TITLE: Are there number systems or rings in which not every number is a product of primes?
QUESTION [14 upvotes]: I am reading through some number theory and abstract algebra books, and in the number theory books they all prove the theorem which states that every integer is a product of primes (irreducibles). In particular, in Hardy and Wright, they prove that every element in $\mathbb{Z}[i]$ and $\mathbb{Z}[\rho]$ where $\rho:=e^{\frac{2}{3}\pi i}$ is a product of primes, where they define primes to be elements who are only divisible by themselves and the units of their ring. They prove it in a way which uses the norm function ($N(a+bi)=a^2+b^2$), and I believe the proof can be easily adapted to include any euclidean domain. 
This property (ability for every element to be written as a product of primes, not necessarily unique) seems so inherent to me, so I am wondering: are there any number systems or rings in which not every element can be written as a product of primes?
Clarification: I am only talking about the non-units of a ring (that aren't 0). Units are not considered to be prime. $\mathbb{Q}$ satisfies this property because there are no non-units (except 0). 
Clarification 2: The answer given by Barry Smith was great in that it gave an example of a ring which was not a field but contained no primes so it had nonunit elements that were not products of primes, but I was really looking for a ring which contained primes/irreducibles and also contained elements that could not be expressed as a product of primes/irreducibles. My intuition states that this does not exist straightforwardly from the definition of prime/irreducible. The reason I question my intuition is because my number theory book exploits the norm function, which is a function that not all rings contain.

REPLY [3 votes]: This answer deals only with the fact that $5$ remains prime in every extension $\Bbb Q(2^{1/2^n})$. I hesitated a bit to give a proof so close to Barry’s, but since it is incrementally more elementary, I decided to go ahead.
My argument depends only on the fact that if $p$ is a prime in $\Bbb Z$, and $K$ is an extension of $\Bbb Q$ of degree $n$, in whose ring of integers the primes above $p$ are $\mathfrak p_1,\cdots, \mathfrak p_m$, each $\mathfrak p_i$with its ramification index $e_i$ and residue field extension degree $f_i$, then
$$
n=\sum_{i=1}^me_if_i\>\>.
$$
I’ll show that for $p=5$, adjoining the $2^n$-th root of $2$ must require the residue field to be $\Bbb F_{5^{2^n}}$, that is, an extension of the prime field $\Bbb F_5$ of degree $2^n$. Since that’s the degree of the field $K_n$ of $2^n$-th roots of $2$, we must have $m=1$, and furthermore the unique ramification index $e$ must be $1$. As a result of the last fact, $\mathfrak p$ is generated by $5$, so that $5$ remains prime in $K_n$.
Now for my claim that the residue field has $5^{2^n}$ elements when you adjoin a $2^n$-th root of $2$ to $\Bbb F_5$. In the prime field, $2$ is a primitive fourth root of unity, so that to adjoin its $2^n$-th root, you’re adjoining a primitive $2^{n+2}$-th root of unity. For finite fields $k=\Bbb F_{5^N}$ of characteristic $5$, the following are equivalent:
$$
2^{n+2}\big||k^*|\Leftrightarrow
|k|\equiv1\pmod{2^{n+2}}\Leftrightarrow
5^N\equiv1\pmod{2^{n+2}}\,.
$$
But you easily show that for any integer $a$ and any $r\ge2$, if $a=1+2^nu$ with $u$ odd, then $a^2=1+2^{n+1}u'$ with $u'$ odd as well. In particular $5^{2^M}=1+2^{M+2}u$ for an odd number $u$. Therefore, to get the primitive $2^{n+2}$-th roots of unity in a finite field of characteristic $5$, you exactly need an extension of the prime field of degree $2^n$, as I claimed.