Can you show that $M \otimes_A A_f \cong M_f$ for some $A$-module $M$?

@user251222 I'm not sure I understand why this is related to the question, but using the universal property of tensor products and the concrete constructions of $A_f,M_f$ with fractions, I can prove $(m,\frac rs)\mapsto \frac{rm}s$ is the inverse of $\frac ms\mapsto m\otimes_A\frac 1s$... Incidentally, is there a slicker proof?

The point of my comment is that tensor products of algebras can be constructed by taking tensor products of the underlying module and finding a reasonable algebra structure. The obvious algebra structure here works. Since $f$ takes $x$ to an element in $A$ in the left- vertical arrow and the top right arrow inverts $x$, you are forcing this element in $A$ to be inverted in the pushout.

@user251222 that was the reasoning which led me to believe this square is a pushout, but I don't think there's any need to refer to tensor products or modules - clearly an arrow out of $k[x,x^{-1}]$ amounts to picking an invertible element. Am I right in saying this square is a pushout in the category of $k$-algebras?

No, you are taking a pushout of $k[x]$-algebras. Pushouts of algebras are given by the tensor product. If you know anything about schemes, then this corresponds to the pullback.

@user251222 sorry, I'm still confused. Are you saying the square is a pushout in the category of $k[x]$-algebras? We can look at $k[x,x^{-1}]$ as the localization $k[x]_x$ which fits in with your tensor product point of view.

Yes. That is correct

@user251222 hello! Thanks for your patience

you're welcome

Great. Now, the arrows from the top left corner make the algebras in question all $k[x]$-algebras, but I think the colimit itself is taken in the category of $k$-algebras.

Why?

This is a completely different object.

[

Because $A_f$ is a $k$-algebra which verifies the universal property of the pushout in $k$-algebras: this pushout amounts to a universal inversion of $f\in A$, no?

$A_f$ is a $k[x]$ algebra

You are looking at a diagram of $k[x]$-algebras

yes, but then you are not taking a pushout

I disagree - we don't need to actually add the arrow to the diagram - its mere existence means we're living in $k$-algebras

(sorry for being dense)

And everything is a $\mathbb{Z}$-algebra, but we are not taking tensor products of $\mathbb{Z}$-algebras

Once you understand what affine schemes are all about, you should revisit this question. Taking the tensor product $A\otimes k k[x,x^-1]$ is a completely different space than $A \otimes_{k[x]} k[x,x^-1]$

I'm familiar with affine schemes. Could you explain the geometric intuition?

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

A family of hyperbolas?

yes

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

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

What is the pullback?

The canonical morphism is projecting to the $t$ axis?

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

But geometrically, yes

Ok, with you so far. Then taking the pullback... we get... (thinking)

Hint: Nullstellensatz and tensor products

I'm lost :(

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

For starters $xy-t$ looks irreducible...

Maximal ideals are in one to one correspondence with points

That much I know :)

Okay. Then what does a maximal ideal correspond to?

An irreducible polynomial

yes, but what else?

(Hint: think categorically)

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

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

I'm not following :(

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

ohh okay, I thought that was overthinking it - that algebras are Barr exact

Okay, sure

I have no idea what barr exact means, but okay

(It means equivalence relations are effective, which in this setting boils down to saying kernels correspond to ideals)

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.

Can I interrupt really quickly for just a second?

Sure

I think I intuitively realized why the pushout is of $k[x]$-algebras from an algebraic perspective. We need the top left corner to be $k[x]$ so that arrows will be determined by elements.

Does that make sense or is it just handwaving and missing the point?

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

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

Ah, right - the top left corner is part of the diagram of which we take a colimit...

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

Got it

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

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

This is all there is.

I'll definitely learn this stuff properly. Thank you. Any chance for that geometric explanation?

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

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

(sorry, what is a normal crossing divisor? English is not my native language)

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

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

they are just divisors whose singularities look like transverse intersections

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

Okay, I think I understand

Okay cool.

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

Okay, that much was clear to me

Just confused about why we must take the pushout of $k[x]$-algebras

The biggest formal hint, now that I think about it, is really the tensor product notation

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

The product is completely different

which is the tensor product of k-algebras

Writing $k[x,x^{-1}]$ as $k[x]_x$ we should write $k[x]_x\otimes _{k[x]}A$, not $k[x]\otimes _k A$

Ok, it's starting to make sense now from the affine scheme perspective as well

I see why you pointed out the tensor product - it guides intuition

Okay, I think I'll take some time to work things out tomorrow (it's night here).

Thank you very very very much for all your time. If you feel like posting a short answer, I'll gladly accept it.