16
$\begingroup$

Let $k[X_0, X_1, \ldots, X_n]_d$, or briefly $k[X]_d$, be the $k$-vector space whose elements are the zero polynomial and homogeneous polynomials of degree $d\geq 1$. I found the following formula for the dimension $$dim(k[X]_d)= \binom{n+d} {n}$$

but in my book there is no justification for this equality. Could someone explain me, possibly in an intuitive way, why that binomial coefficient is the dimension of that vector space?

$\endgroup$

1 Answer 1

4
$\begingroup$

Added later: Except the formula is correct, as shown in the comments below; I had mis-read it; when $d = 2$ and you have 3 variables, $n = 2$, since the variable indexing starts at zero. And 4-choose-2 is in fact six, as expected.

[what follows is incorrect]

A nice basis for that space consists of all monomials in the $n$ variables with total degree $d$.

Wait...what about degree $d = 2$ and $n = 3$ variables. Listing the basis elements, I see $$ x^2, y^2, z^2, xy, xz, yz, $$ which is only $6$ dimensions, but 5 choose 3 is 10. Seems as if there might be a mistake in your formula, or perhaps I'm not understanding how it's supposed to be applied. (See below for resolution of this apparent contradiction.)

I think (or thought) perhaps your formula is incorrect.

$\endgroup$
5
  • $\begingroup$ It looks like the problem of choosing $d$ elements outside a set of $n$ elements, but this would lead to $n$ choose $d$. Why instead we have $n+d$? $\endgroup$ Commented Mar 3, 2014 at 15:35
  • $\begingroup$ Close: it's picking $d$ elements, with possible repetitions, from $n$. But the 5 choose 3 case suggests that there might be an off-by-one error somewhere in here. Even the $d = 1, n = 2$ case seems wrong, since 3 choose 1 overestimates the answer, which should be "2". $\endgroup$ Commented Mar 3, 2014 at 15:42
  • $\begingroup$ the number of variables is $n+1$, the first variable is $x_0$ $\endgroup$ Commented Mar 3, 2014 at 15:43
  • 10
    $\begingroup$ D'oh! Thanks for helping with my reading comprehension! Here's a solution ("stars and bars"): place $n$ bars and $d$ stars in a row, like $*|*||$. Read each bar as a divider between variables, and each * as an accumulated exponent. So our example is $x_0^1 x_1^1 x_2^0 x_3^0$, while $|||**$ would be $x_3^2$. Clearly each stars-and-bars pattern corresponds to a homogeneous monomial. And the number of stars and bars patterns is the number of ways to place $d$ stars in a list of $n+d$ items. Done! $\endgroup$ Commented Mar 3, 2014 at 15:51
  • 2
    $\begingroup$ Dear @John: if you have three variables, then $n=2$. Indeed since variables are numbered from $0$ to $n$ their number is $n+1$. Hence in your example $\binom{n+d} {n}=\binom{2+2} {2}=\binom{4}{2}=6$, just as you computed. So Danae's formula is correct: please, modify your claim that it isn't. $\endgroup$ Commented Mar 3, 2014 at 20:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .