Elements in a localization - category theoretic approach - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T02:29:03Z http://mathoverflow.net/feeds/question/86923 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86923/elements-in-a-localization-category-theoretic-approach Elements in a localization - category theoretic approach Martin Brandenburg 2012-01-28T21:31:36Z 2012-02-18T03:40:31Z <p>This question is about the elements in a localization $S^{-1} A$ of a commutative ring $A$. Is it possible to derive $\frac{a}{1} = 0 \in S^{-1} A \Rightarrow \exists s \in S : sa = 0$ only using the universal property of $S^{-1} A$? In order to make this question clear enough I will have to digress a little bit.</p> <p><strong>Examples.</strong> a) Let's start with a related example. If $C$ is a category of algebras of some type, then every continuous functor $C \to \mathrm{Set}$ is representable (special case of SAFT). For example for $C=\mathbf{Ab}$ the functor $\mathrm{Hom}(A,-) \times \mathrm{Hom}(B,-)$ is representable for all abelian groups $A,B$, showing the existence of the coproduct $A + B$. I claim that we can find a <em>description of its elements via the universal property</em>. Namely, choose the coproduct injections $i : A \to A + B$ and $j : B \to A + B$. Then $C:=\mathrm{im}(i) + \mathrm{im}(j)$ is an abelian group such that $i,j$ factor through $C$ and still satisfy the universal property - this shows $A = \mathrm{im}(i) + \mathrm{im}(j)$, i.e. every element has the form $i(a) + j(b)$ for $a \in A, b \in B$. I claim that $a,b$ are unique: By the universal property there is some $f : A + B \to A$ with $fi = \mathrm{id}$ and $fj=0$. It follows $f(i(a) + j(b))=a$ and therefore $a$ is unique, similarly $b$.</p> <p>Conclusion: We defined $A + B$ via some universal property (whose existence follows from general category theory) and found the structure of its elements - only by applying the universal property. We didn't need to know any specific constructed model of $A + B$ in advance for this!</p> <p>Of course for abelian groups and coproducts this isn't really exciting. But what about more complicated (algebraic) structures (e.g. groups, rings, affine spaces, lie algebras, etc.) ? There is a general construction of $A + B$ with generators and relations of $A$ and $B$ (see e.g. <a href="http://arxiv.org/abs/0704.2030" rel="nofollow">Durov</a>, 4.16.14/15), but this does <em>not</em> give us a criterion when two given elements of $A + B$ are equal. It gets even more complicated for other universal properties aka representable functors: For example you can write down a group defined by $10$ generators and $27$ explicitly (<a href="http://en.wikipedia.org/wiki/Word_problem_for_groups" rel="nofollow">Wikipedia</a>) with an unsolvable word problem. Let me mention two (of the many) positive results in this direction:</p> <p>b) Let $M,N$ be modules over a commutative ring $k$ and define the tensor product $M \otimes_k N$ as the classifying object of $k$-bilinear maps on $M \times N$. If $E$ is a generating set of $M$, then the universal property implies that every element has the form $\sum_{e \in E} e \otimes n_e$ for some $n_e \in N$ (which vanish for almost all $e$). There <em>is</em> a criterion when this element is zero and it can be proved with the universal property of the tensor product, without using its explicit construction (Pierre Mazet, Caracterisation des Epimorphismes par relations et generateurs, <a href="http://archive.numdam.org/ARCHIVE/SAC/SAC_1967-1968__2_/SAC_1967-1968__2__A2_0/SAC_1967-1968__2__A2_0.pdf" rel="nofollow">online</a>).</p> <p>c) If $H$ is a subgroup of a group $G_1$ as well as of a group $G_2$, then it is clear how to represent elements in the amalgamated sum $G_1 *_H G_2$, which is defined as a pushout. The uniqueness of the representation is shown in Serre's <em>Trees</em>, Section 1.2, by application of the universal property to the symmetric group over all reduced words. There is no need to impose a group operation on the set of reduced words (which would be very tedious) - after this proof you get it for free!</p> <p><strong>Localization.</strong> Let's consider a commutative ring $A$ and a map $i : S \to |A|$ into the underlying set of $A$. Then we can consider the subfunctor of $\mathrm{Hom}(A,-)$ which is given by homomorphisms $g : A \to B$ such that $gi$ factors through $B^*$. By general theorems a representing object exists and is usually denoted the localization $S^{-1} A$ when $i$ is understood. If $S'$ denotes the multiplicative closure of the image of $i$, then $S'^{-1} A = S^{-1} A$ (they satisfy the same universal property), thus we always may assume that $S$ is just a multiplicative closed subset of $A$.</p> <p>Let $\tau : A \to S^{-1} A$ be the universal homomorphism which maps $S$ to units. Then $S^{-1} A = \{\tau(a) \tau(s)^{-1} : a \in A, s \in S\}$ (they satisfy the same universal property). Let es write $\frac{a}{s} = \tau(a) \tau(s)^{-1}$. Thus every element in $S^{-1} A$ is some fraction $\frac{a}{s}$. Now when are two such fractions equal? Of course we know this from the usual construction using equivalence classes of pairs $(a,s)$, but I want to avoid this construction and derive it only with the universal property.</p> <p>It is clear that $s'a = sa'$ implies $\frac{a}{s} = \frac{a'}{s'}$. From this one easily derives more generally that $ts'a = tsa'$ also implies $\frac{a}{s} = \frac{a'}{s'}$; here $a,a' \in A$ and $s,s',t \in S$.</p> <p><strong>Question.</strong> Is there a proof of the converse, i.e. $\frac{a}{s} = \frac{a'}{s'} \Rightarrow \exists t \in S : ts'a = tsa'$, which only uses the universal property of the localization?</p> <p>First observe that $\frac{a}{s} = \frac{a'}{s'}$ iff $\frac{sa'-sa'}{1}=0$. Thus it is enough to show $\frac{a}{1} = 0 \Rightarrow sa=0$ for some $s \in S$, i.e. that the kernel of $\tau : A \to S^{-1} A$ equals $I=\cup_{s \in S} \mathrm{Ann}(s)$. Since $S$ is multiplicative, $I$ is an ideal, and we may replace $A$ by $A/I$, where localization commutes with quotients because of universal properties. Thus we may assume that $I=0$, i.e. that $S$ consists of <em>regular elements</em>. Our task is then to show that $A \to S^{-1} A$ is injective. But a priori we only know by the universal property that the kernel of $A \to S^{-1} A$ equals the intersection of all kernels of homomorphisms $A \to B$ which map $S$ to units. Of course we have to use this and construct some specific $A \to B$, but I hope that we can either avoid some nasty element construction of $B$ or that we can just use a ring which is built up out of $S^{-1} A$, but can be used to show that the kernel is zero - therefore to find a kind of self-referential proof. The examples above suggest that this might be possible after all.</p> <p>Another approach is to use $S^{-1} A = A[\{X_s\}_{s \in S}]/(s X_s - 1)$ (which is clear since both sides satisfy the same universal property). Here the polynomial algebra should be <em>defined</em> by its universal property, as well as the quotient ring. Assuming that we know the structure of its elements, then one can also show that the kernel of $\tau$ is as desired; see the <a href="http://www.ma.utexas.edu/users/voloch/Homework/fractions.pdf" rel="nofollow">note</a> "Rings of fractions the hard way" by José Felipe Voloch. I think this is quite interesting, but it basically just gives <em>another</em> construction of the localization in terms of polynomial algebras and quotients, and then computes the kernel of $\tau$ from this specific construction. So this is not really what I'm after.</p> <p>Moosbrugger suggest to talk about the category of $A$-modules. Here is a basic observation which follows from the universal property: If $R$ is a possibly noncommutative $A$-algebra, then there is at most one algebra homomorphism $S^{-1} A \to R$, and it exists iff all elements of $S$ become invertible in $R$. This follows easily from the universal property of $S^{-1} A$ applied to the center of $R$. Now let $M$ be an $A$-module and apply the above to $R=\mathrm{End}_A(M)$. It follows that the category of $S^{-1} A$-modules is equivalent to the category of $A$-modules, on which the elements of $S$ act as isomorphisms.</p> <p>Therefore a <strong>possible formalization</strong> of my question might be the following: Let $S \subseteq A$ as above and let $B$ be a commutative $A$-algebra such that scalar restriction $\mathrm{Mod}(B) \to \mathrm{Mod}(A)$ is fully faithful and whose image consists of those $A$-modules on which the elements of $S$ act as isomorphisms. Can we then compute the kernel of $A \to B$ (without refering to the explicit construction of the localization $S^{-1} A$, but only using this statement about module categories, which of course yields $B \cong S^{-1} A$ by Morita).</p> <p><strong>Motivation.</strong> I hope that a categorical proof makes the usual construction of the localization (via equivalence classes of pairs) redundant. Note also that it is rather nasty to prove all the details (equivalence relation, well-defined addition, well-defined multiplication, universal property) with the usual construction. Also the definition $(a,s) \sim (a',s') \Leftrightarrow \exists t : ts'a=tsa'$ is not motivated at all there. A categorical proof should show in particular that this is the right choice, and not some random definition which turns out to be correct only after some computation. But of course I don't claim that a categorical construction of the localization is the easiest or best one.</p> <p>Although the universal property is probably the most important aspect of localization (which we get, as already mentioned, by abstract nonsense), the criterion for the equality of elements is essential even for basic properties in commutative algebra, for example in order to prove that every integral domain <em>embeds</em> into its field of quotients.</p> <p>On the other hand, localization could be seen just as a toy example for other, more involved examples, where it is not clear at all how to understand some representing object of some functor, which exists my general nonsense. </p> http://mathoverflow.net/questions/86923/elements-in-a-localization-category-theoretic-approach/86939#86939 Answer by Denis-Charles Cisinski for Elements in a localization - category theoretic approach Denis-Charles Cisinski 2012-01-29T02:04:46Z 2012-02-13T17:03:07Z <p>If you want to understand $S^{-1}A$ for any $S\subset A$, you may write $S$ as a filtered union of its finite subsets $S_i$, and it is clear from the universal properties that $$S^{-1}A=\varinjlim_i \ S_i^{-1}A$$ Therefore it is sufficient to consider the case where $S$ consists of a finite set of elements of $A$. If $f$ denotes the product of all the elements of $S$ in $A$, it is clear that inverting $f$ is the same thing than inverting each element of $S$. Therefore, we may assume that $S$ contains a unique element $f$. It is then obvious (in terms of universal properties) that $S^{-1}A$ is canonically isomorphic to the (filtered) colimit of the diagram indexed by non negative integers $$A\overset{f}{\to}A\overset{f}{\to}A\overset{f}{\to}\cdots\overset{f}{\to}A\overset{f}{\to}A\overset{f}{\to}\cdots$$ (where $f$ stands for `multiplication by $f$'). Puting back all of the above reductions/descriptions together is the categorical way of writing the usual description of $S^{-1}A$ with elements. (As an exercise, you may turn all this into a global construction, i.e. as one filtered colimit of a diagram which is objectwise just $A$, but in which the transition maps are given by multiplication by a finite product of elements of $S$.)</p> <p>The problem is now reduced to the understanding of colimits of diagrams of $A$-modules of the shape $$M_0\overset{s_1}{\to} M_1 \overset{s_2}{\to} M_2\to\cdots M_n\overset{s_{n+1}}{\to} M_{n+1}\to\cdots$$ Using the universal properties, one can see that $\varinjlim_n M_n$ is the cokernel of the map $$1-s:\bigoplus_n M_n\to \bigoplus_n M_n$$ where $s$ sends an element $x$ of $M_n$ to $s_{n+1}(x)$. In particular, we have a canonical epimorphism $$\bigoplus_n M_n\to \varinjlim_n M_n$$ This implies that any element of $\varinjlim_n M_n$ comes from an element of $M_n$ for $n$ big enough. This presentation of $\varinjlim_n M_n$ also shows that if an element $x$ of $M_n$ becomes zero in $\varinjlim_n M_n$, there exists a positive integer $m$ such that $s_{n+m}\ldots s_{n+1}(x)=0$. Applying this to the diagram $$A\overset{f}{\to}A\overset{f}{\to}A\overset{f}{\to}\cdots\overset{f}{\to}A\overset{f}{\to}A\overset{f}{\to}\cdots$$ we see that $A[f^{-1}]$ admits the usual description.</p> <p>N.B. In an abstract context, proving things about localizations consists to use the categorical description above and to use some special properties of filtered colimits (which are often exact, for instance), so that you don't need any description in terms of elements. For instance, the flatness of $S^{-1}A$ comes from the fact that it is a filtered colimit of free $A$-modules (of rank $1$). One can also compute the kernel of the map $\tau:A\to S^{-1}A$. To keep things simple let us do this in the case where $S$ consists of a single element $f$. Then $ker(\tau)$ is the colimit of $A$-modules $$\varinjlim_n \; \; ker(A\overset{f^n}{\to}A)$$ (because filtered colimits are exact in the category of $A$-modules).</p> <p>In conclusion, it seems possible to describe $S^{-1}A$ using categorical arguments for $A$-modules, but I don't see how to obtain such a description using only the theory of rings.</p> http://mathoverflow.net/questions/86923/elements-in-a-localization-category-theoretic-approach/88804#88804 Answer by Manny Reyes for Elements in a localization - category theoretic approach Manny Reyes 2012-02-18T03:40:31Z 2012-02-18T03:40:31Z <p>I've had this question on my mind for some time now. While I appreciate its spirit, I confess that I've had a hard time imagining what a satisfactory answer would be. But I'm going to be brave and propose a solution anyways. Let me know if it misses the mark.</p> <p>I'm interpreting the question as Andreas Blass did in the <a href="http://meta.mathoverflow.net/discussion/1269/reopen-a-question-with-an-answer-which-was-only-accepted-due-to-a-bounty" rel="nofollow">meta-discussion</a>: given $a \in A$ not annihilated by anything in $S$, exhibit a ring $B$ and a homomorphism $A \to B$ that inverts $S$ but doesn't annihilate $a$. In fact, this homomorphism $A \to B$ will be outright injective.</p> <p>This is a module-theoretic approach that you can find in greater detail in <a href="http://books.google.com/books?id=vXh1MtvQr_8C&amp;lpg=PR5&amp;ots=d20jIrvsvM&amp;dq=goodearl%2520warfield%2520noncommutative%2520noetherian%2520rings%2520second%2520edition&amp;pg=PA108#v=onepage&amp;q=goodearl%2520warfield%2520noncommutative%2520noetherian%2520rings%2520second%2520edition&amp;f=false" rel="nofollow">Theorem 6.2 of <em>An introduction to noncommutative Noetherian rings (2nd ed.)</em></a> by Goodearl and Warfield. (That theorem constructs the ring of fractions at a right Ore set of regular elements of a noncommutative ring. Things are a little easier here because $A$ is assumed to be commutative.)</p> <p>As Martin has already pointed out, we may assume that $S$ consists of regular elements. Let $E(A)$ be the injective hull of the module $A_A$ (being a noncommutative algebraist, I can't help but use right modules) and let $M = \{x \in E(A) : xs \in A \textrm{ for some } s \in S\}$, which is a submodule of $E(A)$. Define $B = \mathrm{End}_A(M)$. Scalar multiplication by an element of $A$ is an endomorphism of $M$, and this gives us a ring homomorphism $\phi \colon A \to B$. This morphism is injective because the element $1 \in A \subseteq M$ serves to "separate elements" of $\phi(A)$. It remains to show that $\phi$ inverts $S$. </p> <p>First notice that $E(A)$, hence $M$, is $S$-torsionfree. For let $0 \neq m \in E(A)$. There exists $a \in A$ such that $0 \neq ma \in A$ because $A$ is essential in $E(A)$. Since $A$ is $S$-torsionfree, for any $s \in S$ we have $(ms)a = (ma)s \neq 0$. It follows that $ms \neq 0$.</p> <p>Next notice that any element of $B = \mathrm{End}_A(M)$ is uniquely determined by its value at $1 \in A \subseteq M$. For suppose $f,g \in B$ with $f(1) = g(1)$. Let $m \in M$, and fix $s \in S$ such that $ms \in A$. Then $f(m)s = f(ms) = f(1)\cdot(ms) = g(1)\cdot(ms) = g(ms) = g(m)s$. Since $M$ is $S$-torsionfree, we conclude $f(m) = g(m)$. Hence $f = g$.</p> <p>Finally, let $s \in S$, and let's prove that $\phi(s)$ is invertible in $B$. By the previous paragraph, it suffices to construct $h \in B$ such that $h(s) = 1$. For then $h \circ \phi(s) \colon M \to M$ sends $1 \mapsto s \mapsto 1$, so this map must be equal to $\mathrm{id}_M$.</p> <p>Because $s$ is regular, the ideal $sA$ is a free module, so there is a module homomorphism $h_0 \colon sA \to M \subseteq E(A)$ sending $s \mapsto 1 \in A \subseteq M$. By injectivity of $E(A)$, this extends along the embedding $sA \subseteq A \subseteq E(A)$ to an endomorphism $h' \colon E(A) \to E(A)$ that satisfies $h'(s) = h_0(s) = 1$. To see that $h'$ restricts to an endomorphism of $M$, let $m \in M$ and fix $t \in S$ such that $mt \in A$. Then $ts \in S$ satisfies $$h'(m)(ts) = h'(mt)s = h'(1)(mt)s = h'(s)(mt) = 1 \cdot mt = mt \in A,$$ proving that $h'(m) \in M$. Thus $h = h'|_M \in B$ is the desired morphism.</p> <p>(Goodearl and Warfield's analysis actually proves that $B = S^{-1}A$ and that $B_A \cong M_A$ as modules. This requires only a little more work.)</p>