1.1 – Dasheng Wang

1.1

(1) Let $\varphi: R\to R'$ be a ring homomorphism and $\mathfrak m\in \text{Spm} (R')$ . Can we conclude $\varphi^{-1}(\mathfrak m)\in\text{Spm} (R)$ ?

Yes. Since $R'/\mathfrak m$ is a field, $R/\varphi^{-1}(\mathfrak m)$ is a field, $\varphi^{-1}(\mathfrak m)$ is a maximal ideal.

(2) Prove that an integral domain is a field as soon as it contains only finitely many elements. Deduce that in a finite ring every prime ideal is maximal.

Suppose $R$ is an integral domain and $r\in R$ , since $R$ is finite, $r^k=r$ for some $k>1$ , then $r^k-r=r(r^{k-1}-1)=0$ . Since $R$ is an integral domain, $r^{k-1}-1=0$ , then $r^{k-1}=rr^{k-2}=1$ , i.e. $r^{k-2}$ is an inverse of $r$ , so every nonzero element of $R$ is invertible, $R$ is a field. Suppose $I$ is a prime ideal of $R$ , then $R/I$ is an integral domain, hence a field, so $I$ is maximal.

(3) Prove the Chinese Remainder Theorem: Let $R$ be a ring with ideals $\mathfrak a_1,\ldots,\mathfrak a_n$ satisfying $\mathfrak a_i+\mathfrak a_j=R$ for $i\neq j$ . Then there is a canonical isomorphism

$R/\bigcap_{i=1}^n\mathfrak a_i=\prod_{i=1}^n R/\mathfrak a_i$

where the cartesian product of the rings $R/\mathfrak a_i$ is viewed as a ring under componentwise addition and multiplication.

Suppose $I$ , $J$ are two ideals of $R$ such that $I+J=R$ , then for any $r\in R$ , $r=r_i+r_j$ for some $r_i\in I$ , $r_j\in J$ , so $r\equiv r_j(\text{mod}\;I)$ , $r\equiv r_i(\text{mod}\;J)$ . Let $\varphi:R\to R/I\times R/J$ , then $r\in\ker \varphi$ if and only if $r_j\in I$ and $r_i\in J$, i.e. $r\in (I\cap J)$ . Therefore, $R/\ker\varphi=R/(I\cap J)\cong R/I\times R/J$ . Since if $I+J=R$ , $(I+J)+K=R$ , the isomorphism can be generalized to arbitrary number of ideals.

(4) Consider a ring $R$ and ideals $\mathfrak a_1,\ldots,a_r\subset R$ satisfying $\mathfrak a_i+\mathfrak a_j=R$ for $i\neq j$ . Show that, in this case, the inclusion $\prod_{i=1}^n\mathfrak a_i\subset \cap_{i=1}^n\mathfrak a_i$ is an equality.

If $a\in\cap_{i=1}^n\mathfrak a_i$ , then by Problem 3, $a/\cap_{i=1}^n\mathfrak a_i\cong\prod_{i=1}^n 0_{a/\mathfrak a_i}\cong 0_{\prod_{i=1}^na/\mathfrak a_i}$ , hence $a\in \prod_{i=1}^n\mathfrak a_i$ .

(5) Let $R$ be a principal ideal domain. (a) Give a characterization of Spec $R$ and of Spm $R$ . (b) Give a characterization of the ideals in $R/(a)$ for any element $a \in R$ .

(a) For a principle ideal $(a)$ , $(a)\in \text{Spec} R$ if and only if $a$ is a prime element. Every nonzero prime ideal in a PID is a maximal ideal, hence Spm $R=$ Spec $R\cup{0}$ .

(b) If $(a)$ is a prime ideal, $R/(a)$ is also a PID. If $(a)$ is not a prime ideal, $R/(a)$ is not an integral domain.

(6) Let $R_1, \ldots, R_n$ be rings and consider the cartesian product $R_1\times\ldots\times R_n$ as a ring under componentwise addition and multiplication. Show: (a) Given ideals $\mathfrak a_i \subset R_i$ for $i = 1,\ldots,n$ the cartesian product $\mathfrak a_1\times\ldots\times \mathfrak a_n$ is an ideal in $R_1\times\ldots\times R_n$ . (b) Each ideal in $R_1\times\ldots\times R_n$ is as specified in (a). (c) There is a canonical bijection

$\text{Spec}(R_1\times\ldots\times R_n)\xrightarrow{\ \sim\ }\amalg_{i=1}^n\text{Spec}R_i$

and a similar one for spectra of maximal ideals.

(a) Let $a,b\in \mathfrak a_1\times\ldots\times \mathfrak a_n$ and $r\in R_1\times\ldots\times R_n$ , then $(a-b)\in \mathfrak a_1\times\ldots\times \mathfrak a_n$ , $ra\in \mathfrak a_1\times\ldots\times \mathfrak a_n$ since every component is closed under these operations.

(b) Suppose $a,b$ are two elements in an ideal $I$ of $R_1\times\ldots\times R_n$ and $r\in R_1\times\ldots\times R_n$ , then since $ra=(r_1a_1,\ldots,r_na_n)\in I$ , $(b_1-a_1,\ldots,b_n-a_n)\in I$ , $a_i$ is an ideal of $R_i$ for $i=1,\ldots,n$ .

(c) Suppose $p=p_1\times\cdots\times p_n$ is a prime ideal of $R=R_1\times\cdots\times R_n$ , then we show that either $p_i$ is a prime ideal of $R_i$ or $p_i=R_i$ . Without loss of generality take $i=1$ . Let $x,y \in R_1$ with $xy \in p_1$ . Then $(x,0,\ldots, 0)\cdot (y,0,\ldots, 0) \in p$ , one of $(x,0,\ldots, 0)$ and $(y,0,\ldots, 0)$ is in $p$ , hence one of $x$ and $y$ is in $p_1$ . Therefore $p_1$ is prime or $p_1 = R_1$ . Next we show that at most one of the $P_i$ is not equal to $R_i$ . Suppose not. Without loss of generality suppose that $p_1\neq R_1$ and $p_2 \neq R_2$ . Then $(1,0,\ldots,0), (0,1,0,\ldots, 0) \in R\setminus p$ , but their product is in $p$ , so $p$ cannot be prime. Therefore, $R_1\times \cdots\times p_i\times\cdots\times R_n\mapsto \coprod_{i=1}^{i-1}R_i\sqcup p_i\sqcup\coprod_{i+1}^nR_i$ gives the bijection.

3 responses to “1.1”

dashengwang

May 25, 2022

Please leave a comment if you find a mistake.

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Niclas

January 19, 2025

Hey Dasheng Wang, deine Begründung bei Aufgabe 4 scheint mir nicht ganz richtig zu sein. Was genau meinst du denn mit der Notation a / cup_{i}^{n} mathfrak{a_i}?

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Niclas

January 19, 2025

Hey Dasheng Wang, deine Begründung bei Aufgabe (6) (b) ist mir auch nicht ganz klar. Du musst zeigen, dass jedes Ideal I in dem Produktring R_1 times dots times R_n das Produkt von Idealen in den Ringen R_j ist. Es gilt I = pi_1(I) times dots times pi_n(I), wenn pi_j colon R_1 times dots times R_n to R_j die j-te Projektion ist. Viele Grüße

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3 responses to “1.1”

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