The standard way of defining the localization of a commutative ring is as follows: given a multiplicatively closed subset $S\subset R$ the localization is defined first by considering the set $R\times S/\sim$ where $$ (r,s) \sim (r',s') \text{ if there exists a } u \in S \text{ such that } u(rs' - r's) =0 $$ the rest is just equipping this set with a ring structure, but my question lies here: why do we want the $u$? Don't we not want nilpotents in our denominator?
53010
Let $A$ be a ring and $S$ a multiplicative closed set. Then the localization of $A$ with respect to $S$ is defined as the set $S^{-1}A$ consisting of equivalence classes of pairs $(a, s)$ where to such pairs $(a,s), (b,t)$ are said to be equivalent if there exists some $u$ in $S$ such that  $$u(at-bs)=0$$ Now, in the  about the localization of a ring, it says that the existence of that $u\in S$ is crucial in order to guarantee the transitive property of the equivalence relation.   I've seen the proof that the equivalence relation defined above is indeed an equivalence relation, but I fail to see how crucial the existence of $u$ is. For example, why doesn't it work if we simply say that two pairs $(a,s),(b,t)$ are equivalent iff $at - bs = 0$? I tried to come up with a counterexample for such case, but failed in the attempt.
64912
Let $R$ be a ring such that   $$ x^2-x \in C(R) \quad \forall x \in R,$$   where $C(R)$ is the center of $R$. Prove that $R$ is commutative.   I cannot find any route to proceed. How can I start? Please give hints instead of full solution.
6434
Possible Duplicate:          Given a commutative ring $R$ and two nilpotent elements $r$, $s$ there exists an $n \in \mathbb{N}$ such that  $$ (r+s)^n = 0.$$  I want to prove that in order to show that the nilpotent elements of a commutative ring are closed under addition (to show that the nilpotent elements form an ideal within a commutative ring). Is that a good way to do it? I'm a little stuck.
2563
I'm currently reading up on Associated Primes and localization. I came across the following theorem. Let $M$ be an $R$ module.  If $M = 0$ then $Ass(M)$ is empty. The converse is true if $R$ is a Noetherian ring. I understood the proof of this result.   Now I'm trying to come up with an example of $M$ being a non-zero $R$ module with $R$ NOT Noetherian but $Ass(M)$ is empty. Is there an easy example to show that the converse fails to hold when $R$ is not Noetherian? Thank you so much for any assistance! :)
111209
Let $A$ be a commutative ring with finitely many minimal prime ideals $\{p_1,\dots,p_n\}$. Let $A_{p_1,\dots,p_n}$ be the localization of $A$ away from the minimal primes, i.e. $S^{-1}A$ where $S = A-\bigcup_{i=1}^np_i$.   Question 1 (Answered): A paper I am reading claims that $A_{p_1,\dots,p_n} \cong \prod_{i=1}^n A_{p_i}$. I have established by hand that the natural map from left to right is injective, but surjectivity seems harder. What's the best way to see this isomorphism?  Also, the paper uses the notation $\text{Frac}(A)$ for $A_{p_1,\dots,p_n}$ (actually, $\text{Frac}(A)$ is defined to be the ring obtained by inverting in $A$ all non-zero-divisors in $A_\text{red} = A/\sqrt{0}$, but it's easy to see that this is the same thing). This seems odd to me, since, as far as I know, the standard meaning of $\text{Frac}(A)$ is the localization of $A$ at all non-zero-divisors in $A$. Note that if $A$ is not reduced, then there could be zero-divisors which are not in any of the minimal primes, and hence are inverted in $A_{p_1,\dots,p_n}$ (example: $y$ in $\mathbb{Q}[x,y]/(x^2,xy)$).   Question 2 (Still open): Is this definition of $\text{Frac}(A)$ common? Is there a good reason for preferring this definition of $\text{Frac}(A)$ over the other in certain situations (when $A$ is non-reduced)?  One more observation: I've noticed that with this definition of $\text{Frac}(A)$, we have that a map of rings $A\to B$ extends to a map $\text{Frac}(A)\to \text{Frac}(B)$ if and only if the inverse image of every minimal prime ideal in $B$ is a minimal prime ideal of $A$. Geometrically, this says that the map $\text{Spec}(B)\to\text{Spec}(A)$ maps generic points of irreducible components to generic points of irreducible components. Maybe this is relevant...
78017
The center of a ring $R$ is $\{c\in R : cr=rc $ for every $ r \in R\}$. Prove that the center of a ring is a subring. What is the center of a commutative ring?  Is my solution right?  solution  You just need to prove that the centre is a ring within itself.    Associativity of both addition and multiplication is inherited from R, and distributivity of multiplication over addition is inherited from R.  Show that it is a group under addition:  Take a and b in the centre and r in R.  Then  $(a + b)r = ar + br = ra + rb = r(a + b)$   hence it is closed under addition.  Show that $0$ (additive identity) is in the centre.  $0 = 0.r = r.0$  so $0$ is in the centre.   For a in the centre, there exists $-a$ in R such that $a + (-a) = 1 = (-a) + a$  Show that $-a$ is in the centre.  Let r in R  $0 = 0.r = (a + (-a))r = ar + (-a)r $ and  $0 = r.0 = r(a + (-a)) = ra + r(-a) $  as $ ra = ar$, it follows that $(-a)r = r(-a) $ Hence inverses exist in the centre.  So the centre is a group under +.    Show that the centre is closed under multiplication:  For a,b in the centre and r in R,#  $(ab)r = a(br) = a(rb) = (ar)b = (ra)b = r(ab) $   thus multiplication is closed.  And show that 1 (the multiplicative identity) is in the centre.  $1.r = r = r.1$ hence 1 is in the centre.   So the center is a subring of R.   The center of a commutative ring is the ring itself. (By definition the centre is the ring of all commutative elements.)
15788
I'm trying to learn the theory of algebraic groups using "modern" algebraic geometry, going past the traditional, very standard books by Borel, Springer, and Humphreys, all of which try to avoid the use of group schemes. One of the motivations for using schemes is that in the classical setting, "nilpotents are not allowed," a sentence I don't understand. For example, on page 18 in Milne's Algebraic Groups (which is still in draft form), he writes, "Since we allow nilpotents in the structure sheaf, the points of an algebraic group with coordinates in a field, even algebraically closed, do not convey much information about the group. Thus, it is natural to consider its points in a $k$-algebra. Once we do that, the points capture all information about the algebraic group."   This is clearly not meant to be totally precise. But I still wonder what he means. I'm thinking of the curves $x^2=0$ and $x=0$ as one example which is sometimes used to motivate the definition of scheme (the behavior at 0 is different in those two examples).   Here are my questions, presumably related: By "nilpotents in the structure sheaf," does he mean nilpotents in the coordinate ring? What does "allowing nilpotents" mean, and what is wrong with not allowing nilpotents? Why doesn't allowing nilpotents force the points to "convey much information" about the group? What information does thinking of algebraic groups as functors capture?  I should add that I'm mostly used to thinking of algebraic groups as functors. For example, $SL_n$ is thought of as a functor that takes a ring $R$ to the matrix group $SL_n(R)$.  (I can post this as more than one question if it would be better, but I don't know enough about the answer I'm looking for to know whether or not that is a good idea.)
72862
Refer to localization of rings in Lang's Algebra p. 108. In particular, let $A$ be a commutative ring, $S$ a subset of $A$ that is a submonoid of the multiplicative monoid structure of $A$, containing the identity. Then  $S^{-1}A$ is the set of elements $\frac{a}{s}$ with $a \in A$ and $s \in S$.  We have $\frac{a_1}{s_1}=\frac{a_2}{s_2}$ if there exists $\bar{s} \in S$ such that $\bar{s}(s_2a_1 - s_1 a_2)=0$. Now consider the injection of $A$ into $S^{-1}A$ given by $x \mapsto \frac{x}{1}$. This is a ring homomorphism. Let $J$ be an ideal of $A$. Ring homomorphisms preserve ideals and so the injection of $J$ into $S^{-1}A$, call it $J^{*}$, will be an ideal. My problem is that i don't see why for any $\frac{a}{s} \in S^{-1}A$ and any $\frac{j}{1} \in J^{*}$ we will have that $\frac{aj}{s}$ will be in $J^{*}$, i.e. why there must exist an element $\kappa \in J$ such that $\frac{aj}{s}=\frac{\kappa}{1}$.
12858
Say I have a commutative ring $R$ with a maximal ideal $m$. Then $m/m^k$ is a maximal ideal in $R/m^k$ for any $k$. Is it the only maximal ideal, i.e. is $R/m^k$ a local ring?    This is a well known result for $k = 1$, as $R/m$ is a field. It seems to be true in other cases, e.g. $p\mathbb{Z}\subset \mathbb{Z}$ for prime $p$ and for $(x,y) \subset \mathbb{F}[x,y]$, the maximal ideal generated by $x$ and $y$ in a polynomial ring over the field $\mathbb{F}$.   Equivalently, if an element $x\notin m$, is $x + m^k$ an invertible element in $R/m^k$?
117895
Proposition in abstract algebra. The book gives this as a proposition and asks the reader to do it as a exercise. It seems it should be easy.(Assuming every ring has a 1)  Let R be a subring of S. If S\R is closed under multiplication then R is integrally closed in S  I went by cotradicton let s be in R\S then. We get that $$ s^n+a_1s^{n-1}..sa_{n-1}+a_{n}=0 $$ From  and the a's are in R. Now I am thinking this $$ s^n=(-1)(a_1s^{n-1}..sa_{n-1}+a_{n}) $$ Now multiplying both sides by s again $$ ss^n=(-s)((a_1s^{n-1}..sa_{n-1}+a_{n}) $$ Now by closure $ss^n$ is in S\R which means the rest $(a_1s^{n-1}..sa_{n-1}+a_{n})$ is in S\R which means the a's are in it which is a contradiction Does this look correct?
72037
I am doing some exercises on commutative algebra and came along the following expressions, which were not elaborated on. Is someone familiar with them?  The first is for $p$ a prime number $\mathbb{Z}_{(p)}$, where the exercise is to determine all ideals. Do you think this is just $\mathbb{Z}/p\mathbb{Z}$?  The second one is for $P$ a prime ideal the expression $P_P$, where $P_P$ lies in $R_P$ for $R$ a ring.  I never came across this notation before so I have to ask here.
35405
In a preparation question for an exam, I am asked to give an example of a ring $A$ such that the nilradical $\operatorname{Nil}(A)$ is strictly smaller that the Jacobson radical $J(A)$. Here's how I solved the problem:  It is enough to find some ring $A$ with a prime ideal $p$ which is strictly contained in some maximal ideal $m$. Then the localization $A_m$ is a local ring with maximal ideal $m$, and by the correspondence between ideals of $A_m$ and ideals of $A$ contained in $m$ we have that $J(A)=m\supsetneq p\supset\operatorname{Nil}(A)$.  An example is $A=\mathbb{Z}[x]$, $p=(x^2+1)$ and $m=(x^2+1,2)$.  Now I was wondering: are there also any "elementary" examples satisfying the condition above? By this I mean basically examples where I don't have to localize the ring.
26946
Let $R$ be a commutative Noetherian ring with unit and let $I$ be a fixed ideal. I am sorry if the following turns out to be a very silly question.  1) Suppose $\operatorname{Ass}(R/I)$ are all minimal, then what can we say about $R/I$?  One conclusion is that $I$ has a unique primary decomposition. If $R/I$ is CM, then $\operatorname{Ass}(R/I)$ are all minimal, but I don't think the converse is true. Can someone provide a counterexample?  2) Suppose $\operatorname{Ass}(R/I)$ are all minimal and have the same height. I still dont think we can conclude that $R/I$ is CM. Can someone provide a counterexample? Theorem 2.1.6 of Bruns and Herzog, CM Rings, gives a result in this direction but it needs additional hypothesis. It says that if every ideal $I$ is generated by height $I$ elements and $\operatorname{Ass}(R/I)$ are all minimal, then $R$ is CM.  It would be nice if someone can say something nice about the ring in the special situation described in 1) and 2).  Thanks.
106906
I am studying ring theory and I am having a hard time understanding a part of a proof of a certain theorem. Note that in my lecture notes, a ring is by definition commutative with $1$. Let $R$ be a ring, then the set of units of $R$ is denoted by $R^*$. Consider the following theorem:  Let $R$ be a principal ideal domain (PID) and $\pi\in R$ irreducible. For any $a,b\in R$ with $\pi|ab$ we have $\pi|a$ or $\pi|b$.  The following is the first part of the proof from my lecture notes:  Consider the ideal $(a,\pi)$. Since $R$ is a PID this ideal is generated by a single element, which we call g. Note that $g|\pi$. Hence either $g$ is a unit, or $g=\epsilon\pi$ with $\epsilon\in R^*$. In the second case, since $g|a$ we also get $\pi|a$ and our assertion is proven.  The second part of the proof covers the case in which $g$ is a unit; I understand what's going on there (under the assumption that $g$ is indeed a unit) so I've omitted it. Please note that this is the proof word by word, i.e. I have quoted it from my lecture notes. First of all, is it true that, since $(g)=(a,\pi)$ we have $\{rg\ |\ r\in R\}=\{r_1a+r_2\pi\ |\ r_1,r_2\in R\}$ and by choosing $r_1=0$ and $r_2=1$ we have $rg=\pi$ for some $r\in R$, thus $g|\pi$? If this is true, then by similar reasoning one can proof that $g|a$ (which is stated in the last sentence of the proof). Next, the proof says that $g$ is either a unit or $g=\epsilon\pi$ for some unit $\epsilon$. By definition, we see from the factorisation above that either $g$ or $r$ is a unit. The first case corresponds to the statement in the proof that $g$ can be a unit. But if $g$ is not a unit, why is $g=\pi\epsilon$ for some unit $\epsilon$? Lastly, how does $\pi|a$ follow from all that has been said in the proof? EDIT: As requested by a user in the comments, here is the second part of the proof in my lecture notes, along with the first one for easy reading:  Consider the ideal $(a,\pi)$. Since $R$ is a PID this ideal is generated by a single element, which we call g. Note that $g|\pi$. Hence either $g$ is a unit, or $g=\epsilon\pi$ with $\epsilon\in R^*$. In the second case, since $g|a$ we also get $\pi|a$ and our assertion is proven. Now suppose  that $g$ is a unit and hence $(a,\pi)=R$. In this case there exist $x,y\in R$ such that $ax+\pi y=1$. Multiply by $b$ to get $abx+\pi by=b$. Both terms on the left are divisible by $\pi$. Hence the sum $b$ is also divisible by $\pi$ which proves our theorem.
53008
Let $A$ be a ring (commutative with $1$), let $S$ be a multiplicatively closed subset of $A$, i.e $S$ is contained in $A$ , $1\in S$ and $a,b\in S$ implies $ab\in S$, for every $a,b\in A$. Consider the following relation defined in $A\times S$: $$(a,s)\equiv(b,t) \iff \exists u\in S:u(at-bs)=0$$ I proved this relation to be reflexive and symmetric, but i have difficulties in proving its transitivity. Can someone suggest me a trick?
111911
Let $A$ be a commutative ring, $S$ a multiplicatively closed subset of $A$. For any ideal $\mathfrak a$, let $S(\mathfrak a)$ denote the inverse image of $S^{−1}\mathfrak a$ under the localization map $\varphi: A → S^{−1}A$. We denote $\mathfrak p^{(n)}:= S_{\mathfrak p}(\mathfrak p^n)$ where $S_{\mathfrak p}:= A \setminus\mathfrak p$.      Prove that $\mathfrak p^{(n)} $ is a $\mathfrak p$-primary ideal, and if $\mathfrak p^n$ has a primary decomposition, then $\mathfrak p^{(n)}$ is a $\mathfrak p$-primary component. (Atiyah and Macdonald, Introduction to Commutative Algebra, Chapter 4, Exercise 13 i) and ii).)   My approach: to prove that $\mathfrak p^{(n)}: = \varphi^{-1}(S_{\mathfrak p}^{-1}\mathfrak p^n)$ is primary. Let $xy \in \mathfrak p^{(n)}, x \notin \mathfrak p^{(n)} $. I need to show that there exists an $m\in \mathbb{N}$ such that $y^m \in \mathfrak p^{(n)}$. From $xy \in \mathfrak p^{(n)}, x \notin \mathfrak p^{(n)}$ it follows $\varphi(x)\varphi(y) \in S_{\mathfrak p}^{-1}\mathfrak p^n, \varphi(x) \notin S_{\mathfrak p}^{-1}\mathfrak p^n$. At this step I do not know how to show that $\exists m\in \mathbb{N}$ s.t. $\varphi(y)^m \in S_{\mathfrak p}^{-1}\mathfrak p^n$.  Please give me some hints. Many thanks
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?