Please help me with this problem . The problem was used in one of the questions in my examination and I failed to understand.  Problem as follows:     If $A$ is a commutative ring in which $m_1,\ldots,m_n$ are maximal ideals in $A$.    Then if $m_1 m_2...m_i/m_1 m_2....m_{i+1}$ can be regarded as $A/m_{i+1}$ module , how can we treat $m_1 m_2...m_i/m_1 m_2....m_{i+1}$  as a vector space over $A/m_{i+1}$?   Waiting for some help..
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)?
14965
Let $R$ be commutative ring with identity that contains a field $K$ as a subring. If$ $R is a finite dimensional vector space over the field $K$, prove that every prime ideal in $R$ will be maximal.   My idea was to prove the integral domain $R/p$ (if $p$ is a prime ideal) was a field as for any ideal $P$, $R/P$ will be a field iff $P$ is maximal in $R$. But how can I use the fact that $R$ is finite dimensional over $K$? I don't understand.
73365
Let $A$ be a commutative Noetherian ring and let $I$ and $J$ be ideals of $A.$  Suppose that $I\subseteq J$ and that $A/I \cong A/J$ as rings.  I want to prove that $I=J.$    Observations so far:  1) If we drop the hypothesis $I\subseteq J,$ then the result is false:  $$\mathbb{Q}[X,Y]/(X)\cong \mathbb{Q}[X,Y]/(Y) \text{ but } (X)\neq (Y).$$  2) If $A/I\cong A/J$ as $A$-modules, then $I=J$:  $$I=\mathrm{Ann}_A(A/I)=\mathrm{Ann}_A(A/J)=J.$$  I'm now a bit stuck as to how to proceed. Any hints would be most helpful!  Many thanks :)
20206
Let $A$ be a commutative ring with 1, set $V(\frak a)=$ the set of prime ideals of $A$ that contains $\mathfrak a$, and write $\operatorname{Supp}(M)$ for the support of the $A$-module $M$. We always have $\operatorname{Supp}(M)\subseteq V(\operatorname{Ann} M)$; if $M$ is finitely generated then the other inclusion also holds. Assume now $M$ is finitely generated. It is ok to show that if $M_{\frak p}=0$ then $\operatorname{Ann} M\cap A\setminus\frak p\ne\varnothing$, I was just wondering if the following procedure for showing this inclusion is flawed:  Suppose $\frak p$ contains $\operatorname{Ann} M$. Since $M$ is finitely generated we have the equality $(\operatorname{Ann} M)_{\frak p}=\operatorname{ Ann} (M_p)$, hence the annihilator of $M_{\frak p}$ is contained in maximal ideal of $A_{\frak p}$ and is therefore different from (1), so $M_{\frak p}\ne0$.
81108
Given a Noetherian ring $R$ and a proper ideal $I$ of it.  Is it true that $$\bigcap_{n\ge 1} I^n=0$$ as $n$ varies over all natural numbers? If not, is it true if $I$ is a maximal ideal? If not, is it true if $I$ is the maximal ideal of a local ring $R$? If not, is it true under additional assumptions on $R$ (like $R$ is regular)?
98966
Let $A$ be a commutative ring, $I$ an ideal of $A$ and $M$ an module over $A$. Is it true that $\operatorname{Ann}(M/IM) = \operatorname{Ann}(M) + I$?    One inclusion is certainly true, but I don't know about the other one. If the above statement does not hold in general, does it perhaps for finitely generated modules?
106259
Dummit and Foote, 7.4.33(a): Let $R$ be the ring of all continuous functions $[0,1] \to \mathbb{R}$ and let $M_c$ be the kernel of evaluation at $c \in [0,1]$, i.e. all $f$ such that $f(c) = 0$. Show that if $M$ is a maximal ideal in $R$ then $M = M_c$ for some $c$.    I am aware of proofs that use the compactness of $[0,1]$; however I have been told that a simpler proof exists using only the isomorphism theorems for rings along with basic facts about ideals. I have tried on my own to prove it this way but still have no success; is it possible?  (My attempts go something like: since $M$ is maximal, the quotient ring $R/M$ is a field. Then it would be nice to use the fact that $R/M$ has no zero divisors, but I can't quite make it work. Alternatively, consider $M \cap M_c$.)
27785
So I basically have to prove what is on the title.      Given $R$ a commutative rng (a ring that might not contain a $1$), with the property that $I+J=R$, (where $I$ and $J$ are ideals) we have to prove that $IJ=I\cap J$.   One inclusion is easy. If $x\in IJ$, then $x=\sum a_ib_i$ where $a_i\in I$ and $b_i\in J$. Thus for any fixed $i$, we have that since $a_i\in I$, we have that $a_ib_i\in I$, and the same argument shows that $a_ib_i\in J$, thus $\sum a_ib_i\in I$ and $\sum a_ib_i\in J$, this means that $x=\sum a_ib_i\in I\cap J$, and thus $IJ\subset I\cap J$.   I am having troubles proving the other inclusion. Any comments?  Thanks
61393
I have a question which should be pretty basic commutative algebra, but I can't find a reference and I'm stuck on proving a result myself, so here it goes:  Let $\varphi: S \to R$ be a morphism of commutative rings and let $M$ be a finitely generated $S$-module. Then, we can "extend the scalars of $M$" by considering $M \otimes_S R$, which is naturally an $R$-module. I would like to get a description of $\text{Supp}(M \otimes_S R)$ (as an $R$-module) involving $\text{Supp}(M)$ and $\varphi$.  Maybe this helps: by Eisenbud's Commutative algebra with a view toward algebraic geometry, Corollary 2.7, the problem can probably be solved by finding a description of $\text{Ann}(M \otimes_S R)$ and it is not difficult to see that $\varphi(\text{Ann}(M))R \subset \text{Ann}(M \otimes_S R)$, but I fail to prove the other inclusion.
97142
Please give me a hint for this problem.  Let $R$ be a ring, and $I$ a right ideal of $R$. Show if $M$ is a left $R$-module, then $$ f  \   \colon (R/I) \otimes_{R} M \to  M/IM$$  defined by $ f((r+I) \otimes m) = rm + IM$ is an isomorphism of commutative groups.   And show that if $L$ is a left ideal, then  $$ (R/I) \otimes _{R} (R/L) \cong R/(I+L)$$ as commutative groups.
119369
Let $R$ be a commutative ring and let $I$ be a prime ideal of $R$. Prove that $S = R \setminus I$ is a multiplicative closed subset containing $1$.  Please can anyone help me out here?
13698
I am given that $R$ is a commutative ring, $A$ is an ideal of $R$, and $N(A)=\{x\in R\,|\,x^n\in A$ for some $n\}$.  I am studying with a group for our comprehensive exam and this problem has us stuck for two reasons.  FIRST - We decided to assume $n\in\mathbb{Z}^+$ even though this restriction was not given. We decided $n\ne 0$ because then $x^0=1$ and we are not guaranteed unity. We also decided $n\notin\mathbb{Z}^-$ because $x^{-1}$ has no meaning if there are no multiplicative inverses. Is this a valid argument?  SECOND - We want to assume $x,\,y\in N(A)$ which means $x^m,\,y^n\in A$ and use the binomial theorem to expand $(x-y)^n$ which we have already proved is valid in a commutative ring and show that each term is in A so $x-y$ is in $N(A)$. The biggest problem is how to approach the $-y$ if we are not guaranteed unity. Does anyone have any suggestions?  Thank you in advance for any insight.
76779
Let $R$ be a ring such that for all $a,b$ in $R$, $(a^2-a)b=b(a^2-a)$. Then $R$ is commutative.   I am having difficulties making the connection. Any help would be appreciated.
54581
Let $R$ be a commutative ring and $M$ a finitely generated $R$-module. Let $s$ be the maximum number of linearly independent elements of $M$, while $t$ is the minimum number of a system of generators of $M$. Show that $s\leq t$.  It is immediate in the case of vector spaces, but my first attempt (explicit computation) didn't work here due to the fact that there are coefficients (elements of $R$) without an inverse. I tried in so much other ways spending a lot of time, and I should be very glad if you cold give me a precise, complete and rigorous proof. Thank you in advance.
28247
I was having some trouble with the following exercise from Atiyah-Macdonald.     Let $A$ be a subring of $B$ such that $B$ is integral over $A$. Let $\mathfrak{n}$ be a maximal ideal of $B$ and let $\mathfrak{m}=\mathfrak{n} \cap A$ be the corresponding maximal ideal of $A$. Is $B_{\mathfrak{n}}$ integral over $A_{\mathfrak{m}}$?   The book gives a hint which serves as a counter-example. Consider the subring $k[x^{2}-1]$ of $k[x]$ where $k$ is a field, and let $\mathfrak{n}=(x-1)$. I am trying to show that $1/(x+1)$ could not be integral over $k[x^{2}-1]_{\mathfrak{n}^{c}}$.  I have understood why this situation serves as a counterexample. But I am essentially stuck at trying to draw a contradiction. A hint or any help would be great.
40012
Say I have a module $M$ over a ring $R$ that is   $1.$ finitely-generated, i.e. $\exists$ finite $X \subset M$ such that $M = \sum_{i=1}^{n} Rx_{i}, x_{i}\in X$ (but $X$ is not necessarily free)  $2.$ free, i.e. $\exists$ a basis $Y \subset M$ a.k.a. $Y$ is free and generates M, (but $Y$ is not necessarily finite).  Now I want to show, that there exists a finite basis for $M$. For vector spaces, this would obviously be trivial, but for some  $R$-Modul, I'm not sure how to proceed.   Furthermore, is every minimal generating subset of $M$ a basis?
110838
Let A be a commutative ring with unity ring, M be a maximal ideal and $b\notin M$ . Show that there exists $q \in M, s\in A$ such that for every $a \in A$ we have $a=sba+qa$. Show that this is true only when $M$ is a maximal ideal.   Suppose if we take $A=2\mathbb Z$ and $M=4\mathbb Z$, $a=6.$ Then by the given statement $\exists$ $s\in A$, $q\in M$ s.t. $6=6(2s+q)$ $\implies$$s=\frac{1-q}{2}\implies q \text { is odd}$. But $q \in 4\mathbb Z.$  Please someone give some hints. how can solve this.  Thank you..