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May
21
answered A question regarding etale morphisms of affine varieties
May
21
comment A question regarding etale morphisms of affine varieties
I suppose you mean to either have $p$ surjective, or "locally isomorphic" instead of "isomorphic".
May
20
comment Does every category have a functor?
Your answer is false if taken literally: $1$ does not embed into the empty category, and the category $1$ only has the identity endofunctor.
May
20
comment Does every category have a functor?
Of course, if $D$ is empty...
May
20
comment Question about comultiplication
This is formally dual to the following problem: Suppose $A$ is an algebra over the field $k$. Show that the multiplication map $\mu : A \otimes A \to A$ is an algebra map if and only if $A$ is commutative.
May
19
comment ZF Extensionality axiom
The converse is a logical tautology, however.
May
19
comment Finishing the proof that $\textrm{Spec}K[X,Y]\setminus\{(X,Y)\}$ is not an affine scheme
$f^*$ is an isomorphism, but whether or not it is the identity depends on the explicit construction of the structure sheaf etc.
May
19
comment Finishing the proof that $\textrm{Spec}K[X,Y]\setminus\{(X,Y)\}$ is not an affine scheme
No, you start with $f$ and then calculate $f^*$.
May
19
comment Finishing the proof that $\textrm{Spec}K[X,Y]\setminus\{(X,Y)\}$ is not an affine scheme
You can do that if you like, but that is not necessary.
May
19
answered Finishing the proof that $\textrm{Spec}K[X,Y]\setminus\{(X,Y)\}$ is not an affine scheme
May
19
comment Does $\,x>0\,$ hint that $\,x\in\mathbb R\,$?
It looks like 1.5cm to me...
May
18
comment Degree of effective Cartier divisor
Perhaps the degree is supposed to be a function, rather than a number? After all, the LHS of the equation appearing in Theorem 8.3 is also a function...
May
18
revised Uniformly solvable families of polynomials
edited tags
May
18
answered Uniformly solvable families of polynomials
May
18
comment Uniformly solvable families of polynomials
It's basically a translation of your problem into algebraic geometry. Rather than thinking of $f(a_1, \ldots, a_n; x)$ as a family of one-variable polynomials, think of it as just one; then it defines a hypersurface $V$ in $U \times \mathbb{A}^1$ where $U$ is an open subset of $\mathbb{A}^n$, and what you want to do is to study the properties of the projection $V \to U$ in terms of the generic fibre.
May
18
comment Topology of the Segre product vs. the product topology
I think you've missed the point, though: it's not important that the product of varieties very rarely has the product topology; rather, it's more important to know that it doesn't happen in all cases!
May
18
comment Uniformly solvable families of polynomials
Very interesting! Perhaps it would be better to consider the question in the setting of commutative algebra: assume $f$ is monic, irreducible, and separable; let $A$ be the subring of the coefficient field generated by $\mathbb{Q}[a_0, \ldots, a_k]$ and the coefficients of $f$, and consider the $A$-algebra $A [x] / (f)$. Then the automorphism group of $A [x] / (f)$ (as an $A$-algebra) will also act on each of the specialisations $A [x] / (f) \otimes_A A / \mathfrak{m}$, for each maximal ideal $\mathfrak{m}$ of $A$, and the "generic" situation is obtained by considering $\mathfrak{p} = (0)$.
May
18
awarded  Constituent
May
18
awarded  logic
May
18
awarded  Good Question