Relation between definition of determinant and UFD

Question

I've asked the question about inductively displaying Vandermonde determinant as a product of polynomials and one of the answers I got had an important piece of information within:

From the definition of a determinant (sum of products), the expansion must be a polynomial in $x_1,x_2,\cdots x_n$, of degree $0+1+2+\cdots n-1=\dfrac{(n-1)n}2$, and the coefficient of every term is $\pm1$.

Below the answer there is an interesting comment:

Note: This requires some knowledge about the polynomial ring in variables -- either that it is a UFD, or at least that common multiples of distinct $x_i - x_j$'s (with $i<j$) are multiples of their product.

I do not have a good knowledge of ring-like structures (especially polynomials), but I understand the very basics, so I decided to make a separate question in case explanation might not be brief.

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I know that Unique Factorization Domain (UFD) is an integral domain containing subset of non-zero non-identity elements that can be written as a product of irreducible elements. So I made a little hypothesis:

1. The complete expansion of determinant is written in the product of $p_{ij} = x_i - x_j$s with $i < j$. Thus maybe these elements are part of UFD since they can not be reduced anymore?
2. Perhaps the fact that least common multiples of distinct $p_{ij} = x_i - x_j$s with $i < j$s (which belong to UFD, according to point above) are multiples of their product? If so, how exactly?

Question:

Is my hypothesis above accurate by any measures? How exactly does UFD relate with representation of determinants as polynomials? Am I incorrectly understanding this concept?

Thank you and I apologize if I made basic mistakes with group-theory based statements in this answer.

Answer 1

1. The $x_i-x_j$ belong to the ring of polynomials in $n$ indeterminates $k[x_1,x_2,\dots, x_n]$, where $k$ is the field of coefficients. It is known that the ring of polynomials in a single indeterminate over a field is a principl ideal domain (P.I.D.),, like $\mathbf Z$. In several indeterminates, it is only a unique factorisation domain (U.F.D.), which mean any polynomial can be factored as product of irreducible polynomials, and this factorisation is unique, up to the ordEr of the factors and a unit factor.
2. The $x_i-x_l\;(i<j)$ are irreducible (essentially for degree reasons) and distinct, so their l.c.m. is their product (just like with integers). Here it happens that this product has exactly the expected degree of the determinant.