User matthew morrow - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T03:20:36Z http://mathoverflow.net/feeds/user/5830 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86388/vanishing-of-nisnevich-cohomology-of-k-theory-over-a-one-dimensional-local-ring Vanishing of Nisnevich cohomology of K-theory over a one-dimensional local ring. Matthew Morrow 2012-01-22T17:36:52Z 2012-01-22T17:36:52Z <p>Let <code>$A$</code> be a one-dimensional, Noetherian, local ring, and let <code>$\mathcal{K}_n$</code> denote the sheafification of <code>$K_n$</code> (degree <code>$n$</code> <code>$K$</code>-theory) on the Nisnevich site of <code>$X:=\mbox{Spec }A$</code>. Then is it true that <code>$H^1(X_{\mbox{Nis}},\mathcal{K}_n))=0$</code>? Since <code>$X_{\mbox{Nis}}$</code> has cohomological dimension $1$, this is the same as asking whether <code>$\mathcal{K}_n$</code> is acyclic on <code>$X_{\mbox{Nis}}$</code>.</p> <p>If $K$-theory satisfies descent on $X$ in the Nisnevich topology (which I think it does if $A$ is regular, but I'd be interested to know if this is true more generally), then the descent spectral sequence <code>$H^p(X_{\mbox{Nis}},\mathcal{K}_q)\Longrightarrow K_{q-p}(X)$</code> gives short exact sequences </p> <p><code>$$0\to H^1(X_{\mbox{Nis}},\mathcal{K}_n)\to K_{n-1}(A)\to H^0(X_{\mbox{Nis}},\mathcal{K_{n-1}})\to 0$$</code></p> <p>and so my question becomes equivalent to the following: is $K_{n-1}(A)\to H^0(X_{\mbox{Nis}},\mathcal{K}_{n-1})$ injective (and hence an isomorphism)?</p> <p>Well, any Nisnevich cover of $X$ must include a point isomorphic to $\mbox{Spec }F$, where $F$ is the field of fractions of $A$ (assume $A$ is a domain for simplicity for a moment), so if $A$ satisfies Gersten's conjecture, i.e. $K$-theory of $A$ embeds into $K$-theory of $F$, then it seems to follow that $K_{n-1}(A)\to H^0(X_{\mbox{Nis}},\mathcal{K}_{n-1})$ is indeed injective. This appears to answer my original question in the affirmative when $A$ is a discrete valuation ring (at least, assuming Gersten for dvrs).</p> <p>When $A$ is non-regular I therefore seem to be asking for some sort of generalisation of Gersten's conjecture, in the Nisnevich topology. I've asked a few experts but without success, so I'm handing the question to MO, since I'd be amazed if this question had not already been considered! Thank you!</p> http://mathoverflow.net/questions/28435/automatically-extract-a-bibitem-not-bibtex-from-mathscinet/28439#28439 Answer by Matthew Morrow for Automatically extract a bibitem (not BibTeX!) from MathSciNet? Matthew Morrow 2010-06-16T22:19:39Z 2010-06-16T22:19:39Z <p>Hi Andreas. When you run BibTeX on a .bib file then it produces a .bbl file which more-or-less contains your bibliography in \bibitem format. So I recommend exporting everything BibTex style from MathSciNet, creating a .bib file, running BibTeX, and then making any remaining changes to the resulting .bbl file (which you can then just copy and paste into your .tex file).</p> <p>I suppose some people even know how to write BibTeX style files, so that the resulting .bbl file contains <em>exactly</em> the formatting you want.</p> <p>Despite my advice, when I need \bibitem bibliographies, I usually just copy citations from the MathSciNet clipboard (in Ascii format) and manually insert the additional formatting. I'm not sure if this is more, or less, lazy than the .bbl business...</p> http://mathoverflow.net/questions/27197/local-complete-intersections-which-are-not-complete-intersections/27210#27210 Answer by Matthew Morrow for Local complete intersections which are not complete intersections Matthew Morrow 2010-06-05T23:28:14Z 2010-06-05T23:58:14Z <p>EDIT: This is wrong. I haven't deleted it in order that the subsequent comments make sense.</p> <p>You will never <strong>see</strong> an example, for the following reason: given a local complete intersection $V$ inside $\mathbb{A}_k^n$, you can always find a global complete intersection $W$ inside $\mathbb{A}_k^n$ such that the reduced varieties associated to $V$ and $W$ are the same.</p> <p>Proof:</p> <p>Suppose $I$ is an ideal of $k[X_1,\dots,X_n]$ such that the variety $V(I)$ is a local complete intersection. This forces all the local rings of $V(I)$ to be Cohen-Macaulay, hence equidimensional. So the irreducible components of $V(I)$ all have the same codimension in $\mathbb{A}_k^n$; lets call this codimension $r$.</p> <p>Since $k[X_1,\dots,X_n]$ is Cohen-Macaulay, the height of $I$ (which is $r$) is the same as its depth, meaning that $I$ contains a regular sequence $f_1,\dots,f_r$ of length $r$. By considering heights we see that the minimal primes over the ideal $J=\langle f_1,\dots,f_r\rangle$ are the same as the minimal primes over $I$. Therefore $J$ and $I$ have the same radical, which implies the claim (with $W=V(J)$). QED</p> <p>So if you are trying to draw counterexamples, you have to worry about whether that line on the paper has nilpotent elements in the structure sheaf...</p> http://mathoverflow.net/questions/27168/injective-objects/27170#27170 Answer by Matthew Morrow for injective objects Matthew Morrow 2010-06-05T16:56:41Z 2010-06-05T16:56:41Z <p>Yes. If $X$ is a Noetherian scheme, then the category of quasi-coherent sheaves on $X$ has enough injectives. See Hartshorne's Algebraic Geometry, exercise III.3.6(a).</p> http://mathoverflow.net/questions/26558/on-semi-local-schemes/26562#26562 Answer by Matthew Morrow for On semi-local schemes Matthew Morrow 2010-05-31T11:23:50Z 2010-05-31T11:23:50Z <p>Your construction number (1) seems completely natural and correct to me, for the following reason:</p> <p>If $A$ is a (Noetherian, commutative, unital) ring and $I$ is an ideal of $A$, then the localaization of $A$ away from $I$ is $A_I=S^{-1}A$, where</p> <p><code>$$S=\{a\in A: a\mod{I}\mbox{ is not a zero divisor is }A/I\}$$</code></p> <p>If $I$ is prime, then this is what you expect.</p> <p>Next, suppose that $I$ is radical and that $\mathfrak{p}_i$ are the finitely many minimal prime ideals containing it (so $I$ is the intersection of the $\mathfrak{p}_i$). Then the zero divisors in the reduced ring $A/I$ are exactly the union of the minimal prime ideals of the ring, implying that $S=\bigcup_i\mathfrak{p}_i$. From this it follows that the localization $A_I$ is a semi-local ring whose maximal ideals correspond to the $\mathfrak{p}_j$. Moreover, $A_I$ will exactly by the direct limit you describe in (1), if you take $X=\mbox{Spec}A$ and <code>$C=\{\mathfrak{p}_i\}$</code>.</p> http://mathoverflow.net/questions/23670/can-the-cross-section-x-section-be-removed-from-the-ax-kochen-proof-in-diophan/23686#23686 Answer by Matthew Morrow for Can the cross-section (x-section) be removed from the Ax-Kochen proof in "Diophantine problems over local fields"? Matthew Morrow 2010-05-06T11:07:07Z 2010-05-11T10:21:23Z <p>I am not a model theorist but have studied this stuff for applications in number theory, and this is what I think is true:</p> <p>First some notation. I like my Henselian fields to be denoted $F$, the value group $\Gamma(F)$, and the residue field $\overline{F}$. Up until [edit: I am wrong about the history here. See Scanlon's answer above] E. Hrushovski and D. Kazhdan's work (Integration in Valued Fields, 2005, arXiv:math/0510133v3), everyone assumed the existence of a cross section $\Gamma(F)\to F^{\times}$. But H. and K. introduced a new object to get around this, namely $$RV(F)=F^{\times}/1+M_F,$$ where $M_F$ is the maximal ideal of the ring of integers $O_F$ of $F$. Note that the valuation $F^{\times}\to\Gamma(F)$ descends to $RV(F)$, where it has kernel $O_F^{\times}/1+M_F=\overline{F}^{\times}$. In other words, there is an exact sequence $$1\to\overline{F}^{\times}\to RV(F)\to\Gamma(F)\to 1$$ and we think about $RV(F)$ as "wrapping together" (to quote their paper) the residue field and value group.</p> <p><em>If</em> you have a cross section then this sequence will split, giving an isomorphism $RV(F)\cong\overline{F}^{\times}\times\Gamma(F)$, but this splitting is <em>not necessary</em> for what follows.</p> <p>I am reasonably certain that it follows from their paper that if $F$, $F'$ are two suitable valued fields for which $RV(F)$ and $RV(F')$ are elementarily equivalent (in a suitable language), then $F$, $F'$ are elementarily equivalent. I'm not sure what happens if you replace 'elementarily equivalent' by 'isomorphic'.</p> <p>I await corrections from the model theorists here, and apologise if I have inadvertently attributed someone else's work to Hrushovski and Kazhdan.</p> <p>p.s. It took me <em>ages</em> to work out that "x-section" means "cross-section"!</p> http://mathoverflow.net/questions/23950/how-does-f-o-x-measure-ramification-and-grothendieck-riemann-roch/23955#23955 Answer by Matthew Morrow for How does f_* O_X measure ramification and Grothendieck-Riemann-Roch Matthew Morrow 2010-05-08T19:48:39Z 2010-05-08T19:48:39Z <p>The following does not exactly answer your question, but you may find it interesting. It is the Riemann-Hurwitz formula for surfaces.</p> <p>Let $\phi:S_1\to S_2$ be a finite morphism between smooth, projective surfaces (over an algebraically closed field of characteristic zero) of degree $n$, and let $B\subseteq S_2$ be the set of $y\in S_2$ such that $\phi^{-1}(y)$ does not contain $n$ points (i.e. $B$ is the ramification locus). Zariski's purity theorem states that $B$ is pure of dimension one; let $B_1,\dots,B_r$ be its irreducible components, and let $n_i$ be the degree of the morphism $\phi|_{\phi^{-1}(B_i)}:\phi^{-1}(B_i)\to B_i$. Then</p> <p>$$\chi(S_1)=\chi(S_2)\deg \phi-\sum_{i=1}^r(n-n_i)\chi(B_i)+\sum_{y\in B}\left(|\phi^{-1}(y)|-n+\sum_{i=1}^r(n-n_i)m_i(y)\right)$$</p> <p>where $m_i(y)$ denotes the number of local branches of $B_i$ at $y$. Here $\chi$ is the $\ell$-adic Euler characteristic of the surface ( topological Euler characteristic if $k=\mathbb{C}$), which can be translated into a Chern class if you prefer.</p> <p>The proof is B. Iversen, 'Numerical invariants and multiple planes', Amer. J. Math. 92 (1970), 968-996. When $k=\mathbb{C}$, you can prove it by thinking of the topological Euler characteristic as a measure on constructible sets (e.g. O. Ya. Viro, Some integral calculus based on Euler characteristic); then the formula is equivalent to Fubini's theorem ($\int\int dxdy=\int\int dydx$) for the graph of $\phi$.</p> http://mathoverflow.net/questions/23940/why-free-topological-groups-on-tychonoff-spaces/23951#23951 Answer by Matthew Morrow for Why free topological groups on Tychonoff spaces? Matthew Morrow 2010-05-08T19:26:08Z 2010-05-08T19:26:08Z <p>Let $X$ be a topological space. If $F(X)$ is $T_0$ then I think $F(X)$ is isomorphic (as topological groups) to $F(Y)$, where $Y$ is the Tychonofficiation (see below) of $X$. So it is enough to study topological free groups on a Tychonoff space.</p> <h3>Explanation:</h3> <p>First let me remind myself about some notation. <em>Completely regular</em> means that any point can be separated from a closed set not containing it by a continuous real-valued function. <em>Tychonoff</em> then means completely regular and $T_2$(=Hausdorff). Any topological group is completely regular; and for topological groups $T_0$ is equivalent to $T_2$.</p> <p>Suppose $X$ is an arbitrary topological space, and let $Y$ be its "Tychonoffication" (!). That is, set theoretically $Y$ is the quotient of $X$ by the equivalence relation $x\sim x'$ if and only $f(x)=f(x')$ for all continuous $f:X\to\mathbb{R}$; each such $f$ descends to $Y$ and we give $Y$ the weak topology induced by all these real-valued maps. This makes $Y$ into a Tychonoff space which I think satisfies the following universal property: any continuous map from $X$ to a Tychonoff space factors uniquely through $Y$. Also, the natural map $X\to Y$ induces an isomorphism $C(Y)\stackrel{\cong}{\to} C(X)$, where $C(-)$ denotes the ring of real-valued continuous functions.</p> <p>Assuming $F(X)$ is $T_0$, then the natural map $F(X)\to F(Y)$ seems to be an isomorphism of topological groups. It is enough to construct an inverse. Since $F(X)$ is $T_0$, it is even $T_2$, and therefore it is Tychonoff. So the natural map $X\to F(X)$ factors through $Y$ and induces $F(Y)\to F(X)$, which surely does the trick?</p> <h3>What about that $T_0$ assumption?</h3> <p>Lots of people are only interested in Hausdorff topological groups, so it seems reasonable to only study spaces $X$ for which $F(X)$ is $T_0$ (hence $T_2$). Otherwise you could replace $Y$ by the "complete-regularization" of $X$ (i.e. $X$ equipped with the weak topology induced by $C(X)$) and repeat the argument, but it doesn't work so nicely.</p> <h3>Edit:</h3> <p>While I was typing my answer, you asked about this Tychonoffication business!</p> http://mathoverflow.net/questions/23891/when-is-the-push-forward-of-the-structure-sheaf-locally-free/23903#23903 Answer by Matthew Morrow for When is the push-forward of the structure sheaf locally free Matthew Morrow 2010-05-07T22:08:51Z 2010-05-07T22:51:47Z <p>Assuming that $f$ is finite (I'm not quite sure if you are), then $f_*\mathcal{O}_X$ is locally free if and only if $f$ is a flat morphism.</p> <p>Like Peter Bruin says, you are asking the following: given a finite ring homomorphism $A\to B$, when is $B$ a locally free $A$-module? But there is a natural characterization of this! A finitely generated $A$-module is locally free if and only if it is flat, because a finitely generated module over a local ring is free if and only if it is flat.</p> <p>Example 3 is a special case of this, because <em>any</em> surjective morphism to a regular one dimensional scheme is automatically flat (because any injection of a dvr into another ring is flat).</p> <p>Edit: Worth noting that if $X,Y$ are regular and $f$ is finite and surjective, then $f$ is flat.</p> http://mathoverflow.net/questions/23573/do-morphisms-locally-decompose-into-finite-surjective-followed-by-smooth-update Do morphisms locally decompose into finite surjective followed by smooth? (update: Is every projective variety over a finite field a finite cover of $\mathbb{P}^d$ for some $d$?) Matthew Morrow 2010-05-05T11:45:07Z 2010-05-06T14:43:21Z <p>I have looked through all my standard algebraic geometry texts and tried many tricks using Zariski's main theorem and Noether normalization, but remain stuck by the following:</p> <p>Let $\pi:X\to S$ be a morphism of finite type between integral, Noetherian schemes and let $x$ be a point of $X$. Does there exist an open neighbourhood of $X$ which admits a finite, surjective morphism onto a smooth $S$-scheme?</p> <p>In this generality I think that the answer is 'no', though I do not have a counterexample. What if we impose additional assumptions such as $\pi$ being flat or proper (or even projective)?</p> <p>A related question, an affirmative answer to which would imply the same for the previous question in the projective case, is the following: if $X$ is a projective scheme over a local ring $A$, then does $X$ admit a finite surjection to $\mathbb{P}_A^d$ for some $d\ge 1$?</p> <p>(I am imagining that $A=\mathbb{Z}_p$, so please do not assume that the residue field of $A$ is infinite!)</p> <p>Thank you!</p> <hr> <h2>Update</h2> <p>With Brian's help (thank you), the interesting remaining problem is the following: does every projective variety $V$ over a finite field $k$ admit a finite surjective morphism to $\mathbb{P}_k^d$ for some $d$? I have a gnawing suspicion that the answer is 'no'.</p> <p>It is useful to remember Noether normalisation in this case: if $I$ is a non-zero ideal of $k[X_1,\dots,X_n]$, then one can find a finite morphism $k[Y_1,\dots,Y_{n-1}]\to k[X_1,\dots,X_n]/I$ by sending $Y_i$ to $X_i-X_n^e$ for some big enough $e\ge 1$. Unfortunately, projectivising this construction produces a morphism $\mathbb{P}_k^n\setminus C\to\mathbb{P}_k^{n-1}$ where $C$ is quite a large closed subscheme of $\mathbb{P}_k^n$ (unless I have made a mistake); so if $V$ is our variety inside $\mathbb{P}_k^n$, then it is difficult to ensure that $V$ doesn't meet $C$. Therefore we can't successively project down to smaller dimensional spaces. (In contrast with the case when $k$ is infinite, for then we use changes of variables looking like $Y_i=X_i-\alpha_iX_n$ and the resulting morphism between projective spaces is defined everywhere except for one point, which we can assume doesn't lie on $V$).</p> http://mathoverflow.net/questions/23561/nice-definition-of-discriminant-as-alluded-to-in-an-answer-of-qing-liu/23569#23569 Answer by Matthew Morrow for "Nice" definition of discriminant as alluded to in an answer of Qing Liu Matthew Morrow 2010-05-05T11:24:38Z 2010-05-05T11:24:38Z <p>The "magistral paper of Takeshi Saito" mentioned by Qing Liu is the following: 'Conductor, Discriminant, and the Noether formula of Arithmetic surfaces', Duke Math Journal, vol. 57, no. 1. See here if you have a subscription to Duke: projecteuclid.org/euclid.dmj/1077306852 The definition of the discriminant is at the top of the second page. Or see section three of Liu's notes which he referenced in his reply to which you provided a link.</p> <p>But for what it's worth, here is a quick answer (basically copied from Saito's paper):</p> <p>Let $T$ be a nice scheme (spectrum of a field is enough for us), and $g:Y\to T$ a relative curve (i.e. geometrically connected, proper, relative dimension one). When $Y$ is smooth over $T$, there is a functorial isomorphism $\Delta:\det Rg_*(\omega_{Y/T}^{\otimes 2})\to(\det Rg_*\omega_{Y/T})^{\otimes 13}$. This is due to Deligne (unpublished letter to Quillen....).</p> <p>Now let $\mathcal{O}_K$ be a Henselian discrete valuation ring with algebraically closed residue field, put $S=\mbox{Spec}\mathcal{O}_K$, and let $X\to S$ be a relative curve which is regular. We have an invertible $\mathcal{O}_K$-module</p> <p>$V=\mbox{Hom} (\det Rg_*(\omega_{X/S}^{\otimes 2}),(\det Rg_*\omega_{X/S})^{\otimes 13})$.</p> <p>and Deligne's result, applied to the generic fibre of $X$, produces a natural element $\Delta\in V\otimes_{\mathcal{O}_K} K$. The discriminant of $X$ is defined to be the order of $\Delta$ (that is, the unique $d\in\mathbb{Z}$ such that $\mathcal{O}_K\Delta=\mathfrak{p}^d V$).</p> <p>I do not know if there is a similar notion of discriminant for higher dimensional $S$-schemes.</p> http://mathoverflow.net/questions/28435/automatically-extract-a-bibitem-not-bibtex-from-mathscinet/28439#28439 Comment by Matthew Morrow Matthew Morrow 2010-06-16T22:37:14Z 2010-06-16T22:37:14Z Ah, yes, you are right. If I had ever needed to turn the <i>whole</i> .bib file into a .bbl I wouldn't have known what to do (apart from writing a document full of \nocite{this} and \nocite{that}) - your suggested package listbib fixes that! http://mathoverflow.net/questions/27424/a-ring-such-that-all-projectives-are-stably-free-but-not-all-projectives-are-free Comment by Matthew Morrow Matthew Morrow 2010-06-08T09:12:00Z 2010-06-08T09:12:00Z (typo: $0=\mathbb{Z}$). http://mathoverflow.net/questions/27424/a-ring-such-that-all-projectives-are-stably-free-but-not-all-projectives-are-free Comment by Matthew Morrow Matthew Morrow 2010-06-08T09:11:13Z 2010-06-08T09:11:13Z Very nice question! I was asking people in the dept this a couple of weeks ago, and I we couldn't answer it. My interest was the following: by homotopy invariance of K-theory we know that $K_0(k[T_1,\dots,T_n])=0$, and I wondered if this implied Serre's conjecture (even though elementary, direct proofs of Serre's conjecture are now known). Seems not. http://mathoverflow.net/questions/27197/local-complete-intersections-which-are-not-complete-intersections/27210#27210 Comment by Matthew Morrow Matthew Morrow 2010-06-06T08:58:05Z 2010-06-06T08:58:05Z Hailong, thanks very much for the reference! I've actually been reading an old paper by Hochster on Cohen-Macaulay rings (and I've just noticed from your webpage that he was your supervisor!). Thanks again. http://mathoverflow.net/questions/27197/local-complete-intersections-which-are-not-complete-intersections/27210#27210 Comment by Matthew Morrow Matthew Morrow 2010-06-05T23:56:54Z 2010-06-05T23:56:54Z Yes, 25 minutes after posting I realised my error and have returned to edit my 'answer'. But I thought that the notion of analytic spread and set-theoretic complete intersection were understood well in the affine case, though not the projective. Are they not? http://mathoverflow.net/questions/27197/local-complete-intersections-which-are-not-complete-intersections/27199#27199 Comment by Matthew Morrow Matthew Morrow 2010-06-05T22:49:27Z 2010-06-05T22:49:27Z I was worrying about a very similar problem concerning complete intersections recently, and the twisted cubic provided a counterexample to what I hoped was true. Very depressing when a counterexample is contained in the second exercise in Hartshorne's Alg. Geom... http://mathoverflow.net/questions/26844/some-sort-of-descent-theory Comment by Matthew Morrow Matthew Morrow 2010-06-02T20:29:24Z 2010-06-02T20:29:24Z (ah, or direct sums of copies of this constant sheaf! Or maybe there are more interesting examples which I am missing.) http://mathoverflow.net/questions/26844/some-sort-of-descent-theory Comment by Matthew Morrow Matthew Morrow 2010-06-02T20:27:37Z 2010-06-02T20:27:37Z By 'stable', do you mean that the Frobenius acts as the identity on $\mathcal{F}$? Surely the only such sheaf on $\overline{X}$ with this property is the sheaf which is constantly equal to $\mathbb{F}_p$. http://mathoverflow.net/questions/26856/a-question-about-the-assassinator-associated-primes-and-the-support-of-a-mod Comment by Matthew Morrow Matthew Morrow 2010-06-02T20:02:27Z 2010-06-02T20:02:27Z Do you mean that $\mathfrak{p}$ consists of <i>zerodivisors</i>? of $M/xM$? http://mathoverflow.net/questions/26557/what-was-galois-theory-like-before-emil-artin/26561#26561 Comment by Matthew Morrow Matthew Morrow 2010-05-31T11:35:55Z 2010-05-31T11:35:55Z There is also Dickson's first book (according to Wikipedia) &quot;Linear Groups with an Exposition of Galois Field Theory&quot;, first published in 1901. Ancient! Oh whoops. Actually, the book doesn't seem to feature any Galois theory, but rather &quot;Galois Field&quot;=&quot;Finite Field&quot;. It's a book about linear groups over finite fields. http://mathoverflow.net/questions/23670/can-the-cross-section-x-section-be-removed-from-the-ax-kochen-proof-in-diophan/24212#24212 Comment by Matthew Morrow Matthew Morrow 2010-05-11T09:39:11Z 2010-05-11T09:39:11Z @Thomas S. Thanks for correcting my history! http://mathoverflow.net/questions/24144/is-there-an-uncountable-non-discrete-hausdorff-toronto-space Comment by Matthew Morrow Matthew Morrow 2010-05-10T21:51:26Z 2010-05-10T21:51:26Z This at least used to be one of the major open problems in set-theoretic topology. See page 15 of the &quot;Open problems in topology book&quot; (now 20 years old) www1.elsevier.com/homepage/sac/opit/book.pdf I have no idea if there has been any progress in the past 20 years. http://mathoverflow.net/questions/23940/why-free-topological-groups-on-tychonoff-spaces/23951#23951 Comment by Matthew Morrow Matthew Morrow 2010-05-10T16:01:51Z 2010-05-10T16:01:51Z Phew! I'm glad it works :) http://mathoverflow.net/questions/23940/why-free-topological-groups-on-tychonoff-spaces/23951#23951 Comment by Matthew Morrow Matthew Morrow 2010-05-10T14:30:08Z 2010-05-10T14:30:08Z This is the problem I was seeing which caused me to say &quot;it doesn't work so nicely&quot; in my answer (in the case when $F(X)$ is not $T_0$, which I now know is the same as $X$ being functionally Hausdorff). But I'm not sure what you are asking now. Taking into account this result of Thomas (which I didn't know), my answer says that if $X$ is functionally Hausdorff then $F(X)=F(Y)$, where $Y$ is $X$ equipped with the weak $C(X)$ topology ($Y$ is also the Tychonoffication of $X$). http://mathoverflow.net/questions/24082/what-are-the-prime-ideals-of-kx-y/24088#24088 Comment by Matthew Morrow Matthew Morrow 2010-05-10T12:38:50Z 2010-05-10T12:38:50Z This process gives all the prime ideals of $k[[x,y]]$ whose contraction to $k[x,y]$ is non-zero. Unfortunately, there are also LOTS of prime ideals $Q$ of $k[[x,y]]$ whose contraction to $k[x,y]$ is zero, and these are a bit weird, e.g. the localization $k[[x,y]]_Q$ contains $k(x,y)$.