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1 vote
1 answer
72 views

If $p$ is a prime and $m,n\in\mathbb{N}$, prove that $\binom{pm}{pn}\equiv\binom{m}{n}$ mod $p$.

Hint: Compare the binomial expansions of $(1+x)^{pm}$ and of $(1+x^m)^p$ in $\mathbb{F}_p[x]$. If $R$ is a commutative ring, then the set of all polynomials with coefficients in $R$ is denoted by $R[x]...
Junk Warrior's user avatar
1 vote
0 answers
71 views

Why do two ways of expanding the same formal polynomial lead to matching coefficients?

There are a few proofs in which the technique is to expand the product of some formal polynomials in $\mathbb{R}[x_1,x_2,\ldots,x_k]$ in more than one distinct way and then we can match up the ...
Favst's user avatar
  • 3,415
1 vote
1 answer
34 views

how to reduce $(1-\alpha)^{T-i}$ into a sum

I'm given the following proof: \begin{align} & \sum^T_{i=k} \alpha^i(1-\alpha)^{T-i} \binom{T}{i} \\&=\sum^T_{i=k} \alpha^i \binom{T}{i} \sum_{j=0}^{T-i}(-\alpha)^j\binom{T-i}{j} \\&=\...
MoneyBall's user avatar
  • 877
1 vote
1 answer
49 views

Binomial expansion of $(1+X)^b$ of the form $\sum\limits_k b^kf_k(X)$ with $f_k(X)$ polynomial

For each $b \in \mathbb{N}, (1+X)^b=\sum\limits_{n=0}^b{{b}\choose {n}}X^{n}$ is a polynomial function of $X$. How to write $(1+X)^b$ in terms of $b$ as a closed expression of the form $$(1+X)^b=\sum\...
MathStudent's user avatar
  • 1,816
3 votes
1 answer
463 views

Gaussian polynomial identities

I'd appreciate any hints for showing that these identities are true for Gaussian polynomials. I've tried to approach the problem using basic algebra but it gets messy very quickly and I've gotten ...
User's user avatar
  • 371
0 votes
0 answers
44 views

How to calculate the $k$-dimension of a subspace of a polynomial ring? [duplicate]

Let $k$ be an infinite field and $R:=k[x_1,...,x_n]$ the polynomial ring in $n$ indeterminates. Why is the $k$-dimension of $U$ given by $\begin{pmatrix} n+m-1 \\ m\end{pmatrix}$, when $U$ is the ...
LSt's user avatar
  • 540
10 votes
1 answer
1k views

Vandermonde identity in a ring

Let $R$ be a commutative $\mathbb{Q}$-algebra. For $r \in R$ and $n \in \mathbb{N}$ we can define the binomial coefficient $\binom{r}{n}$ as usual by $\binom{r}{0}=1$ and $\binom{r}{n+1}=\frac{r-n}{n+...
Martin Brandenburg's user avatar