The problem is the following.      Let $R,S$ be rings such that $R\neq 0 $, $R\subset S$ and $S$ is finitely generated as an $R$-module, with $t$ generators. Let $\mathfrak{m}$ be a maximal ideal in $R$. Show that there are at most $t$ maximal ideals in $S$ lying over $\mathfrak{m}$.   My attempt to solve  First of all, we have that $S$ is integral over $R$. My idea was to suppose that there are $\mathfrak{m}_1,\ldots,\mathfrak{m}_{t+1}$ distinct maximal ideals in $S$ lying over $\mathfrak{m}$, and get a contradiction from that. It's possible to show that, for each $i=1\ldots t+1$, there is $a_i\in\mathfrak{m}_i$ such that $a_i\notin\mathfrak{m}_j$ for all $j\neq i$. We have that $a_1\cdot\ldots\cdot a_{t+1}\in\mathfrak{m}$ and $a_1\cdot\ldots\cdot a_{t+1}\in\mathfrak{m}_1\cap\ldots\cap\mathfrak{m}_{t+1}$. Also, by the chinese remainder theorem, we have $$\frac{S}{\mathfrak{m}_1\cap\ldots\cap\mathfrak{m}_{t+1}}\cong \frac{S}{\mathfrak{m}_1}\oplus \ldots\oplus \frac{S}{\mathfrak{m}_{t+1}}.$$  With this observations I tried to get some contradiction, like $S$ is not integral over $R$, or the Going Up property fails, or $S$ has more than $t$ generators and so on. Nothing is working, so I don't know what to do. I really appreciate a help here.   Thanks.
121070
I want to understand why the following is true:     Let $S \subseteq R$ be commutative rings with $1$ and assume that $R$ is finitely generated as an $S$-module by at most $k$ elements. For every maximal ideal $M$ of $S$ there are at most $k$ maximal ideals of $R$ lying over $M$.   So let $T$ be a maximal ideal of $R$ lying over $M$ then $T \cap S = M$. Now it can be shown that $R/T$ is a finitely generated $S/M$-module. Hence $R/T$ is Artinian and thus it has finitely many maximal ideals. From here I have two questions:  1) Why from this follows that there can be only finitely many $T$ lying over $M$? I suspect they are using some correspondence which I don't see.  2) Why exactly at most $k$ ?  Can you please explain?
9858
Well, the problem I'm trying to solve is this:      Let $A$ be the ring of all continuous functions from $\mathbb{R} \to \mathbb{R}$. Show that $M = \{f \in A: f(0)=0\}$ is a maximal ideal of $A$.   I tried to show that if $J$ is an ideal of A that properly contains M, so $J = A$: we know that there is a function $g \in J$ with $g(0) \neq 0$. So, we have that $$J \supseteq \langle g, M\rangle=\{ag+f;a \in A, f\in M\}$$   I tried to show that 1 $\in$ $\langle g,M\rangle$. However I'd have to find some function $h(x)\in A$ such that $h(x)(ag+f) = 1$, but this function $h$ has to be different from 0 for all x $\in \mathbb{R}$, so I don't know how to find this function.  My second idea to solve this problem was prove that the quocient $A/M$ is a field. We have that $$A/M = \{h(x) + M; h(x) \in A\},$$ but again I have the problem to show that all elements of $A/M$ are units.  If someone could help, I would be really grateful (:
5528
This result appears to be ubiquitous as an algebra exercise. How do you prove this result?     Let $A$ be an integral domain with field of fractions $K$, and let $A_{\mathfrak{m}}$ denote the localisation of $A$ at a maximal ideal $\mathfrak{m}$ considered as a subring of $K$. Prove that $$A = \bigcap_{\mathfrak{m}} A_{\mathfrak{m}}\,,$$ where the intersection is taken over all maximal ideals $\mathfrak{m}$ of $A$.
40203
Let $I$ be an ideal of $R$ and let $M$ be an $R$-module. Prove that the formula $(r + I) \cdot m = r \cdot m$ for $r + I \in R/I$, and $m \in M$ makes $M$ into an $R/I$-module if and only if $i \cdot m = 0$ for all $i \in I, m \in M$.   I don't quite understand why we must have $i \cdot m = 0$ instead of $I \cdot m \in M$. This is for proving the only if part.
26118
I'm doing this exercise.     Let $A$ be an integral domain, then prove that $$A = \bigcap_{\mathfrak{p} \in \text{Spec(A)}} A_{\mathfrak{p}} = \bigcap_{\mathfrak{m} \in \text{MaxSpec(A)}} A_{\mathfrak{m}}$$ where the intersection is taken in the quotient field of $A$ and $A_{\mathfrak{p}}$ is the localization of $A$ at the prime ideal $\mathfrak{p}$, similarly for $A_{\mathfrak{m}}$ .   What I have done: we have $$A \subseteq \bigcap_{\mathfrak{p} \in \text{Spec(A)}} A_{\mathfrak{p}} \subseteq \bigcap_{\mathfrak{m} \in \text{MaxSpec(A)}} A_{\mathfrak{m}}$$ so it suffices to prove that $$\bigcap_{\mathfrak{m} \in \text{MaxSpec(A)}} A_{\mathfrak{m}} \subseteq A$$  Suppose $\frac{a}{b} \in \bigcap_{\mathfrak{m} \in \text{MaxSpec(A)}} A_{\mathfrak{m}}$ . I want to show that $b$ is invertible.  If $b$ is not invertible then there exists $ \mathfrak{m} \in \text{MaxSpec(A)} $ such that $b \in \mathfrak{m}$. By hypothesis $$\frac{a}{b} = \frac{r}{s}$$ with $s \not\in \mathfrak{m} $.   $A$ is a domain so $sa = rb \in \mathfrak{m} $ and thus $ a \in \mathfrak{m} $. Then I don't know how to proceed , any hint ?
80299
I want to prove the following result:      Let $R$ be a ring and $M$ a maximal ideal in $R$. If $P$ is a prime ideal in $R[x]$ that (strictly) contains $M[x]$, then $P$ is a maximal ideal in $R[x]$.    I have an idea how to prove that, but I'm not quite sure if the argument is totally valid; maybe someone has a cleaner proof. My argument is like that:   $R[x]/M[x]$ is isomorphic to $(R/M)[x]$, that is a PID (since $R/M$ is a field). So, any prime ideal in this ring is maximal; If $P$ contains $M[x]$, then $P$ corresponds to a (prime??) ideal in $R[x]/M[x]$ (this needs more clarification); The prime ideal (that corresponds to $P$) in $R[x]/M[x]$ is then maximal, and this guarantees that $P$ is maximal in $R[x]$.   I'll appreciate any comment.
78025
Let $R$ be a commutative ring with identity. I've been trying to prove the following:     If $S \subset R$ is a maximal multiplicative set, then $R \setminus S$ is a prime ideal of $R$.   I have spent some time in it, but nothing is coming to my mind. Any idea/solution will 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.
93467
This is a problem in Atiyah Macdonald, Commutative Algebra. Problem 5.22: $S$ is a subring of an integral domain $R$. $R$ is a finitely generated $S$ algebra. If the Jacobson radical of $S$ is 0, then the Jacobson radical of $R$ is 0.  I followed the hint provided in the book as follows.   The purpose is to show for each element of $R$, there is a maximal ideal of $R$ avoiding it. Pick a $0\neq v\in R$. Since we have injective maps $S\to R$ and $R\to R_v$, we embed $S$ in $R_v$ as a subring and this $R_v$ is finitely generated as well over $S$.   It is clear that there is $0\neq s\in A$ such that we can extend the ring homomorphism $\phi:S\to\Omega$ to $\tilde{\phi}:R_v\to\Omega$ where $\Omega$ is some algebraically closed field. Here $\Omega$ is chosen to be the algebraically closed field of $S/m$ where $s\not\in m$ as jacobson radical of $S$ is 0. It is clear that $v$ under the map $R_v\to\Omega$ is not zero or we will have $1$ being sent to $0$.   However, I am stuck at showing $Ker(R_v\to\Omega)\subset R_v$ is maximal ideal since this will lead to $Ker(R_v\to\Omega)\cap B$ is maximal. It is not even clear that this will be maximal. However, I can show if $S$'s Jacobson radical is trivial, then $R=S[x]$'s is trivial. It is possible that $R=S[x]/Q$ where $Q$ is some prime ideal and Jacobson radical of $R$ is the intersection of maximal ideals containing $Q$. I also noticed that quotient and localization do not necessarily commute with infinite intersection. What other options do I have here?
1531
If $R$ is a local ring and $M$ and $N$ are finitely generated $R$-modules such that $M\otimes N=0$ then how does it follow from Nakayama's lemma that either $M=0$ or $N=0$?    This is an exercise in Atiyah and Macdonald. The part I could not show in the hints is $({M{\otimes}_R N)}_{k}=0$ implies $M_{k}{\otimes }_{k} N_{k}=0$, where $k=R/\mathfrak m$ and $\mathfrak m$ is the maximal ideal of $R$.
116297
I'm working on the following problem...     Suppose that $A$ is a Dedekind domain with fraction field $K$.  $L/K$ is a finite separable extension of $A$ of degree $n$, and $B$ is the integral closure of $A$ in $L$ (so that $B$ is also a Dedekind domain).  Suppose that $\mathfrak{p}$ is a nonzero prime ideal of $A$, and $B\mathfrak{p}=\mathfrak{q}^{e_1}\cdots \mathfrak{q}_r^{e_r}$ is the prime factorization of $B\mathfrak{p}$.  Show that $$e_1f_1+\cdots +e_rf_r=n$$ where $f_i=[B/\mathfrak{q}_i:A/\mathfrak{p}]$.    My first question is, why does integrality over $A$ imply that $B$ is also a Dedekind domain? My second question is, well, how to do the problem.  I need an outline so that I know what I'm looking for. What results will be needed here?
105792
Let $\mathfrak a,\mathfrak b$ be ideals in a ring $A.$ The quotient of $\mathfrak a$ and $\mathfrak b$ is $(\mathfrak a:\mathfrak b)=\{x\in A:x\mathfrak b\subseteq \mathfrak a\}$ and if $f:B\to A$ is a ring homomorphism, then the contraction of $\mathfrak b$ is $\mathfrak b^c=f^{-1}(\mathfrak b).$     In one of the exercises in the first chapter of Atiyah-Macdonald (commutative algebra), one is asked to show that $(\mathfrak a:\mathfrak b)^c\subseteq(\mathfrak a^c:\mathfrak b^c).$ I did this perfectly fine, but one of the remarks after the exercises implies to me that this should be an equality, which doesn't seem right.   Explicitly, they remark "Thus extension is closed under the operations of ideal addition and multiplication, while contraction is closed under [intersections, radicals, and quotients]." Indeed, part of the exercise was to show that the extension of the sum is the sum of the extensions, etc. This is the only one (of those mentioned in the remark) in which equality was not shown.  Is the remark a mistake, or is there a typo in the exercise and I'm wrong?
76592
I want to prove that if we have a commutative integral domain $D$ with field of fractions $F\neq D$ then $F$ is not finitely generated as a $D$-module. (In this question it may be the case that $1\not\in D$.)    My original plan was to localize at a maximal ideal and then use Nakayama's lemma but as we don't have a $1$ our localization may not be local (in fact we need not have maximal ideals).  So I'm a bit lost as to where to go?  Thanks for any help.
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.
68593
If $I$ is a maximal ideal of $R$, why is $R/I$ a field?  I'm trying to use the fact that $I$ is maximal to show that $R/I$ only have ideals $\{0\}$ and $R/I$. Can anyone help me with this method. Many thanks.
47042
In commutative rings we have the following  Theorem. $R$ is Noetherian if and only if each prime ideal of $R$ is finitely generated.  From this Theorem I am looking for commutative rings $R$ in which every maximal ideal is finitely generated but $R$ is non Noetherian.  Question: Is there a straightforward example of a commutative ring $R$ so that each maximal ideal is finitely generated, but $R$ is non Noetherian?  Thank You
68811
Let $R$ be any ring. The total quotient ring of $R$, denoted $K(R)$, is $R$ localized at the set of non zero-divisors. I am trying to solve Exercise 3.15 in Eisenbud's Commutative Algebra:  a) Let $R$ be a Noetherian reduced ring and let $U$ be a multiplicative subset. Then $K(R[U^{-1}])=K(R)[U^{-1}]$. b) If $R$ is a Noetherian ring, then $K(R[U^{-1}])=K(K(R)[U^{-1}])$.  For a), I was able to show that all prime ideals in $K(R)$ are maximal. Further, there are finitely many of them. Let these ideals be denoted $P_1,\ldots ,P_k$. By the Chinese Remainder Theorem, we have that $$K(R)=\prod_{i=1}^k K(R/P_i)$$ How should I proceed? Also, can I have a hint to start part b)? Should I use the universal property of the localization somehow?