# What powers of $p$, where $p$ is prime, divide $\binom {p^{\alpha} m}{p^{\alpha}}$?

My question is about a short combinatorial discussion in Herstein's Topics in Algebra, on page 92, preceding one of his three proofs of Sylow's Theorem. I will cite the important part:

If $$p^r | m$$ but $$p^{r+1} \nmid m$$, consider $$\binom {p^\alpha m}{p^{\alpha}} = \frac{(p^{\alpha}m)!}{(p^{\alpha})!(p^{\alpha}m-p^{\alpha})!} = \frac {p^{\alpha}m(p^{\alpha}m-1) \dots (p^{\alpha}m-i) \dots (p^{\alpha}m-p^{\alpha} + 1)} {p^{\alpha}(p^{\alpha}-1) \dots (p^{\alpha}-i) \dots (p^{\alpha}-p^{\alpha} + 1)}.$$ The question is, What power of p divides $$\binom {p^{\alpha} m}{p^{\alpha}}$$? Looking at this number, written out as we have written it out, one can see that except for the term $$m$$ in the numerator, the powers of $$p$$ dividing $$(p^{\alpha}m-i)$$ is the same as that dividing $$p^{\alpha}-i$$, so all powers of $$p$$ cancel out except the power which divides $$m$$. Thus $$p^r | \binom {p^{\alpha} m}{p^{\alpha}}$$ but $$p^{r+1} \nmid \binom {p^{\alpha} m}{p^{\alpha}}$$.

My understanding is as follows. $$\binom {p^\alpha m}{p^{\alpha}} = m *\frac{\binom {p^\alpha m}{p^{\alpha}}}{m}$$, consider the unique factorization of both factors. The right will not contain any nonzero powers of $$p$$, and so the power of $$p$$ dividing our binomial is the largest power in the unique factorization of $$m$$. But there are two gaps in my understanding: why is the second factor neccesarily an integer, and why does the second factor not contain powers of $$p$$? Herstein explains this second question, but I don't get his argument.

• I don't see why $\frac1m\binom{p^\alpha m}{p^\alpha}$ need be an integer. Anyway, Herstein neither asserts this nor needs this. – Lord Shark the Unknown Mar 24 at 18:11

Kummer's Theorem, which is proven in this answer, says that the number of factors of $$p$$ in $$\binom{n}{k}$$ is the number of carries when adding $$k$$ to $$n-k$$ in base-$$p$$.
If $$p^r\mid m$$ and $$p^{r+1}\nmid m$$, then adding $$1$$ to $$m-1$$ will have exactly $$r$$ base-$$p$$ carries. Likewise, adding $$p^\alpha$$ to $$(m-1)p^\alpha$$ will have exactly $$r$$ base-$$p$$ carries. Therefore, there will be $$r$$ factors of $$p$$ in $$\binom{mp^\alpha}{p^\alpha}$$.
• The base-$p$ carries refer to the base-$p$ representation of numbers; not to $p$-adic numbers. The rest of this is detailed in the answer cited above, where it is noted that the number of base-$p$ carries when adding $k$ to $n-k$ equals $\frac{\sigma_p(k)+\sigma_p(n-k)-\sigma_p(n)}{p-1}$ where $\sigma_p(n)$ is the sum of the digits in the base-$p$ representation of $n$. – robjohn Mar 25 at 9:32