User karl schwede - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:28:01Z http://mathoverflow.net/feeds/user/3521 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/133567/counter-example-of-upper-semicontinuity-of-fiber-dimension-in-classical-algebraic/133633#133633 Answer by Karl Schwede for Counter example of upper semicontinuity of fiber dimension in classical algebraic geometry Karl Schwede 2013-06-13T13:15:20Z 2013-06-13T14:34:22Z <p>For what its worth, I can give you a non-Noetherian example, even with both schemes affine, irreducible (and of finite Krull dimension).</p> <p>Set $R = k[x,y,x/y, x/y^2, x/y^3, ...]$ and $S = k[y]$. We have the obvious map $S \hookrightarrow R$ which induces $$X = \text{Spec }R \to Y = \text{Spec }S.$$ Now, away from the origin of $S$, $y$ is invertible and $R[y^{-1}] = k[x,y,y^{-1}]$ has all fibers with dimension $1$. On the other hand, once we set $y = 0$ in $R$, we notice that $x = (x/y) y$ is a multiple, as is $(x/y) = (x/y^2) y$, and so is $(x/y^n)$ for all $n$. This is already a maximal ideal, so the fiber over $y = 0$ is $0$-dimensional. </p> http://mathoverflow.net/questions/133495/inverse-image-in-the-blowup/133501#133501 Answer by Karl Schwede for inverse image in the blowup Karl Schwede 2013-06-12T12:41:05Z 2013-06-12T12:41:05Z <p>It depends on what scheme structure you give $Z$. </p> <p>If $I_1$ defines $Z_1$ and $I_2$ defines $Z_2$ then the blowup of $Z = V(I_1 \cdot I_2)$ does turn $Z_1$ and $Z_2$ into Cartier divisors. This is actually a pretty straightforward exercise from several perspectives (either working out charts or using universal properties), so I'll let you do it. (You don't need $X$ to be regular for this).</p> <p>On the other hand, if you set $Z = V(I_1 \cap I_2)$ then you are out of luck. Let me give you an example. Set $X = \mathbb{A}^2 = \text{Spec}[x,y]$, $Z_1 = V(x, y^2)$ and $Z_2 = V(x^2, y)$. Fix $Z = V( (x, y^2) \cap (x^2, y) ) = V(x^2, xy, y^2)$. The blowup of $Z$, $Z_1$ and $Z_2$ all have one exceptional divisor, but they correspond to different valuations. It follows that the inverse image of $Z_1, Z_2$ are not Cartier divisors in the blowup of $Z$ (if this isn't clear, write down affine charts). </p> <p>It's not so hard to find reduced examples that behave the same way, but you'd need to go to dimension $\geq 3$. </p> http://mathoverflow.net/questions/133253/how-to-define-the-canonical-sheaf-on-singular-varieties/133254#133254 Answer by Karl Schwede for How to define the canonical sheaf on singular varieties Karl Schwede 2013-06-10T02:11:41Z 2013-06-10T02:32:26Z <p>For $X$ normal, saying that canonical divisor is the pushforward of the canonical sheaf of a resolution of singularities is totally fine (it even works in characteristic $p > 0$ if you happen to have a resolution). In particular, if $\pi : Y \to X$ is a resolution of singularities, then $\pi_* K_Y$ is $K_X$ (here we define $\pi_* K_Y$ by simply throwing away any components of $K_Y$ that get contracted to non-divisorial varieties). </p> <p><strong>However:</strong> The pushforward of the canonical <em>sheaf</em> $\omega_Y$ is not the canonical sheaf $\omega_X$ in general. Indeed, $\pi_* \omega_Y = \omega_X$ is very close to requiring that $X$ has <em>rational singularities.</em> (Actually one definition of rational singularities, typically attributed to Kempf, is that $X$ is Cohen-Macaulay and $\pi_* \omega_Y = \omega_X$.</p> <p>A good exercise is to show that if $X = \text{Spec} k[x,y,z]/(x^3+y^3+z^3)$ and $\pi : Y \to X$ is the blowup of the cone point, then $\pi_* \omega_Y = \mathfrak{m} \cdot \omega_X$ where $\mathfrak{m}$ is the maximal ideal of the origin. (<em>Hint:</em> use the adjunction formula and the formula for the canonical divisor when blowing up a point on $\mathbb{A}^3$).</p> <h2>The canonical sheaf</h2> <p>Ok, so what is the right definition of the sheaf $\omega_X$ in general?</p> <p>Well, for a projective variety of dimension $d$, $i : X \hookrightarrow \mathbb{P}^N$, define $$i_* \omega_X := \text{Ext}^{N-d}\big(i_* O_X, O_{\mathbb{P}^N}(-N-1)\big).$$ Since $i$ is a closed embedding, this uniquely determines $\omega_X$. For $X$ quasi-projective, you can define this by localizing. There are generalizations which apply to other schemes of finite type over $k$ defined using the $f^!$ functor for $f : X \to k$ the structural map, but I won't get into that here.</p> <h2>The canonical sheaf is S2</h2> <p>It turns out that for any variety, $\omega_X$ is an S2 sheaf, this means it satisfies Hartog's phenomena (search math overflow). In particular, the sheaf is determined by its codimension-1 behavior. There's an easy way to see this, it turns out that if $h : X \to Z = \mathbb{P^d}$ is a generic projection to a hyperplane of the same dimension, then $$h_* \omega_X = \text{Hom}(h_* O_X, O_Z(-d-1)).$$ Now, it easily follows that this sheaf is S2 since it is reflexive on $Z$ and reflexive sheaves on $Z$ are always S2 (see for example Hartshorne's <em>Generalized divisors on Gorenstein schemes</em>).</p> <p>Why does this matter? Well it means that if $U$ is the regular locus of $X$, and furthermore $X \setminus U$ has codimension 2 on $X$ (which happens for example if $X$ is normal), then if $j : U \hookrightarrow X$ is the inclusion, then $j_* \omega_U = \omega_X$ since both sheaves are S2 and they agree outside a codimension-2 set.</p> <h2>Back to divisors</h2> <p>This also explains our first statement about divisors. Indeed, any divisor, like $\pi_* K_Y$ is determined outside a codimension 2 set, it is determined on $U$ in fact. And so if $X$ is normal, $\pi : Y \to X$ is an isomorphism outside of a codimension-2 set of $X$, and so the canonical divisor on that set works fine as a canonical divisor everywhere. In particular, it can be computed on $Y$ as claimed.</p> <p>For non-normal $Y$, something can be done, but the formula isn't quite so simple (you also have to describe by what exactly you mean by a divisor on a non-normal variety).</p> http://mathoverflow.net/questions/131652/canonical-modules/131662#131662 Answer by Karl Schwede for Canonical Modules Karl Schwede 2013-05-23T22:53:31Z 2013-05-23T22:53:31Z <p>As Graham points out, this ring is Gorenstein so the canonical module is <em>isomorphic</em> to the ring itself. For most hypersurfaces, this is all you can say. However, I think in this case one can say slightly more.</p> <p>The ring is also toric $k[x,y,z]/(x^2 - yz) = k[ab, a^2, b^2]$. Thus we have a canonical way to identify the canonical module/divisor. </p> <p>Recall that $K_X = -\sum \text{[torus invariant prime divisors]}$. For this ring, we are looking for height 1 primes that are toric (monomial). There are two $(ab, a^2) = (x,y)$ and $(ab, b^2) = (x,z)$. The negative sum of the corresponding divisors just corresponds to the intersection of the two ideals. In this case, we get $(ab, a^2) \cap (ab, b^2) = (ab) = (x)$. So the ideal $(x)$ is the canonical module.</p> <p>Determinantal rings also have explicit canonical choices for canonical modules if I recall correctly. Graded rings have graded canonical modules which have some canonical choice of degree.</p> http://mathoverflow.net/questions/131225/relation-between-hi-i-and-hi-j-when-i-subset-j/131391#131391 Answer by Karl Schwede for Relation between $H^i_I(-)$ and $H^i_J(-)$ when $I\subset J$ Karl Schwede 2013-05-21T21:48:35Z 2013-05-22T02:40:38Z <p>There's a map between them, and these do fit into a long exact sequence together. This is explained in the book on local cohomology by Hartshorne, see Lemma 1.8: You can even download this book <strong><a href="http://link.springer.com/book/10.1007/BFb0073971/page/1" rel="nofollow">if your institution has access...</a></strong></p> <p>I will sketch it briefly. Let $V(I) = Y \supseteq Z = V(J)$ and set $W = V(I) \setminus V(J) = Y \setminus Z$, this $W$ is only locally closed in $\text{Spec} R$.</p> <p>We define $H^i_W(\bullet)$ to be the right derived functors of $\Gamma_W(\bullet)$. Here $\Gamma_W(M)$ for any $R$-module $M$ is defined to be the set of elements of $M$ which are zero in $M_{p}$ for all $p \in (\text{Spec }R) \setminus W$.</p> <p>With this notation, for any $R$-module $M$ we have a long exact sequence $$... \to H^i_Z(M) \to H^i_Y(M) \to H^i_W(M) \to H^{i+1}_Z(M) \to ...$$ Note $H^i_Z(M) = H^i_J(M)$ and $H^i_Y(M) = H^i_I(M)$. I doubt this gives you much information unless $Y$ and $Z$ have some special relationship. </p> <p>There are related things you can do too. If for example $Y = Z \cup X$ where $X$ is some other closed subset of $\text{Spec }R$, there's another long exact sequence but this is a special case. Can you tell me what $Y$ and $Z$ are in your case?</p> http://mathoverflow.net/questions/131265/when-is-the-intersection-of-an-isolated-normal-singularity-with-a-generic-linear/131269#131269 Answer by Karl Schwede for When is the intersection of an isolated normal singularity with a generic linear subspace through that singularity normal? Karl Schwede 2013-05-20T20:29:24Z 2013-05-21T13:48:29Z <p>I'm going to assume your singularity is dimension $\geq 3$. Angelo beat me to the answer but he is right, this is not true. But it is true sometimes (including the Cohen-Macaulay case as he implied).</p> <p>A singularity is normal if it is $R1$ and $S2$. In your case, an isolated singularity is normal if the depth at the singular point is at least 2. </p> <p>Now, a general hyperplane section will be $R1$ by Bertini. So we just need to check that the general hyperplane is $S2$. Well, for this we just need the depth to be at least 2 again, and hence we just need the original singularity to have depth $\geq 3$. </p> <p><strong>Conclusion:</strong> <em>If your singularity is $S3$ (in your case just $\text{depth} \geq 3$), then what you want holds after cutting down by ONE hyperplane</em></p> <p><strong>EDIT:</strong> As Angelo pointed out, the actual question didn't cut down by just one hyperplane. In that case you can't just have depth $\geq 3$, you need $X$ to be Cohen-Macaulay.</p> <p>Of course, not all singularities satisfy this, for example a cone over an Abelian surface.</p> <p>You might also look at this preprint which seems to have some related results: <a href="http://front.math.ucdavis.edu/1108.4708" rel="nofollow">Tadashi Ochiai, Kazuma Shimomoto</a></p> http://mathoverflow.net/questions/131235/hartogs-theorem-and-canonical-bundles/131240#131240 Answer by Karl Schwede for Hartogs Theorem and Canonical Bundles Karl Schwede 2013-05-20T16:14:24Z 2013-05-20T17:49:36Z <p>I think the property you want is that the canonical sheaf $\omega_X$ is S2. Note that on a normal affine variety, $\omega_X$ is <em>not</em> necessarily a line bundle (it is if $X$ is a complete intersection though). </p> <p>For simplicity, let's assume $X \subseteq A^{n}$ is of dimension $d$. Then $$\omega_X = Ext^{n-d}(O_X, O_{A^{n}})$$ is a S2 sheaf. This implies that it satisfies Hartog's theorem. Not all sheaves do! For example, the ideal sheaf of a maximal ideal obviously does not (assuming $\dim X \geq 2$).</p> <p>For a reference which discusses the S2 condition and relation to Hartog's phenomenon, see for example</p> <p>Hartshorne, <em>Generalized divisors on Gorenstein schemes</em>. </p> <p>I think Sándor Kovács has also written several good answers explaining this connection on mathoverflow.</p> <p>A proof of the S2ness of $\omega_X$ for varieties can be found in Kollár-Mori, <em>Birational geometry of algebraic varieties</em>. Another proof can be found in Hartshorne's <em>Generalized divisors and biliaison</em>.</p> http://mathoverflow.net/questions/130872/crepant-morphisms-of-varieties/130877#130877 Answer by Karl Schwede for Crepant Morphisms of Varieties Karl Schwede 2013-05-16T21:19:13Z 2013-05-16T22:31:08Z <p>Crepant stands for <em>non-discrepant</em>. It's frequently applied to resolutions of singularities or birational maps (but can be applied more generally).</p> <p>Let's start with the birational case, since that's where the history is. If $f : X \to Y$ is birational, and $K_Y$ is $\mathbb{Q}$-Cartier, then $f^*(K_Y)$ makes sense. In particular, if $nK_Y$ is Cartier, then $f^*(K_Y) = \frac{1}{n} f^*(nK_Y)$ by definition.</p> <p>Write $K_X - f^* K_Y = \sum a_i E_i$ where we pick $K_X$ and $K_Y$ which agree where $f$ is an isomorphism. The $\sum a_i E_i$ is then independent of choices. </p> <p>Now, the numbers $a_i$ are called <em>discrepancies</em>. If there are no discrepancies (ie, all the $a_i$ are zero (for example, if $f$ is a small map), then the map is called <em>crepant</em>. Of course, all $a_i = 0$ if and only if $K_X = f^* K_Y$.</p> <p>Of course, if the pullback of $K_X$ is $K_Y$, then this can be applied to many things. The existence of a crepant resolution of singularities also can be quite useful. Let me give a nonstandard example in characteristic $p > 0$, if $Y$ is Frobenius split and $f : X \to Y$ is crepant, then $X$ is also Frobenius split. Some variant of this appeared in the work of Mehta-van der Kallen and also Mehta-Srinivas.</p> http://mathoverflow.net/questions/120982/base-change-of-trace-for-gorenstein-or-cohen-macaulay-morphisms Base change of trace for Gorenstein or Cohen-Macaulay morphisms Karl Schwede 2013-02-06T16:04:44Z 2013-04-28T04:29:49Z <p>This is basically a question of functoriality for base change of CM morphisms. </p> <p><strong>EDIT:</strong> $\text{ }$ <em>Brian Conrad sent me an email explaining the that this is indeed true, and follows from his book. I'll explain this in my case a little later. First I will state the question.</em></p> <p>Suppose that $f : X \to V$ is an equidimensional (dimension $d$) finite type (reduced, if it helps) Cohen-Macaulay morphism (flat with Cohen-Macaulay fibers). I'm also happy to assume that $V$ is integral, excellent and has a dualizing complex. Additionally suppose that we have $f' : X' \to V$ another equidimensional (dimension $d$) finite-type (reduced) Cohen-Macaulay morphism that factors through $f$ as below. </p> <p>$$f' : X' \xrightarrow{\phi} X \to V.$$</p> <p>Further suppose that $\phi$ is <em>finite</em> (although the question could be asked more generally for proper $\phi$, I'll phrase it for finite $\phi$). If it helps at any point, please feel free to assume that $f$ and $f'$ are Gorenstein morphisms.</p> <p>Recall that $\omega_{f}[d] = f^! \mathcal{O}_V$ and that by Brian Conrad's book [LINK: Google books][1] we know that both $\omega_f$ and $\omega_{f'}$ are compatible with base change. </p> <p>I'd like to conclude that the following natural map is also compatible with base change:</p> <p>$$\phi_* \omega_{f'} \cong R \mathcal{H}om_{O_X}(\phi_* O_{X'}, \omega_f ) \cong \mathcal{H}om_{O_X}(\phi_* O_{X'}, \omega_f ) \to \omega_f$$.<br> The map can be interpreted as evaluation at 1. </p> <p>In other words, I'd like to know that the trace map of $\phi$ is compatible with base change. Furthermore, it would be even good enough to prove this in the $f, f'$ Gorenstein morphisms case. </p> <p>One way to do this would be as follows. If $g : T \to V$ is any other morphism and $f_T : X_T \to T$ and $f_T': X_T' \to T$ are the base changes and $g_X : X_T \to X$ is the projection, is it true that the natural map (denoted [*] below) $$g_X^* \phi_* \omega_{f'} \cong g_X^* \mathcal{H}om_{O_X}(\phi_* O_{X'}, \omega_f ) \to \mathcal{H}om_{O_{X_T}}(\phi_* O_{X_T'}, \omega_{f_T} ) \cong (\phi_T)_* \omega_{f_T'}$$ between abstractly isomorphic sheaves is an isomorphism?</p> <hr> <p><strong>Edit:</strong> Brian Conrad pointed out to me that this is already a special case of what is in his Theorem 3.6.1 (assuming I understand everything right). Essentially the point is that in my case everything is finite type, which makes it all much easier.</p> http://mathoverflow.net/questions/128405/conical-divisor-over-a-mathbb-q-cartier-divisor/128409#128409 Answer by Karl Schwede for Conical divisor over a $\mathbb Q$-Cartier divisor. Karl Schwede 2013-04-22T22:53:22Z 2013-04-24T17:43:56Z <p>I'm going to assume that $L$ induces a projectively normal embedding (so I feel comfortable talking above divisors, or you can take Spectrum of section rings instead of cones). Then this is certainly true for the affine cone which should be all you need, since away from the cone point there is nothing to check. In other words, the index of the divisor on $X$ is the same as that on the cone (away from the cone point).</p> <p>Now, your hypothesis implies that some multiple of $D$ is a multiple of $L$, say $O_X(nD) = L^j$. Set $S = \bigoplus_{i \in \mathbb{Z}} H^0(L^i)$ and $Y = \text{Spec }S$. </p> <p><strong>Claim:</strong> <em>if</em> $D_Y$ <em>is the divisor on $Y$ corresponding to $D$, then</em> $$\Gamma(Y, O_Y(D_Y)) = \bigoplus_{i \in \mathbb{Z}} H^0(L^i \otimes O_X(D)).$$</p> <p><em>Proof of claim:</em></p> <p>I will assume that $D = -P$ is anti-effective and irreducible, as it is straightforward to reduce to this case.</p> <p>First let us explain how to construct $D_Y$. Set $U = Y \setminus${origin}. Then $\rho : U \to X$ is a $\mathbb{A}^1$-bundle over $X$, in particular it is flat. We set $D_U = \rho^* D$ (which makes sense, even for Weil divisors, since it's an $\mathbb{A}^1$-bundle). Then $D_Y$ is the unique divisor on $Y$ which agrees with $D_U$ on $U$. </p> <p>Since $D$ is anti-effective, so is $D_Y$. Then $\Gamma(Y, O_Y(D_Y)) = \Gamma(Y, O_Y(-P))$ is the set of functions in $S$ that vanish along the pullback of $D_Y$. This is clearly a homogeneous prime ideal, and hence easily is seen to coincide with the ideal $\bigoplus_{i \in \mathbb{Z}} H^0(L^i \otimes O_X(D)) \subseteq S$. <em>(Ok, I kind of waved my hands here, but hopefully this part is easy)</em></p> <p><strong>\qed</strong></p> <p>Similarly, the divisor corresponding to $nD$ is $(nD)_Y = nD_Y$ (this equality is just observed by pulling back to $U$ and doing the computation there). $nD_Y$ has corresponding sheaf $$\Gamma(Y, O_Y(nD_Y)) = \bigoplus_{i \in \mathbb{Z}} H^0(L^i \otimes O_X(nD)).$$ Note that because $O_X(nD) = L^j$, we see that $O_Y(nD_Y) = S[-j]$, in other words $O_Y(nD_Y)$ is just $S$ with a different grading. In particular, it is locally free so $nD_Y$ is Cartier.</p> http://mathoverflow.net/questions/127648/reflexive-sheaf/127649#127649 Answer by Karl Schwede for Reflexive sheaf Karl Schwede 2013-04-15T18:04:16Z 2013-04-15T18:04:16Z <p>I think this is not true. For example, what if $X \to Y$ is a resolution of singularities, that's even an isomorphism outside a set of codimension 2. The pullback of reflexive sheaves is definitely not reflexive. </p> <p>For example, consider the blowup of the origin in $R = k[x,y,z]/(xy-z^2) = k[a^2, b^2, ab]$. One of the charts in that blowup is $k[a^2, b/a]$. The pullback of the reflexive sheaf $(a^2, ab)$ in that chart is the tensor product $(a^2, ab) \otimes_R k[a^2, b/a]$. Then the element $$a^2 \otimes (b/a) - ab \otimes 1$$ is clearly torsion and non-zero. </p> <p>What people do frequently do is the reflexive pullback. In other words, pullback and then reflexify. </p> <p>From the point of view of singularities, you might look at a paper of de Fernex and Hacon on singularities in non-Q-Gorenstein rings. There are other sources I can point you towards (say Hassett and Kov\'acs) if you are more interested in moduli types of applications.</p> http://mathoverflow.net/questions/127013/kx-as-a-kxp-module-for-ugly-fields $k[[x]]$ as a $(k[[x]])^p$ module for ugly fields Karl Schwede 2013-04-09T19:14:30Z 2013-04-10T19:05:53Z <p>Suppose that $k$ is a field of characteristic $p$ such that $k$ is not a finite $k^p$-module. For example, $k = \mathbb{F}_p(x_1, x_2, x_3, ...)$. </p> <p>Is it true that $k[[x]]$ is a free $(k[[x]])^p$-module? We know that it is flat by a theorem of Kunz. However, it is certainly not finite, so flat is not the same as free.</p> <p>The naive thing to try (in terms of a basis) is to do the following. Choose {$\lambda_i$} a basis for $k$ over $k^p$. Consider the set {$\lambda_i x^j$} for $0 \leq j \leq p-1$. If $k$ is a finite $k^p$-vector space, this set is easily seen to be a basis for $k[[x]]$ over $(k[[x]])^p$. </p> <p>However, because of our (ugly) field, we can consider the power series:</p> <p>$$\lambda_0 + \lambda_1 x^1 + \lambda_2 x^2 + ... + \lambda_n x^n + ...$$</p> <p>where say $\lambda_0, \lambda_1, ...$ runs over some countably infinite subset of the {$\lambda_i$}. It is easy to see that this cannot be written as a finite $(k[[x]])^p$-linear combination of subset of the $\lambda_i x^j$ (where again $0 \leq j \leq p-1$).</p> <p>But of course, maybe there's some clever way to choose a basis that actually does work?</p> http://mathoverflow.net/questions/126656/linearly-generated-embedding/126725#126725 Answer by Karl Schwede for Linearly generated embedding? Karl Schwede 2013-04-06T18:49:56Z 2013-04-06T18:49:56Z <p><em>This was a comment originally.</em></p> <p>The generation in degree 1 is the same as projective normality at least for normal varieties. </p> <p>For simplicity, assume that $X$ is normal. By Hartshorne, Chapter II, Exercise 5.14, we know that if $X$ is projectively normal (with embedding associated to the complete linear system of $L$), then $$H^0(\mathbb{P}^N, O_{\mathbb{P}^N}(i)) \to H^0(X, L^i)$$ is surjective for all $i$. Therefore $R(X, L)$ is a quotient of $R(\mathbb{P}^N, O(1)) = S$. But $S$ is generated in degree $1$, and so $R(X,L)$ is generated in degree $1$ as well. </p> http://mathoverflow.net/questions/126529/a-criterion-for-freeness-over-a-local-ring/126534#126534 Answer by Karl Schwede for A criterion for freeness over a local ring Karl Schwede 2013-04-04T16:03:53Z 2013-04-04T16:03:53Z <p>I think the answer is no. Suppose $M = B$ is a finite ring extension of $A$, say with an isolated singularity, which is both Cohen-Macaulay except over the closed point of $A$ and normal.</p> <p>Then $M$ cannot be free, since free modules are Cohen-Macaulay. On the other hand, it is free away from the closed point by the Cohen-Macaulay hypothesis. The second hypothesis follows from the S2 assumption.</p> http://mathoverflow.net/questions/48659/normal-macaulayfications Normal Macaulayfications Karl Schwede 2010-12-08T18:10:23Z 2013-04-03T15:30:24Z <p>Given any reduced excellent scheme $X$, there exists a <em>Macaulayfication</em> $\pi : Y \to X$. In other words, there exists a proper birational map $\pi$ from a Cohen-Macaulay scheme $Y$ to $X$. These exist by a result of Kawasaki.</p> <p>Is it possible that we may also pick $Y$ to be normal? I see no reason why this should be true from the construction (blowing up various systems of parameters). Of course, when resolutions of singularities exist, this certainly solves the problem...</p> <p>One might hope that the normalization of a Cohen-Macaulay scheme is Cohen-Macaulay, but I believe this is false (for example given a normal non-CM variety in characteristic zero, you should be able to generically project it to a hypersurface).</p> http://mathoverflow.net/questions/123191/on-a-strongly-f-regular-pair-x-delta/123324#123324 Answer by Karl Schwede for On a Strongly F-regular Pair (X, \Delta) Karl Schwede 2013-03-01T13:28:56Z 2013-03-01T14:39:23Z <p>It isn't true as stated unfortunately. For example, take $X$ to be an ordinary elliptic curve, $\Delta = 0$ and $M = 0$. Then $S^0(X, \tau(X) \otimes O(M)) = H^0(X, O_X)$. However, for any effective Cartier $A > 0$ and any $\varepsilon > 0$, we have $S^0(X, \tau(X, \varepsilon A) \otimes O_X(M)) = 0$ (this can be checked easily with a direct computation).</p> <p><strong>However:</strong> Probably it is true for something like $P^0$, for a definition see <a href="http://arxiv.org/abs/1212.6956" rel="nofollow">Test ideals of non-principal ideals: Computations, Jumping Numbers, Alterations and Division Theorems</a> (there are some modifications one can make to that definition too which might make this easier).</p> <p>Definitely it is true for $P^0$ under suitable positivity assumptions. What can you assume about $M - K_X - \Delta$?. </p> http://mathoverflow.net/questions/121297/what-kind-of-subset-is-specr-p-in-specr/121313#121313 Answer by Karl Schwede for What kind of subset is Spec(R_P) in Spec(R)? Karl Schwede 2013-02-09T14:41:30Z 2013-02-09T17:07:49Z <p>$\text{Spec } R_P$ is by the usual correspondences the set of all primes of $R$ which are contained in $P$. In particular, it is somehow dual to $V(P) =$ the set of primes which are contained in $P$. It is not generally equal to the complement of $V(P)$ plus the point $P$ itself (this is very rare although does occur in all valuation rings as Fred pointed out in the comments). It is not open, $\text{Spec } R_P$ as a subset of $R$ is closed under generization.</p> <p>For example: I like to think of say $\text{Spec }$ of the local ring of the origin in $\mathbb{A}^2$ as the origin itself, some germ of every curve passing through the origin, and the generic point.</p> http://mathoverflow.net/questions/120625/jacobian-ideals-reference Jacobian ideals reference Karl Schwede 2013-02-02T21:46:29Z 2013-02-03T17:56:06Z <p>Suppose that $f : X \to V$ is a flat equidimensional (of dimension $h$) morphism of schemes of finite type and $V$ is excellent (or a variety) For this one can formulate something called the Jacobian ideal of $f$ which measures where $f$ is not smooth. One reference is Section 4.4 of the book by Swanson-Huneke <a href="http://people.reed.edu/~iswanson/book/index.html" rel="nofollow">You can download an earlier draft of the book</a></p> <p>Let me work this out in an explicit affine case. Suppose that $V = \text{Spec} A$ and $X = \text{Spec} A[x_1, \ldots, x_n]/(g_1, \ldots, g_m)$. One can then form the $m \times n$ Jacobian matrix $M_{X/V}$ whose $ij$th entry is ${\partial g_i \over \partial x_j}$. Then the ideal generated by the $h \times h$ minors of $M_{X/V}$ is called the Jacobian ideal of $X$ over $V$, and denote $J_{X/V}$.</p> <p>Does anyone know any good references for this object (I know about the Lipman-Satheye papers, some notes of Hochster, and the above book, but not much else). </p> <p>In particular, I'd love to have references to the following.</p> <p><strong>Question:</strong> Base change for $J_{X/V}$. (ok, this is essentially obvious but a reference would still be great, it also follows from the fitting ideal of the sheaf of differentials description of the Jacobian ideal which is described briefly in the above book).</p> <p><strong>Question:</strong> Say that $V$ is flat, equidimensional and finite type over another excellent scheme $S$ (for example, $S = \text{Spec} k$ for some field $k$). I'd like to relate the Jacobian ideals $J_{X/S}$, $J_{X/V}$ and $J_{V/S}$. In particular, if $V$ is smooth over $S$, I'd love to say that $J_{X/S} = J_{X/V}$.</p> http://mathoverflow.net/questions/116669/pseudoeffective-but-anti-nef-divisor/116683#116683 Answer by Karl Schwede for Pseudoeffective but anti-nef divisor Karl Schwede 2012-12-18T05:04:22Z 2012-12-18T13:09:45Z <p>As Sándor said, this implies that $D$ must be zero (numerically). However, it can happen that $D$ is pseudo-effective while at the same time $-D$ is relatively nef (or even relatively ample) with respect to some map. For example, if you blow up a point on $\mathbb{P}^2$, then the exceptional divisor is effective, but $-E$ is relatively ample. </p> http://mathoverflow.net/questions/115811/english-reference-for-the-grauertriemenschneider-vanishing-theorem/115814#115814 Answer by Karl Schwede for English reference for the Grauert–Riemenschneider vanishing theorem Karl Schwede 2012-12-08T17:08:28Z 2012-12-08T19:48:47Z <p>Dear Rami,</p> <p>You could see Kollar-Mori, <em>Birational geometry of algebraic varieties.</em> (Page 73)</p> <p>or</p> <p>Lazarsfeld, <em>Positivity in Algebraic Geometry I and II</em>. (Page 257) </p> http://mathoverflow.net/questions/115726/artinian-property-of-local-cohomology-module-over-graded-local-ring/115729#115729 Answer by Karl Schwede for Artinian property of local cohomology module over graded local ring Karl Schwede 2012-12-07T17:19:05Z 2012-12-07T17:19:05Z <p>Yes, this even holds for non-graded rings. Indeed, suppose that $R$ is a (Noetherian?) ring and $m \subseteq R$ is a maximal ideal. </p> <p>Then $$H^i_m(M) = H^i_{mR_m}(M_m)$$ essentially by Chapter III, Exercise 2.3(f) (excision) of Hartshorne. </p> <p>Also note that $N \subseteq H^i_{mR_m}(M_m)$ is an $R$-submodule if and only if it is an $R_m$-module. </p> http://mathoverflow.net/questions/114050/training-towards-research-on-birational-geometry-minimal-model-program/114057#114057 Answer by Karl Schwede for Training towards research on birational geometry/minimal model program Karl Schwede 2012-11-21T13:16:20Z 2012-11-21T15:13:04Z <p>I would <em>strongly</em> agree with Arend that you should include <em>Positivity in Algebraic Geometry</em> I &amp; II, by Lazarsfeld, (and that you should ask your potential advisor).</p> <p>However, while you point out that some books have become outdated, it doesn't mean that they aren't worth going through. The book by Kollár and Mori I think is still the standard source for the foundations of the MMP (especially the earlier chapters). You could also see some exercises to supplement that book by Kollár <a href="http://front.math.ucdavis.edu/0809.2579" rel="nofollow">http://front.math.ucdavis.edu/0809.2579</a></p> <p>In particular, I would suggest that that is still the right place to start (possibly combined with Lazarsfeld's books). Debarre's book is also a good starting text.</p> http://mathoverflow.net/questions/112464/multiplicities-of-rational-singularities-in-higher-dimension/112477#112477 Answer by Karl Schwede for multiplicities of rational singularities in higher dimension Karl Schwede 2012-11-15T12:53:42Z 2012-11-18T21:10:06Z <p>That particular bound doesn't hold if I recall correctly, but the following bound does:</p> <p><strong>Theorem :</strong> (C. Huneke and K.-i. Watanabe) <em>The multiplicity of a $d$-dimensional variety with rational singularities and embedding dimension $n$ is at most</em> $${n - 1 \choose d - 1}.$$</p> <p>In the case of a surface, this reduces to the bound you mentioned above. This is an unpublished result of Huneke and Watanabe (currently under review). You could certainly ask them for a preprint.</p> <p><strong>EDIT:</strong> My previous answer said that this was a conjecture, and that Huneke and Watanabe proved something related to this, but I wasn't sure if they actually proved this. It turns out that they did indeed prove this, and I got their permission to post that this was indeed a theorem of theirs.</p> http://mathoverflow.net/questions/112660/free-direct-summand-of-a-module/112744#112744 Answer by Karl Schwede for Free direct summand of a module Karl Schwede 2012-11-18T04:06:41Z 2012-11-18T04:06:41Z <p>Ok, I'm going to answer a different question for which the answer is "yes".</p> <p><strong>Question:</strong> <em>The module $M$ contains no summand of $F$ if and only if $M \subseteq m F$.</em></p> <p><strong>Proof:</strong> Suppose that $M$ did contain an $F$-summand. In other words, there is a surjective map $\phi : F \to R$ such that $\phi|_M$ is also surjective (note if $\psi : M \to R$ is surjective, sending $x \mapsto 1$, then the composition $R \xrightarrow{1 \mapsto x} M \xrightarrow{\psi} R$ gives us a summand). Thus $$m = mR = m \phi(F) = \phi(mF) \supseteq \phi(M) = R$$ which is impossible. </p> <p>Conversely, suppose that $M$ is not contained in $m F$. Choose an element $x \in M$, not in $mF$. If $e_1, \dots, e_n$ is a basis for $F$, then $x = \sum a_i e_i$ for some $a_i \in R$, at least one of the $a_i$ not in $m$. The projection onto that $i$th summand then clearly sends $x$ to a unit, and so by rescaling, I have a map $\phi : F \to R$ which sends $x$ to $1$. In particular, $x$ generates a summand of $F$. And so $M$ contains an $F$-summand. (You can also easily do this part with Nakayama's lemma).</p> http://mathoverflow.net/questions/112244/uniform-bound-of-the-number-of-generators-of-prime-ideals/112384#112384 Answer by Karl Schwede for uniform bound of the number of generators of prime ideals Karl Schwede 2012-11-14T14:59:08Z 2012-11-14T14:59:08Z <p>The following paper seems to indicate that there is no such bound in question 2. </p> <ul> <li>T. T. Moh, <em>On the unboundedness of generators of prime ideals in powerseries rings of three variables</em>. J. Math. Soc. Japan Volume 26, Number 4 (1974), 722-734. <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jmsj/1240435038" rel="nofollow">http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jmsj/1240435038</a></li> </ul> <p>He seems to construct a sequence of prime ideals $P_n$ in $k[[x,y,z]]$ with $n$ generators.</p> <p>On the other hand, in a positive result, in this paper:</p> <ul> <li>M. Boratyński, D. Eisenbud, D. Rees, <em>On the number of generators of ideals in local Cohen-Macaulay rings.</em> J. Algebra 57 (1979), no. 1, 77–81. <a href="http://www.sciencedirect.com/science/article/pii/0021869379902096" rel="nofollow">http://www.sciencedirect.com/science/article/pii/0021869379902096</a></li> </ul> <p>There they show some bounds for 2 -dimensional Cohen-Macaulay rings.</p> http://mathoverflow.net/questions/111905/push-forward-and-strict-transforms/111958#111958 Answer by Karl Schwede for push-forward and strict transforms Karl Schwede 2012-11-10T02:54:34Z 2012-11-10T02:54:34Z <p>No, the is definitely not true. Let me give you an example, suppose that $X = \mathbb{A}^2$ and that $\pi : Y \to X$ is the blowup of $X$ at the origin with exceptional divisor $E$. Suppose that $D$ is the Cartier divisor corresponding made up of $n$ lines through the origin, $n \geq 2$. Then $\widetilde{D}$ is just a disjoint union of $n$ lines on the blown up space, and $\pi^* D = \widetilde{D} + nE$.</p> <p>Suppose now that $\pi_* O_Y(\widetilde D) = O_X(D)$. Then by the projection formula, $$\pi_* O_Y(-nE) = \pi_* O_Y(\widetilde D - \pi^* D) = O_X$$ However, it is obvious that $\pi_* O_Y(-nE) = \mathfrak{m}^n$ is the maximal ideal of the origin raised to the $n$th power. This is a contradiction.</p> http://mathoverflow.net/questions/111040/ideal-of-strict-transform/111146#111146 Answer by Karl Schwede for Ideal of strict transform Karl Schwede 2012-11-01T12:12:59Z 2012-11-01T15:25:36Z <p>Let me fix some notation, let's set $\pi : \widetilde{X} \to X$ be the blowup and set $\bar{I}$ to be the ideal sheaf $(\pi^{-1} I) \cdot O_{\widetilde{X}}$ (note this is an invertible sheaf) and set $\bar{J} := (\pi^{-1} J) \cdot O_{\widetilde{X}}$ (this is probably not invertible). Finally, for clarity, let's set $B = [B_a]_0$, the degree zero piece of $B$ localized at $a$. Note there is a natural map $R \to B$ since $R$ maps to the degree zero piece of the Rees algebra.</p> <p>In terms of your two questions:</p> <h2>Question 1</h2> <p>The ideal sheaf $\widetilde{J}$ defining the strict transform $\widetilde{Y}$ on $\widetilde{X}$ is defined as follows.<br> $$\widetilde{J} = \bigcup_{n = 1}^{\infty} (\bar{J} : \bar{I}^n) =: (\bar{J} : \bar{I}^{\infty})$$ where the colon is taken over $O_{\widetilde{X}}$ (and the infinite power is a formal notation). In particular, as you can see this is a pain to compute. In terms of local coordinates in the notation you wrote, this is just: $$\bigcup_{n = 1}^{\infty} ((J \cdot B) :<em>{B} \langle a^n \rangle</em>{B}).$$ You can find more about this for example in papers on resolution of singularities, I think I first learned this in section 7 of this <a href="http://arxiv.org/pdf/math/0206244.pdf" rel="nofollow">PAPER</a> by Bravo, Encinas and Villamayor.</p> <hr> <h2>Example</h2> <p>Let's do an example. Consider $X = \text{Spec } k[x,y]$ and let $Z = V(x,y)$ be the origin. Let's let $Y = V(x^3-y^4)$, some sort of particularly nasty cusp, so $J = (x^3-y^4)$. </p> <p>There are two affine charts on the blowup. $B = k[x,y/x]$ and $B' = k[x/y,y]$. We first extend $J$ to these two charts. We get $$J \cdot B = (x^3-y^4) \cdot B = (x^3 - (y/x)^4 x^4) \cdot B = = x^3(1 - (y/x)^4 x) \cdot B.$$ and $$J \cdot B' = (x^3 - y^4) \cdot B' = ( (x/y)^3 y^3 - y^4 ) \cdot B' = y^3( (x/y)^3 - y) \cdot B'$$ The ideal sheaf $\bar{J}$ just corresponds to $J \cdot B$ and $J \cdot B'$. The ideal sheaf corresponding to the strict transform corresponds to $(1 - (y/x)^4 x) \cdot B$ and $(1 - (y/x)^4 x) \cdot B$. In other words, strip away the $x^3$ and $y^3$ (respectively) which simply vanish on the exceptional divisor.</p> <hr> <h2>Question 2</h2> <p>I believe this is right. Think about what the kernel of that map is on the $B_a$, certainly you have $J$, but you also have things that are knocked in there by powers of $a$ (due to the localization).</p> http://mathoverflow.net/questions/111104/dimension-and-singularities-of-the-minimal-log-canonical-center/111143#111143 Answer by Karl Schwede for Dimension and singularities of the minimal log canonical center Karl Schwede 2012-11-01T11:52:26Z 2012-11-01T11:52:26Z <p>If $(X, D)$ is a log canonical pair and $Z$ is the minimal log canonical center, then for some appropriate $D_Z$, the pair $(Z, D_Z)$ is KLT. In particular, $Z$ always has rational singularities. Thus I'd say the singularities of minimal LC centers are not so bad. See the papers of Kawamata on subadjunction. Florin Ambro also has a couple surveys on the arXiv.</p> <h2>Dimension</h2> <p>In terms of the dimension, without more information, the dimension of $Z$ can be anything from $\dim X - 1$-dimensional (if and only if $(X, D)$ is PLT) to $0$ dimensional. Certainly if $(X, D)$ is Kawamata log terminal on an open set $U$, then $Z \subseteq X \setminus U$. So you can control dimension in that way.</p> <h2>Multiplicity</h2> <p>If $X$ is smooth of dimension $n$, Stefan Helmke proved that any union of $d$-dimensional log canonical centers has multiplicity $\leq {n \choose d}$. He actually has a number of more precise results as well, see his paper in Duke for more details.</p> <h2>Applications</h2> <p>In many applications, say for the construction of global sections of adjoint divisors, being able to construct 0-dimensional centers would make things very easy (see for Example chapter 10 of Rob Lazarsfeld's book <em>Positivity in algebraic geometry</em>, or the papers of Kawamata on Fujita's conjecture). On the other hand, people have certainly looked for $\dim X - 1$-dimensional centers as well (I believe there is work of Mella, Ambro and Alexeev on this, look up Ladders on Fano varieties). </p> http://mathoverflow.net/questions/110317/doubt-about-normality-and-rational-singularities/110325#110325 Answer by Karl Schwede for Doubt about normality and rational singularities Karl Schwede 2012-10-22T12:12:05Z 2012-10-22T19:09:48Z <p>First, I don't think Miles Reid is dealing with pairs in that paper.</p> <p>Second, I think Miles wants $f$ to be etale in codimension 1 <em>on $Y$</em>. Thus a blowup is not allowed unless the blowup is a small map. You need every divisor on $Y$ to really have image of as divisor on $X$. You can see Miles Reid using this in his proofs.</p> <hr> <p>I'm going to tackle both statements if you just assume etale in codimension 1 on $X$ to show they are false. </p> <p><strong>1.</strong> This is 1.7(II) in Miles Reid's paper.</p> <p>Obviously this is false for a blowup. Take for example $X = \text{Spec } k[x,y,z]/(x^n+y^n+z^n)$ with $n \geq 3$ and let $\pi : Y \to X$ be the blowup of the origin. That's an isomorphism in codimension 1 on $X$, but not on $Y$. Certainly $X$ is not canonical but $Y$ is not.</p> <p>On the other hand, the case when $f$ is finite (and etale in codimension 1) is fairly standard. See for example 5.20 in Koll\'ar-Mori, where the etale in codimension 1 assumption implies that the ramification divisor is zero.</p> <p><strong>2.</strong> Even 2. isn't true if you don't assume that $f$ is etale in codimension 1 on $Y$. You can see <a href="http://mathoverflow.net/questions/43336/blowups-of-cohen-macaulay-varieties/43415#43415" rel="nofollow">THIS answer of Hailong Dao</a> which links to an example of Dale Cutkosky of a normal blowup of $\mathbb{C}[x,y,z]$ which is not Cohen-Macaulay (and thus not Canonical). </p> http://mathoverflow.net/questions/109848/pathological-examples-of-dimension/109875#109875 Answer by Karl Schwede for Pathological Examples of Dimension Karl Schwede 2012-10-17T04:56:22Z 2012-10-17T04:56:22Z <p>Here's a fairly standard one (it's an exercise in Hartshorne). In an integral domain $R$ of finite type over a field, every maximal ideal has the same height (in particular, every closed point has the same dimension). Indeed, it would be natural to define a ring to be equidimensional if every maximal ideal has the same height. Here's a problem with this definition.</p> <p>Suppose now that $R$ is a DVR with parameter $r$. Consider the ring $R[x]$. This ring has one maximal ideal of height one, $\langle xr - 1 \rangle$, and another maximal ideal of height two, $\langle x, r \rangle$. </p> <p>The point being, this is a domain, so its $\text{Spec}$ is presumably equidimensional, of dimension 2 the Krull dimension of $R[x]$. But it has closed points of different heights (although with very different residue fields). Of course, this isn't as pathological as a non-catenary ring, but we can even assume that $R[x]$ is a localization of $k[r,x]$. </p> http://mathoverflow.net/questions/133895/associated-prime-ideal Comment by Karl Schwede Karl Schwede 2013-06-16T17:06:21Z 2013-06-16T17:06:21Z This looks like a homework, but is the question that you don't understand how to interpret $\text{Hom}_R(M,N)$ as a module? Or that you don't know what $\text{Ass}_R$ of a module is? http://mathoverflow.net/questions/133567/counter-example-of-upper-semicontinuity-of-fiber-dimension-in-classical-algebraic/133633#133633 Comment by Karl Schwede Karl Schwede 2013-06-13T15:49:01Z 2013-06-13T15:49:01Z The origin in closed in $X$, is it not? http://mathoverflow.net/questions/133567/counter-example-of-upper-semicontinuity-of-fiber-dimension-in-classical-algebraic/133633#133633 Comment by Karl Schwede Karl Schwede 2013-06-13T15:46:58Z 2013-06-13T15:46:58Z I agree, it doesn't answer your question but I still thought it was amusing. http://mathoverflow.net/questions/133567/counter-example-of-upper-semicontinuity-of-fiber-dimension-in-classical-algebraic/133590#133590 Comment by Karl Schwede Karl Schwede 2013-06-13T14:55:23Z 2013-06-13T14:55:23Z Ah, of course, sorry about that. http://mathoverflow.net/questions/133567/counter-example-of-upper-semicontinuity-of-fiber-dimension-in-classical-algebraic/133590#133590 Comment by Karl Schwede Karl Schwede 2013-06-13T14:37:42Z 2013-06-13T14:37:42Z Hi, I'm still not sure why this is not semicontinuous in the sense of the original question. Consider $n = 1$. Then the set of points <i>of $X$</i> who live in fibers of dimension is $\geq 1$ is just <code>$\mathbb{A}^2 \setminus \{x = 0\}$</code>. This is closed in $X$. The set of points of $X$ who live in fibers of dimension $\geq 0$ is all of $X$ of course. Am I missing something here? http://mathoverflow.net/questions/133495/inverse-image-in-the-blowup/133501#133501 Comment by Karl Schwede Karl Schwede 2013-06-12T15:08:11Z 2013-06-12T15:08:11Z Unfortunately, that is <i>not</i> the case. The product and intersection of ideals can be quite different even in that case. For instance, consider $k[x,y,z]$ and $I_1 = (x,y), I_2 = (y,z)$. Then $$I_1 \cdot I_2 = (xy, xz, yz, y^2)$$ but $$I_1 \cap I_2 = (xy, xz, yz, y).$$ The blowups are different too. http://mathoverflow.net/questions/133253/how-to-define-the-canonical-sheaf-on-singular-varieties/133254#133254 Comment by Karl Schwede Karl Schwede 2013-06-10T05:40:23Z 2013-06-10T05:40:23Z It doesn't come from there exactly. The right way to obtain it is from Hom'ing the exact sequence $$0 \to O_X(-D) \to O_X \to O_D \to 0$$ into $\omega_X$, and noting $Ext^1(O_D, \omega_X) = \omega_D$. The next Ext vanishes which gives you surjectivity on the right. There are no other Ext's for obvious reasons (at least as long as $D$ is Cartier). This isn't birational geometry, it's Grothendieck/Serre-duality theory. Perhaps you could look at Residues and Duality by Hartshorne? http://mathoverflow.net/questions/133253/how-to-define-the-canonical-sheaf-on-singular-varieties/133254#133254 Comment by Karl Schwede Karl Schwede 2013-06-10T03:42:07Z 2013-06-10T03:42:07Z To me, $K_X$ is a divisor, $\omega_X$ is a sheaf, $\omega_X \cong O_X(K_X)$. This is probably the most common convention in birational geometry but not the only one. For your problem, let me give you a hint: do it one $D_i$ at a time (I'm assuming the $D_i$ are Cartier divisors?). Note that we have a short exact sequence: $$0 \to \omega_X \to \omega_X(D_1) \to \omega_{D_1} \to 0.$$ Then do the short exact sequence $$0 \to \omega_{D_1} \to \omega_{D_1}(D_2|_{D_1}) \to \omega_{D_2|_{D_1}} \to 0.$$ Repeat, and keep track of everything, you'll get the answer you want. http://mathoverflow.net/questions/133253/how-to-define-the-canonical-sheaf-on-singular-varieties/133254#133254 Comment by Karl Schwede Karl Schwede 2013-06-10T03:27:21Z 2013-06-10T03:27:21Z Oh, let me explain this perhaps in a better way. Every integral scheme is S1, and hence Cohen-Macaulay outside a set of codimension 2. Thus for some purposes you can just work on the Cohen-Macaulay locus. http://mathoverflow.net/questions/133253/how-to-define-the-canonical-sheaf-on-singular-varieties/133254#133254 Comment by Karl Schwede Karl Schwede 2013-06-10T03:05:23Z 2013-06-10T03:05:23Z Indeed, if $X$ is Cohen-Macaulay, then $\omega_X$ is the dualizing sheaf. But the canonical sheaf is still defined by the same Ext formula even for non-Cohen-Macaulay schemes. If your singularity is regular in codimension-1, then you can still use the pushforward from the regular locus. What exactly is your adjunction problem? Maybe it would be better to state that. http://mathoverflow.net/questions/132757/is-there-a-prime-of-height-i-in-support-of-hi-ir Comment by Karl Schwede Karl Schwede 2013-06-05T15:12:39Z 2013-06-05T15:12:39Z I'm sorry, I was thinking the vanishing locus of the annihilator of the module, not the support. http://mathoverflow.net/questions/132757/is-there-a-prime-of-height-i-in-support-of-hi-ir Comment by Karl Schwede Karl Schwede 2013-06-04T19:19:53Z 2013-06-04T19:19:53Z If $R$ is Cohen-Macaulay and say $I$ is a maximal ideal, then most of these are zero except for $H^{\dim R}_I(R)$, which has support equal to $\text{Spec }R$. In particular, $H^{\dim R}_I(R)$ has primes of height zero, height one, etc. So I don't see why primes in the support of $H^i_I(R)$ have height at least $i$ either... Do you have a particular ring / ideal combination in mind? http://mathoverflow.net/questions/132614/ring-of-even-characteristic Comment by Karl Schwede Karl Schwede 2013-06-03T02:15:24Z 2013-06-03T02:15:24Z I think there are such examples of rings not containing a field with this property. For example: $$(\mathbb{Z}/4)[x,x^{-1},(x+1)^{-1}].$$ Then setting $u = x$, $v = x^2$ and hence $w = 3x+3x^2 = 3x(1+x)$ should do the trick. I can see why you might ask this though. In $\mathbb{Z}/(2n)$, units are always odd, so the sum of three of them is also odd and hence still odd modulo $2n$. However, I expect that there are even a number of easy examples that are finite extensions of $\mathbb{Z}/(2n)$. http://mathoverflow.net/questions/132614/ring-of-even-characteristic Comment by Karl Schwede Karl Schwede 2013-06-03T01:09:24Z 2013-06-03T01:09:24Z Perhaps you want, &quot;does not contain a field&quot;? http://mathoverflow.net/questions/132614/ring-of-even-characteristic Comment by Karl Schwede Karl Schwede 2013-06-03T00:43:00Z 2013-06-03T00:43:00Z Sure, take $R = \mathbb{F}_{2^n}$ or any field of characteristic $2$ with more than 2 elements. Choose a nonzero element $u$, another nonzero element $v$ such that $u + v \neq 0$ and then set $w = u + v = -(u + v)$ because we are characteristic $2$.