Library of Mathexandria

Lemma 1 Let $Z$ be an irreducible subset of $X$. Then the closure $\mathrm{cl}_X(Z)$ in $X$, is an irreducible subset of $X$. Proof) Suppose for contradiction that $\mathrm{cl}_X(Z) = Z_1 \cup Z_2$ where $Z_i$ are closed proper subsets of $\mathrm{cl}_X(Z)$. Then $Z = (Z_1 \cap Z) \cup (Z_2 \cap Z)$ where each $Z_i \cap Z$ is a closed subset of $X$. If either $Z_i \cap Z = Z$, then $Z \subset Z_i \Rightarrow \mathrm{cl}_X(Z) \subset Z_i \subset \mathrm{cl}_X(Z)$ which contradicts our assumption that $Z_i$ are proper subsets of $\mathrm{cl}_X(Z)$. This shows that $Z_i \cap Z$ are proper subset of $Z$. Lemma 2  Let $Y \subset X$ be a subspace and $Z \subset Y$. Then $\mathrm{cl}_Y(Z) = \mathrm{cl}_X(Z) \cap Y$. Proof)  Since $Z \subset \mathrm{cl}_X(Z) \cap Y$, it follows that $\mathrm{cl}_Y(Z) \subset \mathrm{cl}_X(Z) \cap Y$ as $\mathrm{cl}_X(Z) \cap Y$ is closed in $Y$. Now let $\mathrm{cl}_Y(Z) = W \cap Y$ where $W$ is a closed subset of $X$. Then $Z \subset \mathrm{cl}_

Most people know that  \[ (x+y)^n = x^n + y^n \] does not hold for all real numbers $x, y \in \mathbb{R}$ for $n > 1$. However, if one works over a characteristic $p$ ring, we do have \[ (x+y)^p = x^p + y^p \] The above is called the Freshman's dream and follows from Lemma \[p \left | \binom{p}{i} \right . \] for $0 < i < p$. Let $N = \binom{p}{i}$.  Proof) $N = \frac{p!}{(p-i)! i!}$. This implies that $p! = N (p-i)! i!$. For $0 < i < p$, $p \not| (p-i)!$ and $p \not| i!$. Therefore, $p |N$. Now we prove some more general result. Lemma \[ p \left | \binom{p^r}{i} \right . \] for $0 < i < p^r$. Proof)  Let $v_p$ be the $p$-adic valuation. Then \[ v_p(n!) = \sum_{k=1}^r \left \lfloor \frac{n!}{p^k}  \right \rfloor  \] Recall that $\left \lfloor \frac{n}{m} \right \rfloor$ is the number of multiples of $m$ less or equal to $n$. By varying $k$, we would count those are divisible by higher $p$ powers of p.