MoarCake559

If needed, take a break

@Evinda Try writing out what I said on paper and then make a decision for yourself.

bump: Given $k = \mathbb{Q}_p$ and $k_b=k(\sqrt{b})$ for $b \in k^*$, does anyone know what the group of norms of elements $Nk_b^*$ is?

what is the square root of both sides?

c^{\frac{p+1}{2}} = c

Since we know c has a square root, there is a well-defined c^{\frac{p+1}{2}}

and if you multiply c^{\frac{p-1}{2}} by c, then you get c^{\frac{p+1}{2}}

this is one of the definitions for the Legendre symbol

yes

If there is a square root...

@Evinda do you agree that $c^{(p-1)/2} = 1$?

Given $k = \mathbb{Q}_p$ and $k_b=k(\sqrt{b})$ for $b \in k^*$, does anyone know what the group of norms of elements $Nk_b^*$ is?

Does anyone know of a good reference which discusses the classification of absolute values over a number field/ring of integers?

Does anyone know of any good toy examples of Kan complexes?

which is the tensor product of k-algebras

The product is completely different

A_f in the zariski topology corresponds to an open subset of $A$

The algebra is telling you that the pushout of the element in $A$ must be invertible, so it is forcing that to happen.

Okay cool.

In this case Spec(C[x,y]/(xy-0))

they are just divisors whose singularities look like transverse intersections

spec(C[x,y,z]/(xyz)) is an example of one

You definitely should mix algebraic and geometric intuition with your original question

The fiber product of the family of arrows gives you different hyperbolas with a normal crossing divisor above 0.

Fiber products over a point is just the standard set theoretic product for varieties. You have to be careful with the scheme structure.

This is all there is.

The book I sent shows you how pushouts and tensor products of algebras work.

I really wouldn't think of this as taking colimits of diagrams of algebras. Schemes do not behave well with respect to colimits.

Try and find access to springer.com/us/book/9781447148289

The key point is that the construction of pushouts of algebras is from taking tensor products.

No. This is true even if you are taking a map of k-algebras

Sure

Now, you have two arrows of C[t] algebras. If you take the tensor product, and take the functor into affine schemes, all the arrows reverse and you get a pullback.

I have no idea what barr exact means, but okay

It corresponds to a morphism C[x] -> C with kernel the maximal ideal

A ------------> B

------------>

(Hint: think categorically)

yes, but what else?

Okay. Then what does a maximal ideal correspond to?

Maximal ideals are in one to one correspondence with points

Do you know the statement of the nullstellensatz for C[x]?

Hint: Nullstellensatz and tensor products

But geometrically, yes

It's the structure map from C[t] -> C[t,x,t]/(xy - t)

What is the pullback?

and take a point Spec(C) -> Spec(C[t])

Then consider the canonical morphism to Spec(C[t])

yes

do you understand what Spec(C[t,x,y]/(xy - t)) looks like?