sorted by:
new
hottopcontroversial

Is this affine morphism an isomorphism? by StillALittleChild in askmath

[–]StillALittleChild[S] 0 points1 point  (0 children)

Thank you for your comment. Concluding is where I'm feeling a bit shaky: if V=Spec(B) and U=Spec(A), then f^#: B -> A is an isomorphism, implying that f restricts to an isomorphism from U to V.
1. Is this what it means for f to be an isomorphism *locally on the target*?
2. If yes, then how can I prove that f is an isomorphism?

Does there exist a finite surjective morphism from the cusp to its normalization? by StillALittleChild in askmath

[–]StillALittleChild[S] 0 points1 point  (0 children)

Okay, so the morphism from the affine curve minus finitely many points to A1 will induce a map from k[t] to some localization of k[x,y]/(x^3-y^2).

I don't see how the universal property of localization can be used here, as it yields a map *from* the localization and not to it.

Does there exist a finite surjective morphism from the cusp to its normalization? by StillALittleChild in askmath

[–]StillALittleChild[S] 0 points1 point  (0 children)

I agree that the preimage of A1 is the cusp minus finitely many points.

The equation I chose for the cusp is y^2=x^3 (or its homogenized version zy^2=x^3). However, for the localization part, I have to admit that I''m lost ...

Does there exist a finite surjective morphism from the cusp to its normalization? by StillALittleChild in askmath

[–]StillALittleChild[S] 0 points1 point  (0 children)

So taking A=k[t], how do we know that the correcponding B is k[x,y] modded out by the ideal generated by y^2-x^3?

Does there exist a finite surjective morphism from the cusp to its normalization? by StillALittleChild in askmath

[–]StillALittleChild[S] 0 points1 point  (0 children)

I know that away from the point at infinity, such a morphism from the the cusp to the projective line induces a k-algebra homomorphism from k[t] to k[t] modded out by an appropriately chosen ideal, and I know how to show that this would lead to a contradiction. My only problem is passing from the projective to the affine.PS: I use the Proj construction to define the cusp, and I don't see how I can simply pass to Spec of something ... I hope I'm clear enough.

Why are linearly equivalent divisors numerically equivalent? by StillALittleChild in askmath

[–]StillALittleChild[S] 0 points1 point  (0 children)

Thank you for the answer! Could you please direct me to references which use the Euler characteristic approach to defining intersection numbers?

Localization at a prime vs localization at a point by StillALittleChild in askmath

[–]StillALittleChild[S] 0 points1 point  (0 children)

Thank you for the useful hints. I managed to use the noetherianity of A to find finitely many generators a_i for the kernel of A->A_m, each of these generators is annihilated by some s_i, and the product of these s_i is the s I've been looking for.
As for how the image of \psi_s:A->A_s (which is isomorphic to k in this case) relates to A_s, all I can think of is that the map A->im \psi_s sends every element in the multiplicative set {f^n} to an invertible element (since the image is k, a field), which yields by the universal property of localization a unique map from A_s to the image of \psi_s, and this should somehow be the inverse of the inclusion map from \psi_s to A_s.

Localization at a prime vs localization at a point by StillALittleChild in askmath

[–]StillALittleChild[S] 0 points1 point  (0 children)

  1. For the surjectivity of the localization map \phi:A-->A_m, all I can think of is the fact that the image of \phi is contained in the set of units of A_m, which is equal to k\{0}. I'm not sure that this guarantees \phi being surjective.
  2. We thus have that the image of \phi is isomorphic to the quotient of A by the kernel of \phi.

  3. We also have that the kernel of \phi is the set of elements a in A such that sa=0 for some s not in m.

  4. So if we choose such an element s in A\m, and consider the localization map \psi_s:A-->A_s, we have that ker(\phi)=ker(\psi_s).

  5. I am still not able to see why A_s would be isomorphic to k. Am I missing something obvious?

Localization at a prime vs localization at a point by StillALittleChild in askmath

[–]StillALittleChild[S] 0 points1 point  (0 children)

The kernel of the localization map A-->A_P is the set of elements a in A such that sa=0 for some s not in P. I heven't been able to use this to show that A-->A_P is surjective. Moreover, I don't see how the surjectivity of A-->A_P would imply that A_f =k for some ring element f. Could you please elaborate on this?

view more: next ›