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Codimension of a closed irreducible subset and generic points

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}_
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Freshman's Dream

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.
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Max-Spec vs Spec

This is an article on why using Max-Spec and Spec do not matter much if the ring $A$ is a Jacobson ring.  Definition. A ring $A$ is a Jacobson ring if every prime ideal of $A$ is an intersection of maximal ideals. In other words, \[ \mathrm{Nil}(A) = \bigcap_{\mathfrak{p}} \mathfrak{p} = \bigcap_{\mathfrak{m}} \mathfrak{m} = \mathrm{Jac}(A) \]  Proposition.  Let $X =\mathrm{Spec}(A)$ and $X_0 = \mathrm{MaxSpec}(A)$, i.e. the set of closed points of $X$. The $X_0$ is a very dense subset of $X$. We show that some topological properties are preserved. Let's first describe what it means for $\mathrm{Spec}(A)$ to be disconnected. By this means that there exist disjoint closed subsets $V(I)$ and $V(J)$ that cover $\mathrm{Spec}(A)$. In other words, 1) $V((0))=\mathrm{Spec}(A) = V(I) \cup V(J) = V(I \cap J)$.  2) $V((1)) = \varnothing = V(I) \cap V(J) = V(I + J)$. This is same as saying that 1') $\sqrt{I \cap J} = \sqrt{(0)} = \mathrm{Nil}(A)$, 2'
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Question. when does a morphism of schemes send a closed point to a closed point?  1) Integral Morphisms This amounts to saying that $B/A$ integral extension, then $B$ is a field if and only if $A$ is a field. 2) Projective => Proper => Closed => preserves closed point. Reference. mathSE post Let $X$ be a scheme of finite type over a finite (or algebraically closed) field $k$. Let $K$ be a finite extension of $\mathbb{Q}_p$ with residue field $k$.  We would like to define $D_c^b(X, \overline{\mathbb{Q}}_l)$. 1) $D_c^b(X, \mathcal{O}_r)$ where $\mathcal{O}_r$ is $\mathcal{O}/\pi^r \mathcal{O}$.
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Simultaneously principal open set.

For some reason, I have a hard time understanding the proof that in an intersection of two affine open subschemes, there exists an open set that is principal in both affine opens. Here is my attempt to make it slightly more clear. The precise statement that I want to prove is the following. Let $X$ be a scheme and $U=\mathrm{Spec}~A$ and $V=\mathrm{Spec}~B$ be two affine open subschemes of $X$. For all $x \in U \cap V$, there exists $W = D_U(f) = D_V(g)$, a neighborhood of $x$, such that $f \in A$, $g \in B$, and $W \subset U \cap V$. 1) We would like to reduce to a simpler case. $U \cap V$ is an open subset of $V$. Hence there exists a principal open set $D(g) \subset U \cap V$ with $g \in B$ containing $x$. Then we may identify $D(g) = \mathrm{Spec}~B_g$. Therefore, we have $x \in \mathrm{Spec}~B_g \subset \mathrm{Spec}~A$. Suppose we found an open set $W$ containing $x$ that is principal inside both $\mathrm{Spec}~B_g$ and $\mathrm{Spec}~A$, then there exists some $h/g^n \i
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RSA 암호 1. 개요

수학자들 중에서 정수론을 전문적으로 공부하는 사람을 정수론자라고 한다. 워낙 추상적인 개념을 많이 다루는 직업이기에 매번 "그래서 정수론을 배워서 뭐에 써먹는데?" 라는 질문을 자주 받게 된다. 솔직히 말하면 정수론이 실생활에 응용되어 쓰여지는 일은 드물다. 하지만 컴퓨터의 등장 이후, 모듈러 산술을 이용해 암호론에 정수론에 적용되기 시작되었다. 그 중에서 RSA 암호에 대해서 설명해보려고 한다. 정수론에 대해서 이야기 하기 전에 먼저 암호에 대해서 이야기 해보자. 다음 용어를 숙지해두자. 용어정리 평문 : 보호해야 할 메세지 암호문 : 평문을 특정 대상 외에는 이해하지 못하게 변환한 메세지 암호화 : 평문을 암호문으로 바꾸는 과정 복호화 : 암호문을 평문으로 바꾸는 과정 기본적으로 우리는 다음을 하고 싶다. 김씨가 박씨에게 비밀 메세지를 보내고 싶다고 하자. 1. 김씨는 평문을 암호화 하여 암호문을 만든다. 2. 김씨는 암호문을 박씨에게 넘긴다. 3. 박씨는 김씨의 암호문을 받아 복호화를 하여 평문을 읽는다. 두 사람이 다른 사람에게 들키지 않고 비밀 메세지를 주고 받는 방법에 대해서 생각해보자. 일단 가장 간단한 방법은 두 사람이 같은 "키"를 가지고 있는 것이다. 대칭키 암호 (Symmetric-Key Cryptography) 예시 1 자물쇠가 달린 상자로 비유해보자. 김씨와 박씨는 다른 사람은 보지 못하게 비밀 메세지를 교환하고 싶다. 이를 위해서 그들은 자물쇠가 달린 상자를 구매했고 서로 같은 키를 구매했다. 1. 김씨는 비밀 메세지 (평문)을 상자에 넣어 자물쇠를 채운다 (암호화). 2. 김씨는 상자 (암호문)을 박씨에게 넘긴다. 3. 박씨는 김씨의가 보낸 상자를 키를 이용해서 자물쇠를 연다 (복호화). 이와 같이 김씨와 박씨가 같은 키를 쓰는 것을 대칭 키 암호 (Symmetric-Key Algorithm) 이라고 부른다. 만약 박씨가
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Yoshida - Non-abelian Lubin-Tate 7

Now we would like to study the special fiber of $X:=\mathrm{A}$ over $S:=\mathrm{Spec}~W$. We denote the special fiber by $X_s:= X \times_S \mathrm{Spec}~\overline{k}$, i.e. the fiber of the special point (the maximal ideal). Sine this is simply just reducing mod $\pi$, combining with the result from the previous post where we computed the explicit formula for $A$, we obtain that \[ X_s = \mathrm{Spec}~ \overline{k}[[\widetilde{X}_1, \ldots, \widetilde{X}_n]]/ \left ( \prod_{\underline{a} \in k^n \backslash \{ 0 \}} (P_{\underline{a}} ~\mathrm{mod}~ \pi ) \right ) \] Let $Y_{\underline{a}}$ to be the closed subscheme of $X_s$ defined by $P_{\underline{a}} ~\mathrm{mod}~\pi = 0$. Equivalently, this is simply the closed subscheme of $X$ defined by $P_{\underline{a}} = 0$. It can be shown that $Y_{\underline{a}} = Y_{\underline{a'}}$ if and only if $a$ is a scalar multiple of $a'$ by some element in $k^\times$. Therefore, we can label the irreducible component of $X_s
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