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Oct
26
answered conics isomorphic to $\mathbb{P}^1$ over $\mathrm{Spec}(\mathbb{Z})$
Oct
1
revised Enriques Noether Theorem
bouville should be beauville
Oct
1
suggested suggested edit on Enriques Noether Theorem
Jun
7
comment Picard group of a smooth projective curve
Minor nitpick: Certainly, for all $r$, there is a number field $K$ and an elliptic curve $E$ over $K$ of rank $r$. Of course, I know you are referring to the "belief" that there are elliptic curves over $\mathbb Q$ with arbitrarily large rank (although I have the feeling that not everyone seems to agree with this "belief", right?).
Jun
7
comment Algebraic group with no $k$-rational points
Non-trivial examples are also provided by elliptic curves $E$ over $\mathbb Q$ with $E(\mathbb Q) = \{0\}$. There are many of those.
Jun
3
comment What is a “subscheme”?
This also looks relevant: users.mat.unimi.it/users/vezzani/Files/Research/…
May
31
revised Open set in the spectrum of a ring?
Added the words "non-empty" at some places
May
31
suggested suggested edit on Open set in the spectrum of a ring?
May
31
answered Algebraic Compact manifold originates from a proper scheme?
May
5
revised Applications of Belyi's theorem
added 94 characters in body
May
5
answered Applications of Belyi's theorem
Mar
14
comment Finite surjective morphisms between integral algebraic varieties and universal injectivity.
..to conclude that $X(k)\to Y(k)$ is not injective. If $X$ and $Y$ are not normal, you can probably work on the normalizations to get what you want. In fact, the normalization of $X$ maps to the normalization of $Y$ (by the universal property of normalization I mentioned before).
Mar
14
comment Finite surjective morphisms between integral algebraic varieties and universal injectivity.
Assume $X$ and $Y$ are normal. Normalize $Y$ in $L$ to get $Z$ and normalize $Y$ in $K(X)$ to get $X^\prime$. Since $X$ is normal, the morphism $X\to Y$ factors through $X^\prime$. (This is the universal property of normalization, and can be found in Liu's book in Chapter 4.1 I think.) So now you get $X\to X^\prime\to Z\to Y$, where we use that the normalization of $Y$ in $K(X)$ is the same as the normalization of $Z$ in $K(X)$. Now apply what I said to $X^\prime\to Y$ to see that $X^\prime(k)\to Y(k)$ is not injective. If you think a bit more, you can use this (in a slightly stronger form)..
Mar
14
comment Finite surjective morphisms between integral algebraic varieties and universal injectivity.
Assuming $X$ and $Y$ to be normal, I think you can see this by factoring $K(Y) \subset K(X)$ appropriately as $K(Y)\subset L\subset K(X)$, and normalizing $Y$ in $L$ to get $Z$. This only gives a rational map from $X$ to $Z$ though. I will think about it a bit more.
Mar
13
comment Finite surjective morphisms between integral algebraic varieties and universal injectivity.
Suppose that $f:X\to Y$ is not purely inseparable. Then $f$ factors as $g\circ h$ where $g:X\to Z$ is purely inseparable and $h:Z\to Y$ is separable. Note that the degree of $h$ is at least two. (Else $h$ would be an isomorphism and $f$ would be purely inseparable.) By your remark, there exists a point $y$ in $Y$ such that $h^{-1}(y)$ has at least two points. In particular, $f^{-1}(y)$ has at least two points. We conclude that $X(k) \to Y(k)$ is not injective (because the fiber over $y$ has more than two elements). This is what you want to show, right?
Mar
11
comment Noetherian schemes and varieties
The comments above already answer your question. Here's a more general statement. Let $X$ be a noetherian scheme. If $Y\to X$ is a finite type morphism of schemes, then $X$ is noetherian. (To answer your question: take $X$ to be the spectrum of a field...)
Mar
10
comment Isogeny of an elliptic curve
You are absolutely right!
Mar
9
comment projective varieties and locally trivial fibrations
Are you asking why the morphism $X\to Y$ is projective?
Mar
9
comment Isogeny of an elliptic curve
Of course, in your last statement, you are excluding the isogenies given by multiplication by $[n]$ on an elliptic curve. So you should say "$\mathbf Q$-rational isogenies of non-isomorphic elliptic curves", I think.
Mar
9
comment Locally complete intersection in a fiber
Do you really mean "local complete intersection" or do you mean that the $Y$-scheme $Z$ is a complete intersection Zariski locally on $Y$?