User andrew stout - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T09:59:01Z http://mathoverflow.net/feeds/user/5031 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/124149/what-are-the-enforceable-models-of-local-artinian-rings What are the enforceable models of local artinian rings? Andrew Stout 2013-03-10T12:42:53Z 2013-03-11T13:59:40Z <p>I was reading Hodges' "Model Theory" Chapter 8 a propos existentially closed models of $\forall_2$ theories in a countable first order language $L$. He extends the proof of the omitting type theorem to construct an analogue of set-theoretic forcing in Model Theory. This was very interesting to me as it was the first time I had heard this.</p> <p>I am extremely interested in the $\forall_2$ theory $Art_{k}^{l}$ of local artinian rings of length at most $l$ containing the field $k$ in the language of rings. To get an application, we must insist on the language being countable, which will amount to requiring that the field $k$ is countable.</p> <p>My question is the following: what are the <em>enforceable models</em> the theory $Art_{k}^{l}$. This means what models $M$ admit the <em>enforceable property</em> of "$M$ is isomorphic to the <em>compiled structure</em>" ?</p> <p>In particular, can we use this to say something about $Art_{\mathbb{Q}}^{l}$ and/or $Art_{\mathbb{F}_q}^{l}$?</p> <p>Forgive me, if the terminology is a bit wonky. I am asking the question because I am not familiar with forcing in model theory and much less forcing in set theory. Hodges' "Model Theory" is the only reference I have; however, I know that the omitting types theorem is a fundamental result in model theory. I hope someone is familiar with this construction.</p> http://mathoverflow.net/questions/108098/grothendieck-ring-of-varieties-carrying-a-function/123853#123853 Answer by Andrew Stout for Grothendieck ring of "varieties carrying a function" Andrew Stout 2013-03-07T09:19:19Z 2013-03-07T09:25:17Z <p>To answer your question is a bit difficult because anytime you sum elements of something built out of the grothendieck ring of varieties you are doing (or attempting to do) motivic integration. Pairs $\phi = (X, f)$ certainly already exist in the literature (well actually one should consider the tensor product). But, note that you want $f$ to be a function in $n\in \mathbb{Z}$. Thinking of $f$ being algebraic to your base doesn't add anything because that part can be moved over to the 1st factor in the tensor product via the characteristic function on the graph. In fact, I think it is generally accepted that $f$ should be a function <code>$$f : X(K) \to \mathbb{Z}[\mathbb{L}, \mathbb{L}^{-1}, (\mathbb{L}^{i} -1)_{i\in\mathbb{N}}^{-1}]$$</code> where $K$ is a field, and $X(K)$ is definable in the three sorted language of Denef-Pas and where $X$ above lives over $S$ in some suitable sense (factors through some projection $S\times \mathbb{A}^{n}\rightarrow S$).<br> The reason to use the language denef-pas is so that when the summation of a function $\phi(s,n)$ over $n$ exists, then there is in fact a function of the form $I(\phi)$ which can legitimately be called its integral. In other words, the measure exists and generating series will become rational. </p> <p>Perhaps you already know of this article: <a href="http://arxiv.org/abs/math/0410203" rel="nofollow">http://arxiv.org/abs/math/0410203</a> which lays the foundation for both geometric and arithmetic integration and connects it partly to motivic integration over formal schemes. </p> <p>Personally, I think what might be interesting here is trying to develop something non-commutative by taking pairs (X,f) where f is an endomorphism of X. But, to get something non-commutative, you cannot just take an extension of the usual grothendieck ring of varieties -- the multiplicative structure on such pairs needs to correspond with composition of functions. </p> http://mathoverflow.net/questions/123700/crepant-birational-map/123711#123711 Answer by Andrew Stout for Crepant Birational Map Andrew Stout 2013-03-06T06:09:50Z 2013-03-06T06:25:25Z <p>Kontsevich proved in '95 that the hodge polynomial is independent of crepant resolution over a field of characteristic zero. He created the field of motivic integration to prove this fact. Here are the respective hodge polynomials of the two spaces in question where I repress the fact that we are working over the field $k$ just for notational simplicity:</p> <p><code>$$H([\mathbb{P}^n], u,v) = \sum_{i=0}^{n} H([\mathbb{A}^i], u,v) =\sum_{i=0}^{n} (uv)^i$$</code></p> <p>and</p> <p><code>$$H([\prod_{i=1}^{n}\mathbb{P}^{1}], u,v) = \prod_{i=1}^{n} (H([\mathbb{A}^1], u,v)+1) = \prod_{i=1}^{n} (uv+1) = (uv+1)^n = \sum_{i=0}^{n} {n \choose i} (uv)^i$$</code></p> <p>where we denote by $[X]$ the equivalence class of the variety $X$ in the grothendieck ring of varieties. Therefore, there are no crepant birational maps between $\mathbb{P}^n$ and $\prod_{i=1}^{n} \mathbb{P}^1$. </p> <p>Note, actually, here $[X]$ lives in the image of the Grothendieck ring of varieties localized at $[\mathbb{A}^1]$ under the completion homomorphism given by the dimension filtration. But, the hodge polynomial ring homomorphism `'$$H(X,u,v) = \sum_{i,j} (-1)^{i+j} dim(H^i(X, \Omega_{X}^{j}))u^iv^j$$' factors through the ring homomorphism $X \mapsto [X]$. </p> <p>I am only sure that this argument works over a field of characteristic zero for two reasons: 1) there is not currently a resolution of singularities over fields of positive characteristic, and 2) I don't know how to define the hodge polynomial in positive characteristic. </p> <p>Note that this result is sometimes known as Batyrev's Theorem because he originally proved a similar result in '95 involving the beta numbers via p-adic integration and the weil conjectures. </p> http://mathoverflow.net/questions/123597/finite-global-dimension-vs-integral-domain/123599#123599 Answer by Andrew Stout for finite global dimension vs integral Domain Andrew Stout 2013-03-05T07:16:55Z 2013-03-05T07:16:55Z <p>No, but Serre proved that for noetherian local rings having finite global dimension is the same as being regular. </p> <p>So, choose any non-regular local ring which at the same time an integral domain such as the localization of the cuspidal curve at the origin:</p> <p><code>$$k[x,y]_{(x,y)}/(y^2-x^3)k[x,y]_{(x,y)}$$</code></p> http://mathoverflow.net/questions/91138/field-constructions/91143#91143 Answer by Andrew Stout for Field constructions Andrew Stout 2012-03-14T01:44:43Z 2012-03-19T14:47:36Z <p>It should be noted that the characteristic of a field is either prime or zero. If it is zero, then it contains the rational numbers. These are two statements you can probably prove even if algebra isn't your cup of tea. </p> <p>You can study valuation rings of mixed characteristic. A classic example is $\mathbb{Z}_p$ the p-adic integers. This is a ring of characteristic 0 and its fraction field $\mathbb{Q}_p$ is a field of characteristic 0. However, $\mathbb{Z}_p$ has a (unique) maximal ideal generated by $(p)\mathbb{Z}_p$ such that $$\mathbb{F}_p \cong \mathbb{Z}_p/(p)\mathbb{Z}_p$$ which is a finite field of characteristic $p$. </p> <p>There is the famous Ax-Kochen theorem, which is the following </p> <p>$$\Pi_\mathcal{F} \mathbb{Q}_p \cong \Pi_\mathcal{F} \mathbb{F}_p((t))$$</p> <p>where $\mathcal{F}$ is a non-principal ultra-filter on $\mathbb{N}$. This result depends on the <strong>continuum hypothesis.</strong> (J. Ax and S. Kochen, Diophantine problems over local fields I, American Journal of Mathematics,87 (1965), 605–630.)</p> <p>However, there is also an isomorphism </p> <p>$$\Pi_\mathcal{F} \mathbb{Z}_p \cong \Pi_\mathcal{F} \mathbb{F}_p[[t]]$$</p> <p>where $\mathcal{F}$ is a non-principal ultrafilter on $\mathbb{N}$, which does not depend on the Continuum Hypothesis, in "Use of Ultrapoducts in Commutative Algebra" by Hans Schoutens, found here <a href="http://www.springer.com/mathematics/algebra/book/978-3-642-13367-1" rel="nofollow">http://www.springer.com/mathematics/algebra/book/978-3-642-13367-1</a></p> <p><em>Witt Vectors</em> allow you to move from pure characteristic (either zero or positive) to mixed characteristic.</p> <p>Another way of studying different characteristics is through Algebraic Geometry, by thinking of fields of different characteristics as fibers living over primes in $\mathbb{Z}$. </p> <p><strong>Edit:</strong> It occurred to me that an example here might be useful in order to illustrate this point. Given a curve $C$ over a field of characteristic p>0, we know (Winter's theorem) that there exists a discrete valuation ring (for example, $\mathbb{Z}_p$ of mixed characteristic), call it B, and a family of curves over B. Over the generic fiber, we have a GAGA principle, which means that things we can define over $\mathbb{C}$ (e.g., fundamental group), we can define over a field of $char(k)=0$. Now using Winter's theorem we can 'transfer' our definitions from analytic geometry to algebraic geometry over a field of positive characteristic, which gives us new tools when studying number theory. </p> <p>Finally, there is even an "physical" way to interpret moving between fields. I highly recommend this article of A. Connes on the topic: "Characteristic one, entropy and the absolute point," Connes &amp; Consani <a href="http://arxiv.org/abs/0911.3537" rel="nofollow">http://arxiv.org/abs/0911.3537</a></p> http://mathoverflow.net/questions/89152/does-smoothness-descend-along-flat-morphisms/91093#91093 Answer by Andrew Stout for Does smoothness descend along flat morphisms? Andrew Stout 2012-03-13T16:24:30Z 2012-03-14T16:22:22Z <p>Your second question is answered affirmatively by EGA IV_2 Corollary 6.5.2, which references EGA IV_1 $\S0$ 17.3.3(i). Here you only need to assume that $f: X \rightarrow Y$ is a flat morphism of locally Noetherian schemes. </p> <p>In general, smoothness is a stronger condition than regularity. For example, if $k'$ is a non-separable field extension of a field $k$, then the structure morphism $f:\mathbb{P}_{k'}^{1} \rightarrow \mbox{Spec}(k)$ is regular, but it is not smooth. </p> <p>This last point follows because if it were smooth, we should have an exact sequence</p> <p>$$0\rightarrow f^* \Omega_{k'/k} \rightarrow \Omega_{\mathbb{P}_{k'}^1/k} \rightarrow \Omega_{\mathbb{P}_{k'}^{1}/k'}\rightarrow 0$$</p> <p>which is actually</p> <p>$$0 \rightarrow \mathcal{O}_{\mathbb{P}_{k'}^1} \rightarrow \Omega_{\mathbb{P}_{k'}^1/k} \rightarrow \mathcal{O}_{\mathbb{P}_{k'}^1}(-2) \rightarrow 0$$</p> <p>which is absurd as smoothness means that, after taking stalks, the middle module is free of rank 1. </p> <p>If you work in the category of schemes of finite type over a perfect field then they are equivalent (cf., Liu's Algebraic Geometry and Arithmetic Curves, Chapter 4 Corollary 3.33), which answers your first question. </p> http://mathoverflow.net/questions/62218/german-mathematical-terms-like-nullstellensatz/62289#62289 Answer by Andrew Stout for German mathematical terms like "Nullstellensatz" Andrew Stout 2011-04-19T16:00:33Z 2012-03-13T18:46:06Z <p>Einheit = word for unit in algebra. Hence, some use the notation $e\in G$ to denote the element of a group such that $ex = xe = x , \forall x \in G$. Unit is the appropriate translation, yet some algebraist still use the letter $e$ to denote the identity element in a group. </p> http://mathoverflow.net/questions/61935/quasi-compact-maps-in-number-theory Quasi-compact maps in Number Theory Andrew Stout 2011-04-16T17:39:09Z 2011-04-19T09:45:17Z <p>Can someone give me an example of a non-quasi-compact morphism of schemes which arises naturally in the field of Algebraic Number Theory?</p> http://mathoverflow.net/questions/62043/model-theoretic-localization Model Theoretic Localization Andrew Stout 2011-04-17T15:56:23Z 2011-04-18T20:14:49Z <p>This is a re-post on a previous question I asked. My first question was too vague to warrant detailed responses. Really, I have two specific questions to ask. </p> <p>1) Let $\sigma = (A; {0,1}; +, \times)$ be a signature. Form the language $L(\sigma)$ over $\sigma$. Let $T$ be the theory of commutative rings and let $M$ be a model of this theory. We can realize localization in the model $M$ by specifying a class of formulas in our language $$K = {s_x \ \mid \ x \in A - (0)}, \quad \mbox{where}\ s_x = [\exists x, \ x = x]$$ and then for each $x$ defining a formula $s_{x}^{-1} = \exists y, xy = 1$. Adding $s_{x}^{-1}$ to our theory $T$, call it $T_x$ and then taking a model $N$ of $T_x$ with the property that there is a monomorphism $M \rightarrow N$ will realize $N$ as a localization of $M$. My first question is whether or not this is right way, for a logician to think about localization of a commutative ring?</p> <p>2) It seems to me that it should be possible to extend this construction to other languages by specifying an appropriate class $K$ and formula's $s_{x}^{-1}$. In particular, this should work for non-commutative rings. </p> <p>In summary, what can be said about localization in a first order language?</p> <p><strong>Edit</strong> Actually, in 1), I still have a problem. Specifying a monomorphism $M \rightarrow N$ is not accurate because $M$ may not be integral. Actually, I need to specify a map $M \rightarrow N$ by a universal property.</p> http://mathoverflow.net/questions/61937/localization-of-formulas Localization of Formulas Andrew Stout 2011-04-16T17:52:56Z 2011-04-16T22:09:37Z <p>Can someone point me to an article concerning the "inversion" of formulas?</p> http://mathoverflow.net/questions/124404/phd-in-operator-algebras-and-non-commutative-geometry/124410#124410 Comment by Andrew Stout Andrew Stout 2013-03-13T14:57:15Z 2013-03-13T14:57:15Z il est &#224; Vanderbilt http://mathoverflow.net/questions/124381/commutative-algebra-nilpotent Comment by Andrew Stout Andrew Stout 2013-03-13T05:07:50Z 2013-03-13T05:07:50Z binomial expansion. http://mathoverflow.net/questions/124149/what-are-the-enforceable-models-of-local-artinian-rings Comment by Andrew Stout Andrew Stout 2013-03-10T13:33:35Z 2013-03-10T13:33:35Z ah, I should mention the following fact: enforceable models are existentially closed. The existentially closed models of local aritinian rings of length at most $l$ containing the field $k$ are Gorenstein aritinian rings of length $l$ containing the algebraic closure of $k$. This is per H. Schoutens' work on existentially closed models of local aritinian rings. Does requiring the model to be enforceable and not just existential closed, give us even more properties? http://mathoverflow.net/questions/124117/local-gorenstein-artinian-rings-and-k-rational-points-of-a-hilbert-scheme Comment by Andrew Stout Andrew Stout 2013-03-10T09:05:47Z 2013-03-10T09:05:47Z wccanard, I am tired of answering this question. This is what bogged down my last post. I hope you understand. http://mathoverflow.net/questions/124117/local-gorenstein-artinian-rings-and-k-rational-points-of-a-hilbert-scheme Comment by Andrew Stout Andrew Stout 2013-03-10T09:00:30Z 2013-03-10T09:00:30Z Yes, it is dependent on choice of basis! Choose a basis. I mean it in the sense that $Hom_{R}(k,R) \cong k$ is induced functorially from an $f : R \to R$. http://mathoverflow.net/questions/124117/local-gorenstein-artinian-rings-and-k-rational-points-of-a-hilbert-scheme Comment by Andrew Stout Andrew Stout 2013-03-10T08:37:00Z 2013-03-10T08:37:00Z Yes, Angelo, you are right. All I want to say is that it is natural up to ring homomorphisms $R \to R$. http://mathoverflow.net/questions/124117/local-gorenstein-artinian-rings-and-k-rational-points-of-a-hilbert-scheme Comment by Andrew Stout Andrew Stout 2013-03-10T08:23:46Z 2013-03-10T08:23:46Z ...I mean lives in the category of $\mathcal{O}_S(S)$-algebras. http://mathoverflow.net/questions/124117/local-gorenstein-artinian-rings-and-k-rational-points-of-a-hilbert-scheme Comment by Andrew Stout Andrew Stout 2013-03-10T08:15:12Z 2013-03-10T08:15:12Z Let $X, Y$ be $S$-schemes. Then $Hom_{S}(X,Y)$ is a closed subscheme of the Hilbert scheme $Hilb(X\times_S Y/S)$, as such its ring of global sections lives in the category of $\mathcal{O}_S$-algebras. That algebra structure is the same as the one detailed above. How do you know $Hom_R(k,R) \cong k$ is not natural? http://mathoverflow.net/questions/123992/local-gorenstein-rings-of-dimension-zero-and-the-corresponding-action-on-the-res Comment by Andrew Stout Andrew Stout 2013-03-10T01:18:54Z 2013-03-10T01:18:54Z silly to down vote a good question as if I care about reputation points. I am interested in the conjectures themselves. http://mathoverflow.net/questions/123992/local-gorenstein-rings-of-dimension-zero-and-the-corresponding-action-on-the-res Comment by Andrew Stout Andrew Stout 2013-03-10T01:16:05Z 2013-03-10T01:16:05Z yes, i see the confusion, philosophically, there is no way to recover what the morphism corresponding to an element of the representation of $Hom_{k-alg}(R,R)$ unless we keep track of our choice of $k$-basis of $R$. The confusion surrounding addition is really simple to answer $f + g$ is not a ring homomorphism if $f$ and $g$ are ring homomorphisms (i.e., sending $1$ to $1$). A priori, there is no reason to insist that the addition in $Hom_{k-alg}(R,R)$ is given pointwise; in fact, it is only given pointwise in the trival case of $Hom_{k-alg}(k,k) =$ the zero ring. http://mathoverflow.net/questions/123992/local-gorenstein-rings-of-dimension-zero-and-the-corresponding-action-on-the-res Comment by Andrew Stout Andrew Stout 2013-03-10T00:30:50Z 2013-03-10T00:30:50Z Also, fyi, in terms of logic, there is no difference between constants and variables. That distinction is purely a semantic one. http://mathoverflow.net/questions/123992/local-gorenstein-rings-of-dimension-zero-and-the-corresponding-action-on-the-res Comment by Andrew Stout Andrew Stout 2013-03-10T00:13:27Z 2013-03-10T00:13:27Z The case with $Hom_{k-alg}(k,k)$, actually there is no problem here, is that it is the coordinate ring of the empty scheme. http://mathoverflow.net/questions/123992/local-gorenstein-rings-of-dimension-zero-and-the-corresponding-action-on-the-res Comment by Andrew Stout Andrew Stout 2013-03-09T23:46:28Z 2013-03-09T23:46:28Z Regardless, it is the coordinate ring of a closed subscheme of a hilber scheme. and, in this case all of our schemes are finite type over $k$, so there is no need to worry here. http://mathoverflow.net/questions/123992/local-gorenstein-rings-of-dimension-zero-and-the-corresponding-action-on-the-res Comment by Andrew Stout Andrew Stout 2013-03-09T23:07:06Z 2013-03-09T23:07:06Z Btw, I am not frustrated at all. http://mathoverflow.net/questions/123992/local-gorenstein-rings-of-dimension-zero-and-the-corresponding-action-on-the-res Comment by Andrew Stout Andrew Stout 2013-03-09T23:05:04Z 2013-03-09T23:05:04Z I don't think I owe him an apology. I reluctantly showed him how to compute a much harder example. How many ways can I explain that $Hom_{k-alg}(R,R)$ is a $k$-algebra? At this point, I have tried every way possible only to make the idea clear. Perhaps, you should re-evaluate my actions. It would have been easy for me to just say this is a closed subscheme of a hilbert scheme, no?