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Here's another type of counterexample not (as I write) ruled out by the hypotheses of the question: consider an inclusion of fields $K\to L$, with $K$ and $L$ finite extensions of $k$. Now take the sp …

The kernel of an isogeny between abelian schemes is flat. The reason is that it's true over a field, and then you use the fibrewise criterion for flatness. I don't know of any other proof. So there's …

Sounds to me like you don't want to hear the proof, but you want to hear "the point". The point is that a scheme must by definition be covered by affine schemes, and sometimes when doing exercises in …

I don't think this will be true in general. Say $K=\mathbf{Q}$ and $K'=\mathbf{Q}(\sqrt{2})$, and let $X_0$ be $Spec(R')$. Then $X_0$ is regular of dimension 1 and the map down to $S$ is projective a …

IIRC there's an example due to Gabber in the book by Katz and Mazur of a representable moduli problem where objects have automorphisms (I forget the trick---perhaps he rigged it so that every object h …

http://en.wikipedia.org/wiki/Division_polynomials That's not the best wikipedia page. "The division polynomials form an elliptic divisibility sequence." is mentioned well before the far more importan …

Passing comment: Mumford's GIT constructs modular curves as quotients---not of the upper half plane, but of some parameter space of subspaces of projective space, by an algebraic group. As for the las …

I am surprised that Brian got to this one first without making what I thought was another obvious comment: affinoids are Jacobson rings! A function which is zero at all points of an affinoid rigid spa …

Your proof seems wrong to me. I might be misunderstanding some things you wrote, but surely $\mathbf{Q}{}_p=\mathbf{Z}_p[X]/(pX-1)$ is finite type over $\mathbf{Z}_p$, and contains many elements which …

Here's a suggestion. Use polcompositum(FA,FC) (with FA the min poly of A, FC the min poly of C) to find a number field K=Q(alpha) containing roots of both your polynomials, and then use lindep() to fi …

Here's an example where $X(\mathbf{Q})$ is Zariski-dense but the first inequality is not an equality. Let $X$ be an elliptic curve over the rationals, such that the group $X(\mathbf{Q})$ is isomorph …

If U is an open set in X, but U isn't X, then there are non-zero sheaves on X whose support lies outside U. Now add O_X to one of these to get a counterexample.

Look at the subspace of $\mathbf{A}^{n+1}$ cut out by your polynomials. This set is invariant under the diagonal action of $k^\times$. So the functions that vanish on it will be an ideal $I$ (the radi …

On the contrary, some conjectures suggest that the answer is NO! It follows from the Bombieri-Lang conjecture (sometimes known as Lang's conjectures) that a uniform bound should exist. More precisel …

This is inspired by Does "all points rational" imply "constant" for this "cubic" curve over an arbitrary field? . Say $K/F$ is a finite separable extension of fields. Assume $F$ is infinite (or el …