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Given a integral domain $D$, can every non-zero, non-unit element be written as a product of finite number of primes? Since an integral domain is just a generalization of the integers, I feel like this should be possible, but at the same time, things are not exactly so clear or even obvious in algebra so I am just not sure.

Note: an element $p$ is defined to be prime if it is nonzero, non unit and if p divides ab, then p divides a or p divides b.

david h
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    Any element being a product of two primes is something which is not true even in $\mathbb{Z}$. If you mean a finite number of primes, this is still false in general, there are integral domains without prime factorization. See for example: https://math.stackexchange.com/questions/3999158/is-there-an-integral-domain-containing-an-element-having-no-factorization-into-i – Mark Oct 22 '22 at 12:01
  • Yes I meant a finite number of primes. I will make a correction. – david h Oct 22 '22 at 12:39
  • Related: https://math.stackexchange.com/a/1820328/700480 –  Oct 22 '22 at 13:28
  • The same simple proof as in $\Bbb Z$ shows that if a nonzero nonunit has a factorization into primes then that is unique in the sense that every other factorization into atoms (irreducibles) is the same (up to order and associates). So your hypothesis implies both existence and uniqueness of factorizations into atoms, so the domain is a UFD - as explained here in the dupe. – Bill Dubuque Oct 22 '22 at 16:42

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Interesting. Since a general $D$ is no longer necessarily the integers, you have to say what you mean by prime, as well as the fact that $D$ might have its own different definition of multiplication.

Pretty clear that you're referring to the "fundamental theorem of arithmetic".

We do, I believe, have something pretty similar happening in the ring $D[x]$, of polynomials in the single variable $x$, which turns out to be an integral domain itself.

The analog of primes would be irreducible polynomials. Those are polynomials that can't be factored into products of lower degree polynomials.

But, in general, $D[x]$ need not be a UFD, or a unique factorization domain. Thus it won't always be possible to factor into irreducibles uniquely.

However, if we start with a field $\Bbb F$, we do indeed get a UFD for $\Bbb F[x]$.

Turning it back down a notch, I probably should have started with whether or not $D$ itself is a UFD. Then, there's prime ideals, which may be considered to sort of generalize the primes.

Suffice it to say that this story is only the beginning, and that, as you hinted, algebra can be a very elaborate subject. Enjoy your studies!

suckling pig
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    Isn't prime pretty well defined. an element p of a (commutative) ring is prime if p | ab implies p divides a or p divides b. (in other words, p is prime if (p) is a prime ideal) – Saad Haider Oct 22 '22 at 12:51