What do you call this property?

between points, $(x,y)$ and $(1/y,1/x)$

they are related by switching the order of the coordinates and then inverting them

@Ultradark I would just say that - it's certainly concise

okay great

Looking for a little help: I've run in to someone who claims that $k[x-y]_{(x-y)} \to (k[x,y]/(xy))_{(x,y)}$ is a flat map of local rings. On the other hand, I know that every flat module over a local ring is free. I do not believe that this is a free module, therefore it shouldn't be flat, but I don't immediately see how it violates flatness.

I think the statement is every f.g. flat module over a local ring is free? Since otherwise Q_p being flat over Z_p seems to be a counter-example

Sure, I meant finitely generated.

I don't think your map is module-finite

maybe it is

That's what I said, but I they claim that $x,y$ is a generating set

as a module?

yes

that actually seems geometrically plausible

Right? It's the x and y axes projecting on to the line x=y

So either I had better be able to write k[x,y]/(x,y)_{(x,y)} as a direct sum since it's a free module, or it's not flat. I feel like a real dunce for not immediately seeing why it isn't flat, which isn't helping in my attempts to demonstrate this.

@Ultradark if you take the coordinates to be u=log x, v=log y, then your relation becomes that of (u,v) and (-v,-u)

Which corresponds to reflection across the u=-v axis

Of course, this only makes sense if x,y>0

Or, perhaps easier, reflect across diagonal and then reflect through (or rotate 180º around) the origin.

Yeah, I saw that too but liked having just one operation

how did you decide to take the coordinates to be those?

Because log(1/x)=-log(x)

oh

Plus, log-log plots (along with log plots and semi-log plots) are familiar tools in physics

@Semiclassic, I like reflecting across the diagonal because then I'm looking at the inverse function (relation). :)

Log-log plots are kinda magic when you first see them

It’s just a change of variables, of course

Guys need some help : math.stackexchange.com/questions/3465754/…

@KReiser i don't see why this is true - (in fact i don't think you can make 1 - but maybe you meant to take something like 1,x instead)

anyway i think this map is flat - this map is projective and hilb poly is constant

@loch they amended their claim to 1,x being a generating set. How does this square with the proof that an fg flat module over a local ring is free?

I am just so incredibly sure that fg flat over a local ring implies free: fg implies finitely presented, finitely presented flat is equivalent to projective, and projective over a local ring is actually free.

@AlessandroCodenotti hi

Good evening @Leaky

@KReiser even in that case, it's not obvious to me that that's a generating set! maybe i'm being dumb though

@loch but geometrically two-lines is module-finite over one-line right

@LeakyNun yes the morphism is finite

well

i meant to say without all the localizatoin the morphism is finite

and module-finite should be preserved under fibres?

@loch I agree. I am feeling very silly here: I think I ought to be able to swat this whole thing away without too much effort, but I am not getting there.

but the fibre of 0 is just 0 right

yes

so...?

so..

@KReiser i'm somewhat convinced that you can't make 1/(x+1) from 1 and x from messing around with the algebra -- so hopefully that shows that 1,x don't generate the module..

@loch That looks promising. Thank you for your efforts to help!

@Ultradark want to study now?

@loch ah yes, the floor is made of floor

:0)

@shi let's do tomorrow, I'll read the next few pages