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| visits | member for | 1 year, 7 months |
| seen | May 5 at 6:49 | |
| stats | profile views | 2,452 |
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May 3 |
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A question regarding linear disjiontness and the degree of a field extension Yes if $L, K$ are algebraic extensions of $k$ (otherwise, please define linear disjointness). Note that $K'\cap L$ is linearly disjoint with $K$ over $k$, so the compositum $(K'\cap L)K$ has degree over $K$ equal to $[K'\cap L : k]$. But $(K'\cap L)K\subseteq K'$.... (this works in any characteristic). |
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May 3 |
reviewed | Close Advice Commutation algebra |
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May 3 |
reviewed | Approve suggested edit on Is there any prime $p$ such that $(p-1)!+1=p^m$ |
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May 3 |
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Inverse images under universally injective morphisms One can also use Stein factorization : $X\to {\bf Spec}(f_*O_X)\to Y$. It is known that the first map has geometrically connected fibers, and the second has geometrically connected fibers by the universally injective hypothesis. |
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May 3 |
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Inverse images under universally injective morphisms deleted 90 characters in body |
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May 3 |
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Inverse images under universally injective morphisms Heidar Svan: write @QiL'8 instead of QiL in your comment if you want me be aware of your comment. |
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May 3 |
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Inverse images under universally injective morphisms @HeidarSvan: I am very sorry, I confused the morphism with the map $O_Y\to f_*O_X$. Your initial statement is correct. When $\mathrm{Spec}(f_*O_X)\to Y$ is universally injective, it is purely inseparable (see "Algebraic geometry and arithmetic curves", 5.3.13). For any $y\in Y$, $k(y)\to H^0(X_y, O_{X_y})$ is then purely inseparable (op. cit., 5.3.14). So, when you extend to an algebraic closure $\bar{k}$ of $k(y)$, the global sections of the structure sheaf of $X_{\bar{k}}$ is still a local algebra, hence $X_{\bar{k}}$ is connected. |
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May 3 |
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Class group of $k[x,y,z,w]/(xy-zw)$ @xyzzyz: No problem, I just wanted to give the information for arbitrary fields. It is true that all books on toric varieties I know are written for $\mathbb C$. Now can you find out the rational convex polyhedral defining the variety we are interesting here and give its class group by this way ? |
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May 3 |
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Class group of $k[x,y,z,w]/(xy-zw)$ @xyzzyz:Toric varieties are defined for any field and what you said is valid over any field. |
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May 2 |
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Class group of $k[x,y,z,w]/(xy-zw)$ Suppose $n[Z]$ is a principal divisor defined by a rational function $f$ on $X$. Then $f$ is regular and invertible on $U:=X\setminus Z$. Now compute $O_X(U)^{\star}$ (invertible elements, you should find $k^*y^{\mathbb Z}z^{\mathbb Z}$) and observe that the support of $\mathrm{div}(f)$ is then always reducible, hence different from $Z$. |
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May 2 |
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If a maximal ideal $\mathfrak m$ is flat, then $\dim_k (\mathfrak m/\mathfrak m^2) \leq 1$ Your are welcome @Louis. |
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May 2 |
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The genus of a curve with a group structure if your morphism to $\mathbb P^1$ is injective, over an algebraically closed field of characteristic $0$, it will be an isomorphism because the degree of the morphism will be one. See this question math.stackexchange.com/questions/341281. The easiest proof that $\mathbb P^1$ is not an algebraic group I know is again the answer given by Piotr. |
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May 1 |
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Intersection of powers of maximal ideals @YACP: thanks, I will have a look when I will back to home. |
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May 1 |
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Intersection of powers of maximal ideals continued: for smooth subvarieties, the proof can be easier because locally a smooth subvariety is defined by a part of a system of coordinates, so you can imagine the ambient variety is an affine space with coordinates $t_1, \dots, t_n$ and $\mathfrak p$ generated by $t_1, \dots, t_d$ for some $d\le n$. Then computation in this case is easy. |
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May 1 |
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Intersection of powers of maximal ideals @Karl: this should be well-known. I just did some computations by myself. For example, in the case $s=2$, if $f$ belongs to the intersection of $m^2$, then its differential $df$ vanishes at all points $V(\mathfrak p)$. Using the case $s=1$ and the fact that cotangent sheaf of the ambient space is locally free, we see that $df$ has coefficients in $\mathfrak p$. "Integrating $df$" then implies that $f\in \mathfrak p^2$. Higher powers are more complicated especially in positive characteristics. Note that locally complete intersection subvarieties include smooth subvarieties... |
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May 1 |
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Intersection of powers of maximal ideals @YACP: ah sorry, I was referring to your comment. I don't have acccess to Math. Ann. articles for the moment. |
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May 1 |
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If a maximal ideal $\mathfrak m$ is flat, then $\dim_k (\mathfrak m/\mathfrak m^2) \leq 1$ @YACP: thanks for your kind words ! I am kind of burn out at this moment, there are also some other reasons I prefer not to comment on (!) here. |
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May 1 |
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If a maximal ideal $\mathfrak m$ is flat, then $\dim_k (\mathfrak m/\mathfrak m^2) \leq 1$ @Louis: sorry I don't have a link. Try to find the book in your library, the proof I alluded to is in Prop. 3G, p. 21. Projectivity is not needed. And yes, a submodule of a flat module is not necessarily flat. But the proof doesn't rely on this assumption. |
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May 1 |
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If a maximal ideal $\mathfrak m$ is flat, then $\dim_k (\mathfrak m/\mathfrak m^2) \leq 1$ @YACP: I am in comment only mode since a few weeks. |
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May 1 |
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The genus of a curve with a group structure How do you prove the finiteness of Aut(C) using just Riemann-Roch ? Your proof of $P^1$ having no group structure is incomplete. You can't pull-back a group law by $f$ (moreover $f$ is never injective). |
