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This is rather simple, but I am looking for a proof that $\dim(V) = {n + d \choose n}$, where

$$ V = \left\{f \in F[x_1, \dots , x_n] : \deg(f) \leq d\right\}. $$

Here, $F$ is any field (although $F$ can probably be replaced by any commutative ring). It is clear that a basis will consist of monomials of the form $x_1^{\alpha_1} \dots x_n ^{\alpha_n}$, where $\alpha_i$ are nonnegative integers and $\alpha_1 + \dots + \alpha_n \leq d$.

Thank you.

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up vote 3 down vote accepted

Here's a fairly general way to do this. Let $A = \bigoplus A_i$ and $B = \bigoplus B_i$ be graded vector spaces with each graded piece finite-dimensional. Their tensor product $A \otimes B$ has $n^{th}$ graded piece $\bigoplus_{i+j=n} A_i \otimes B_j$. Given a graded vector space as above, its Hilbert series is

$$\chi(A) = \sum_{i \ge 0} t^i \dim A_i.$$

Lemma: $\chi(A \otimes B) = \chi(A) \chi(B)$.

Now, $F[x_1, ... x_n]$ with the standard grading by degree is nothing more than $(F[x])^{\otimes n}$, and by inspection

$$\chi(F[x]) = \sum_{i \ge 0} t^i = \frac{1}{1 - t}$$

from which it follows that

$$\chi(F[x_1, ... x_n]) = \frac{1}{(1 - t)^n}.$$

The coefficients of this generating function are given by the binomial theorem applied to negative exponents. Since you want the sum of the dimension of the first few consecutive graded pieces, you want to multiply the above by $\frac{1}{1 - t}$ one more time, and then you get the appropriate answer.

Of course there are more direct and combinatorial ways to get the answer in this particular case. See the Wikipedia article on stars and bars.

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Thank you for the proof and the "stars and bars" link! – CircleGetsTheSquare May 27 '11 at 8:17
O.K., but see this item on Math Overflow:… – Christian Blatter May 27 '11 at 14:01
@Christian: "fairly general" was meant to distinguish this approach from "simple." It is not the simplest approach, but I think it is a useful general context in which to understand Hilbert series of various graded things. – Qiaochu Yuan May 27 '11 at 14:42

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