Let $R$ be a finite commutative ring. Does $R$ must be a direct product of local rings? The basic example is $\mathbb Z/n\mathbb Z$, but I can not find any counter example. If the answer is no, then in what conditions $R$ is a direct product of local rings?
123479
Proposition. Let $R$ be commutative ring with $1_R$. We assume that $R$ is an Artinian ring and $M_1,\dots,M_n$ its maximal ideals. Then         $R/\mathrm{Jac}(R)\cong (R/M_1)\times \dotsb \times (R/M_n)$.   The ring $R$ is isomorphic to the direct product of a finite number of Artinian local rings.      Proof. 1. It's obvious that $M_i+M_j=R,\ \forall 1\leq i \neq j \leq n$. So, from Chinese Remainder Theorem, we have  $$R/\bigcap_{i=1}^{n}M_i = R/\mathrm{Jac}(R) \cong (R/M_1)\times \dotsb \times (R/M_n),$$ as we wanted.    Since $R$ is an Artinian ring, we have $\mathrm{Jac}(R)^m=\{0_R\}$, for some $m\in \Bbb Z^+$. But,  $$\{0_R\}\subseteq M_1^m\dotsb M_n^m=(M_1\dotsb M_n)^m\subseteq (\bigcap_{i=1}^{n}M_i=\mathrm{Jac}(R))^m=\{0_R\}.$$ So, if we apply CRT we will take  $$R\cong R/\{0_R\}\cong R/M_1^m\dotsb M_n^m\cong (R/M_1^m)\times \dots \times (R/M_n^m).$$   Questions.   1) Are these thoughts complete and correct?  2) Why $R/M_i^m$ are artinian local rings?  3) Could you please elaborate some examples as an application?
118281
[Update: Based on the hints provided by @zcn and @whacka, I believe I have found a solution. See my answer below.]  Below, $R$ is a commutative ring with $1$.  In John J. Watkins' Topics in Commutative Ring Theory, the author observes that if $a$ and $b$ are associates in $R$, then they generate the same ideal: $(a) = (b)$. This is clearly true, because $a = ub$ and $b = va$ for some units $u$ and $v$, and so $a \in (b)$ and $b \in (a)$, whence $(a) \subset (b)$ and $(b) \subset (a)$. Indeed, we didn't even require $u$ and $v$ to be units to reach this conclusion.  The author then claims that if $R$ is not a domain, then the converse may not be true: we may have $(a) = (b)$ even if $a$ and $b$ are not associates. As a "counterexample" he gives $R = \mathbb{Z}/(6)$, with $a = 2 + (6)$ and $b = 4 + (6)$. Clearly these elements generate the same ideal: $(a) = (b) = \{0, 2 + (6), 4 + (6)\}$. The author asserts that $a$ and $b$ are not associates, but this is false: $5 + (6)$ is a unit, and $(2 + (6))(5 + (6)) = 4 + (6)$, so $a$ and $b$ actually are associates.  I figured that the author botched the example but that surely the basic fact must be true: if $R = \mathbb{Z}/(n)$ then there may be nonassociates $a,b \in R$ with $(a) = (b)$. However, I was not able to find a counterexample and eventually resorted to writing a computer program and found that there are no counterexamples in $\mathbb{Z}/(n)$ for $n \leq 100$.  So now I'm starting to think that there is no counterexample in $\mathbb{Z}/(n)$ at all, but I have not yet been able to prove this.  My questions:   If I am correct that $(a) = (b)$ in $\mathbb{Z}/(n)$ implies that $a$ and $b$ are associates, I would appreciate a nudge in the right direction toward a proof. (But if the proof is elementary, please give only a hint, not a solution :-) I'll post my solution once I have one. In either case, could you please point me to a genuine counterexample? If there is one in $\mathbb{Z}/(n)$ for some $n$, that would be ideal (no pun intended), otherwise a counterexample in any commutative ring with $1$ would be great.
54622
Let $I$, $J$ be finitely generated ideals in a ring $A$ (commutative with identity). I know that the intersection need not be finitely generated: can somebody give me an example?   Thanks.
21686
I like to be thorough, but if you feel confident you can skip the first paragraph.  Review: A ring is a set $R$ endowed with two operations of + and $\cdot$ such that $(G,+)$ is an additive abelian group, multiplication is associative, $R$ contains the multiplicative identity (denoted with 1), and the distributive law holds.  If multiplication is also commutative, we say $R$ is a commutative ring. A ring that has no zero divisors (non-zero elements whose product is zero) is called an integral domain, or just a domain.  We want to show that for a domain, the equation $x^2 = 1$ has at most 2 solutions in $R$ (one of which is the trivial solution 1).    Here's what I did:   For simplicity let $1,a,b$ and $c$ be distinct non-zero elements in $R$.  Assume $a^2 = 1$.  We want to show that letting $b^2 = 1$ as well will lead to a contradiction.  So suppose $b^2 = 1$, then it follows that $a^2b^2 = (ab)^2 = 1$, so $ab$ is a solution as well, but is it a new solution?  If $ab = 1$, then $abb = 1b \Rightarrow a = b$ which is a contradiction.  If $ab = a$, then $aab = aa \Rightarrow b = 1$ which is also a contradiction.  Similarly, $ab = b$ won't work either.  So it must be that $ab = c$.  So by "admitting" $b$ as a solution, we're forced to admit $c$ as well.  So far we have $a^2 = b^2 = c^2 = 1$ and $ab = c$.  We can procede as before as say that $(abc)^2 = 1$, so $abc$ is a solution, but once again we should check if it is a new solution.  From $ab = c$, we get $a = cb$ and $b = ac$, so $abc = (cb)(ac)(ab) = (abc)^2 = 1$.  So $abc$ is not a new solution; it's just one.  At this point I'm stuck.  I've shown that it is in fact possible to have a ring with 4 distinct elements, namely $1,a,b$ and $c$ such that each satisfies the equation $x^2 = 1$ and $abc = 1$.  What am I missing?
121528
Let $R$ be a commutative ring with unity, and let $R^{\times}$ be the group of units of $R$. Then is it true that $(R,+)$ and $(R^{\times},\ \cdot)$ are not isomorphic as groups ?    I know that the statement is true in general for fields. And it is trivially true for any finite ring (as $|R^{\times}| \le |R|-1&lt;|R|$, so they are not even bijective).  I can show that the groups are not isomorphic whenever $\operatorname{char} R \ne 2$ , but I am unable to deal with $\operatorname{char} R=2$ case ... Please help. Thanks in advance.
117346
Prove that if $R$ is a commutative ring in which all the prime ideals are finitely generated, then $R$ is Noetherian.   Here is what I been told to do: Suppose that $R$ is not Noetherian, and use Zorn’s lemma to obtain a maximal element $I$ in the collection of all ideals of $R$ that are not finitely generated.  Then use the following proposition: Let $I$ be an ideal of a commutative ring $R$, and let $r ∈ R$. If the ideals $I +rR$ and {$s ∈ R : sr ∈ I$} are finitely generated, then $I$ is a finitely generated ideal.  Can anyone help please. thanks a lot.
108915
The question is from the following problem:     Let $R$ be a ring with a multiplicative identity. If $U$ is an additive subgroup of $R$ such that $ur\in U$ for all $u\in U$ and for all $r\in R$, then $U$ is said to be a right ideal of $R$. If $R$ has exactly two right ideals, which of the following  must be true?   I. $R$ is commutative. II. $R$ is a division ring. III. $R$ is infinite.    I know the definition of every concept here. But I have no idea what is supposed to be tested here.    Why is the ring $R$ which has exactly two right ideals special? What theorem does one need to solve the problem above?   Edit. According to the answers, II must be true. For III, $R$ can be a finite field according to mt_. What is the counterexample for I then?
55017
Let $R$ be a commutative ring with unit and let $m$ be a maximal ideal of $R$. Are there known conditions on $R$ or $m$ such that the $m$-adic completion $\hat{R}$ of $R$ is a local ring.   Since the completion of a Noetherian local ring is again local, I'm primarily interested in cases where $R$ itself is not local.   An example is the polynomial ring $R=k[X_1,...,X_n]$ ($k$ a field) with $m=(X_1,...,X_n)$ where the power series ring $\hat{R}=k[[X_1,...,X_n]]$ is local with maximal ideal $(X_1,...,X_n)$.
78964
I'm relatively new to ring theory so this is probably a simple question.  Its easy to see that a finite ring that is commutative and has no-zero divisors (i.e. an integral domain) must have multiplicative inverses.  I am wondering if we can rearrange these properties and always get implication or produce finite rings with only 2 of the above properties.  Explicitly my questions are:  1) Does a finite ring that is commutative with multiplicative inverses always have no-zero divisors? (EDIT: this one is pretty easy too, let $xy=0$ and assume $x\neq 0$. then $x^{-1}xy=x^{-1}0$. Hence, $y=0$ as desired.)  2) Is a finite ring that has multiplicative inverses and no zero divisors always commutative?  If these are actually simple exercises, I'd just like a hint to get started with the proof or any counterexamples.  Thanks :)
33643
Let $A$ be a commutative ring and let $F$, $E$ be $A$-modules. Then $E$ is the direct limit of its finitely generated submodules. Serge Lang, in his Algebra, p. 604, remarks that a technique to test whether an element of $F \otimes E$ is zero, is to examine its image in $F \otimes N$, as $N$ varies over the finitely generated submodules of $E$. Could somebody please explain why this is a valid technique? What is the image of an element of $F \otimes E$ in $F \otimes N$?
100945
I'm studying recently faithfully flat modules and I'd like to know the following:      Is $R/N$ faithfully flat as $R$-module, where $R$ is a commutative ring with unit and $N$ is the ideal of nilpotent elements of $R$?
46851
The following is exercise number 10 in Hungerford's Algebra, page 190:  Let $R$ be a ring with no zero divisors such that for all $r, s \in R$ there exist $a, b \in R$, not both zero, such that $ar + bs = 0$.  a) If $R = K \oplus L$ (module direct sum) then $K = 0$ or $L = 0$  b) If $R$ has an identity, then $R$ has the invariant dimension property.  I have proved part a), but I'm having difficulty with b). I realize that it suffices to show that $R^m \cong R^n$ implies $m = n$, but I can't seem to use part a) to conclude this.
26902
Let $M$ be an $R$-module and let $I$ be an ideal of $R$ such that $I$ is a subset of $\mathrm{Ann}(M)$. Define a product of an element of $R/I$ by an element of $M$ as follows: $$(r+I)\cdot m=rm.$$ (a) Show that this product is well defined. (b) Show that this product, together with the sum of elements of $M$, turns $M$ into an $R/I$-module.  My answer for (a): Let $r+I,s+I \in R/I$ such that $r+I=s+I$. Then $(r+I)m=(s+I)m$, for all $m \in M$. Then $rm=sm \implies r=s$  (since the inverse of $m$ is in $M$), so the product is well defined.  Is my answer to (a) correct?  How can I use Ann(M) to answer (b)?
50780
Let $R$ be a commutative ring and $M$ and $N$ be two finitely generated $R$-modules. I wanna construct an example s.t. $\operatorname{Hom}_R(M,N)$ is not finitely generated.   It's well-known that if $R$ be a Noetherian ring and $M$ and $N$ be two finitely generated $R$-modules then $\operatorname{Hom}_R(M,N)$ is finitely generated. So in this example $R$ must be a Non-Noetherian ring.
117065
A proposition from Fulton's book 'Algebraic Curves': Let $I$ be an ideal in $k[X_1, ..., X_n]$ ($k$ algebraically closed), and suppose $V(I) = \{ P_1, \ldots, P_N \}$ is finite. Let $\mathcal{O}_i = \mathcal{O}_{P_i}(\mathbb{A}^n)$. Then there is a natural isomorphism of $k[X_1, \ldots, X_n] \mathbin{/} I$ with $\prod_{i = 1}^{N} \mathcal{O}_i \mathbin{/} I\mathcal{O}_i$. But if $k$ is not algebraically closed, is it still true that, if $V(I)$ is finite, $k[X_1, \ldots, X_n] \mathbin{/} I$ is a direct product of local rings? And if so, then some factors are most likely given by localizations at the points of $V(I)$, whereas other factors are not; how to describe these latter factors? The reason for this question is the observation that field extensions of $k$ are fields and thus local rings. For instance, $\mathbb{Q}[X] \mathbin{/} (X^2 + 1)$ is a field, but $V(I)$ is empty and $\mathbb{Q}$ is not algebraically closed. Hence, by the Chinese remainder theorem, it seems to me that, for any $f$ in $k[X]$, $k[X] \mathbin{/} (f)$ is the direct product of local rings anyway. Is it at least that correct?
98611
Let $R$ be a commutative ring with unity and $M$ an $R$-module.   If $M\oplus R^m\cong R^n$ for some integers $m,n \geq 1$ then must $M$ be finitely generated and free?   Can somebody help me with this? Regards.
70983
How to go with proving this : Let $R$ be a ring and let $I$ be an Ideal of $R$.  Prove that the factor ring $R/I$ is commutative iff $rs-sr \in I$ $\forall r$ and $s$ in $R$  any hint how to go with this...