@user101036: you should think of symmetric functions in terms of schur functors. for example, e_n corresponds to the nth exterior power, and h_n corresponds to the nth symmetric power. then plethysm is precisely composition of schur functors

there's a very nice universal story you can tell here. the schur functors are something like the universal endofunctors of Vect-enriched symmetric monoidally cocomplete categories

@ZhenLin or at least, pairs of whole numbers with a complicated and unintutive equivalence relation, to fractions.

@QiaochuYuan thanks, that is very helpful!

Is the fiber of the multiplication map $S\wedge S\to S$ contractible?

(for $S$ the sphere spectrum)

yes

equivalently, the multiplication map is an equivalence

Lol. oh my god...

i'm really killing it with the dumb questions here.

I mean that the base changes are iso of course

there's a cohomology group that contains obstructions to X and Y being isomorphic

I think it's H^1 of Gal(k'/k) with coefficients in Aut(X_{k'})

well, I guess it's a cohomology set in general

but if it were to be a one-point set, you could certainly conclude that X and Y were isomorphic

Right

So if say X_{k'} has no non-trivial automorphisms it is true if I interpret what you wrote correctly

do you have a reference for this btw?

not off the top of my head, unfortunately

If you have a morphism that becomes an isomorphism after base change then the original was already an isomorphism

@user101036: There is a chapter on this (which I have seen go by the name of Galois Descent) in Serre's book on Galois cohomology entitled "Forms"

@user101036 i think one has to be careful here, as @ZhenLin says, with whether or not one means "there is an isomorphism between.." or simply "they ARE isomorphic"

But yeah, as others have said, Galois extensions of fields are, in particular, faithful extensions, so I believe it follows from something like that.

An easy way to see the obstruction is as follows: let P= Iso(X_k',Y_k') this is a non -empty set with a gal(k'/k) action (by conjugation) and and a free transitive action of G:= Aut(X_k') (by compoistion ) Thus P is a principle homogenues G-space

P is trival iff it has a K-point iff X and Y are l-isomorhic

Maps X\to BG give principal G-bundles. Is G admits a double delooping, what do maps X\to BBG classify? Some kind of gerbe or something?

Also - Basterra and Mandell give TAQ(MU/S) as MU\smash bu and Blumberg, Cohen and Schlichtkrull give THH(MU/S) as MU\smash SU_+, and McCarthy and Minasian tell me that THH(MU/S)=P\Sigma TAQ(MU/S) (assuming MU is smooth over S? maybe it isn't???), where P is the free symmetric algebra functor. Does anyone know how these facts can be connected? Is it easy to show (if it's true) that the free symmetric algebra on a shift of MU\smash bu is MU\smash SU_+?

(where bu is the infinite delooping of the infinite loop space BU, and is the double suspension of ku)

I think it's unlikely that MU is smooth over S

aren't we supposed to think of S -> MU as a nilpotent extension? those things are never smooth

B^2 G is supposed to classify gerbes, but no one can tell me what a gerbe is!