and $n$.
Remark: If instead we are working in $\mathbb{Z}[\sqrt{d}]$, then the case $d\equiv 1\pmod{4}$ does not require special treatment.
