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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.

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.

ShellRox
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    Being a UFD is a property of the whole ring, not just of a subset. It is not enough to know that the $x_i-x_j$ are uniquely factorizable into a product of irreducibles; what you need is that the determinant of the Vandermonde matrix is uniquely factorizable into a product of irreducibles. – darij grinberg Aug 19 '19 at 17:08
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    The UFD property has no specific relation to determinants; it is just used (in a technical role) in this particular proof of the Vandermonde determinant. Other proofs of the Vandermonde determinant work fine without it. Even this proof can be put on a different footing, if one so desires: Instead of working in the polynomial ring $\mathbb{Z}\left[x_1, x_2, \ldots, x_n\right]$, you can work in the polynomial ring $A\left[x_1, x_2, \ldots, x_n\right]$ over any commutative ring $A$. This polynomial ring will not always be a UFD, but ... – darij grinberg Aug 19 '19 at 17:09
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    ... it is still true that every polynomial that is divisible by $x_i - x_j$ for all $i < j$ must be divisible by the product $\prod\limits_{i < j} \left(x_i-x_j\right)$ as well. This is Theorem 1.2 of my note Regular elements of a ring, monic polynomials and “lcm-coprimality”. But it takes some work to prove, as well. My main point was that there is a non-obvious step required for this proof, one way or another. – darij grinberg Aug 19 '19 at 17:11
  • @darijgrinberg Thank you for the clarification. So essentially the particular proof of Vandermonde determinant shows that considering that polynomial of "sum of products" representing Vandermonde determinant can be factorized to products of irreducible $x_i - x_j$s due to the Theorem 1.2 that you displayed, correct? But I'm not sure if I understand your second comment, isn't every polynomial representation of Vandermonde determinant element of UFD (since it is product of irreducible $x_i - x_j$s)? – ShellRox Aug 19 '19 at 17:30
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    Yes, you're getting the proof right. But a product of irreducible elements is not necessarily uniquely decomposable into a product of irreducible elements in general. Sometimes, there will be several non-equivalent factorizations. – darij grinberg Aug 19 '19 at 17:32
  • Many of the comments you received are wrong or misleading. No use of UFD is needed. All that is needed is the Factor Theorem and the fact that $,\Bbb Z[x_1,\cdots,x_n],$ is an integral domain so $,x_i-x_j,$ is cancelllable if $,i\neq j.,$ – Bill Dubuque Aug 19 '19 at 17:45
  • Btw, you greatly decrease your chances of receiving diverse quality answers if you accept an answer quickly, because many folks don't browse threads if they have already been marked with an accepted answer. It's always best to wait at least a few days before accepting. The longer you wait the better exposure your question will receive. – Bill Dubuque Aug 19 '19 at 17:52
  • @ShellRox Further, the simple Factor Theorem approach doesn't take "some work" to prove. Rather it takes only a paragraph or two. – Bill Dubuque Aug 19 '19 at 18:14
  • @BillDubuque Thank you for the responses. I was slightly confused about UFD since I thought polynomial representation of Vandermonde matrix belonged to UFD (but seems it is not by the latest comment of Darij). – ShellRox Aug 19 '19 at 18:38
  • @BillDubuque Also thank you for the suggestion, I thought I already received sufficient information (since I do not want to dive to deep into ring theory, due to absence of my experience), but I'll be happy to accept any new higher quality answer that this post gets. – ShellRox Aug 19 '19 at 18:44
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    @ShellRox You can probably figure out the Factor Theorem approach yourself (hint: proceed inductively as in this proof of the Bifactor Theorem). Since this question is specifically about the UFD approach it would be off-topic to answer here about that. – Bill Dubuque Aug 19 '19 at 18:54

1 Answers1

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  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.
Bernard
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