Uhhm, I have a question that might be interesting but hard to prove lol
So suppose I have a number field $K$ and a nonzero ideal $\mathfrak{a} \subseteq \mathcal{O}_K$ with the property that $\mathfrak{a}^m = (a)$ for some $a \in \mathcal{O}_K$. Can I find a finite extension $L/K$ such that $(m, h_L) = 1$?
(and such that $h_L \neq 1$)
The question stems from an exercise to prove that there exists an extension $L/K$ such that $\mathfrak{a}\mathcal{O}_L$ is already principal. I can take $K(\sqrt[m]{a})$ and that should be fine but it's kind of a boring answer, so I thought about taking the induced group hom. on the class groups of $K$ and $L$, then the image of $[\mathfrak{a}^m]$ under that hom is trivial, and if $(m, h_L) = 1$ then the image of $[\mathfrak{a}]$ itself is trivial :P
(I could also take the Hilbert class field of $K$ but for the purposes of the exercise I'd have to prove that exists rofl)
If you have finite models of arbitrarily large size you have an infinite one by compactness and if the size of the models is bounded there's finitely many ways to interpret each symbol
If you have an infinite language most of the symbols will be interpreted in the same way so I don't think that's an issue
Ah ok so infinitely many can be done by having $n$-ary relations for every $n$ in the language
I don't see how to get past countable though, for every fixed $k$ there's a finite number of interpretations of every symbol and at most countably many symbols that can have different interpretations
@LeakyNun given K/F is a finite extension, Hom_F(K, Falg) is equal to |Spec(K otimes Falg)| since the former counts the number of geometric points in Spec K lying over a specific geometric point in Spec F.
if K is purely inseparable finite extension of F, you can decompose it as a tower of 1-generated inseparable extensions of degree p, so note that if K = F[x]/(x^p - a) then K otimes Falg is isom to Falg[x]/(x^p), which is has a unique prime. So if K is purely inseparable finite over F, |Spec(K otimes Falg)| = 1
fiber is multiplicative so the only part that contributes for a general finite extension K/F is the separable part E/F
Also in general is kinda hard to get $\aleph_1$-many things, for example if the language is countable then a theory can have finitely many, $\aleph_0,\aleph_1$ or $\mathfrak c$ countable models, but whether $\aleph_1$ is possible is provably is open
so an F-algebra homomorphism $K \to \overline{F}$ is in 1-1 correspondence with a morphism $\text{Spec} \overline{F} \to \text{Spec} K$ of $\text{Spec} F$-schemes, where $\text{Spec} \overline{F}$ has been given a $\text{Spec} F$ scheme structure by choosing an embedding $F \to \overline{F}$
These therefore correspond to commutative triangles of maps $\text{Spec} \overline{F} \to \text{Spec} K$, $\text{Spec} \overline{F} \to \text{Spec} F$ and $\text{Spec} K \to \text{Spec} F$.
@LeakyNun It's the fiber over the point, right? If $f : Y \to X$ is a map and $x : \{*\} \to X$ is a point in $X$, the fibered product $\{*\} \times_X Y$ is precisely $f^{-1}(x)$
It's a neat way to think about things, I think, and I don't think I am missing a subtlety here
From the above, we're really counting morphisms $\text{Spec} \overline{F} \to \text{Spec}(\overline{F} \otimes_F K)$ of $\text{Spec}\overline{F}$-schemes. These just correspond to $\overline{F}$-algebra homomorphisms $\overline{F} \otimes_F K \to \overline{F}$, so prime=maximal ideals of $\overline{F} \otimes_F K$, really.
Now naively it's easy to see why your result holds, because for every degree of inseparability there's a nil algebra factor in $\overline{F} \otimes_F K$
and for every degree of separability there's a $\overline{F}$
But I'll write down a serious proof if this is too much pretension :P
For each normed vector space $(X,\|\cdot\|)$, let $\Lambda_X:X\to X^{**}$ be the canonical isometric linear map defined as $\Lambda_X(x)(f)=f(x)$, for all $x\in X$ and $f\in X^*$. Assume $X$ is complete with respect to $\|\cdot\|$, so a Banach space.
How can you show $\Lambda_X:X\to X^{**}$ is surjective iff $\Lambda_X:X^*\to X^{***}$ is surjective?
Oh sure, I just didn't understand what it meant to have just finished your second year of a maths degree, but not having done maths in a very long time (which I'd normally take to mean like 2 years or something)
@Adam Also I don't know who that is, and I don't smoke weed (and in particular, I don't get the reference) [Also, you said you wouldn't be offended, which is the only reason I answered...]
i've been using Artin for linear algebra over the last couple weeks but i'm tempted to look at Axler; i remember his proofs, while not being very hands-on, really made me feel like i was getting something
yeah sort of the same with me doing galois theory from Dummit-Foote, the thing is i dont think i "understood" when i did these problems a long time ago
i usually go on detours thinking about related questions when im doing these problems
@LeakyNun Let me write down a very precise story for the separability stuff without the scheme-theoretic stuff anyway. Suppose $F$ is a field of characteristic $p$ such that $\alpha \in F \setminus F^p$. Then $x^p - \alpha$ is irreducible in $F[x]$ because in the algebraic closure of $F$, $x^p - \alpha = (x - \alpha^{1/p})^p$, so by unique factorization in $\overline{F}[x]$, any factor must be of the form $(x - \alpha^{1/p})^n$ where $n < p$ but this cannot be in $F[x]$ because $\alpha^{n/p} \notin F$ since if not, we take $m$ such that $mn \equiv 1 \pmod{p}$, and take the $m$-th power to o…
Finally, let $\varphi : K \to \overline{F}$ be an $F$-homomorphism. Then $\varphi(\beta_1)^p = \beta_1^p$, but the only root of $x^p - \beta_1^p = (x - \beta_1)^p$ in $\overline{F}$ is $\beta_1$, so $\varphi(\beta_1) = \beta_1$. By induction, $\varphi(\beta_i) = \beta_i$ as $\varphi(\beta_i)^p = \beta_i^p$. So $\varphi$ fixes all the $\beta_i$'s, which means $|\text{Hom}_F(K, \overline{F})| = 1$
But yeah now for any extension $K/F$, take the decomposition $K/E/F$ where $E/F$ is separable and $K/E$ is purely inseparable, then $|\text{Hom}_F(E, \overline{F})|$ is the degree of $E/F$ and any $F$-homomorphism $E \to \overline{F}$ uniquely extends to an $F$-homomorphism $K \to \overline{F}$ by whatever said above
So $|\text{Hom}(K, \overline{F})|$ is indeed $|E : F|$
@LeakyNun $E/F$ is finite separable, so choose a primitive element $E = F(\alpha)$. $F$-automorphisms of $\overline{F}$ act transitively on the roots of the minimal polynomial of $\alpha$ by before, so
The minimal polynomial has $[E : F]$ many roots by separability
@LeakyNun This is observing that once you choose an $F$-embedding $\varphi : L \to \overline{K}$ you can extend that in $\text{Hom}_L(M, \overline{K})$-many ways to an $F$-embedding $M \to \overline{K}$.
i trick myself into thinking i understand the material because i'm better at doing handwavy conceptual exercises but then i can't actually do computations
yeah they're not exactly like those abstract-nonsensey arguments that make you think you understand algebra but then you can't actually work with elements
analysis-1 was taught by an algebraic number theorist who works with the indian math olympiad camp so it was just a lot of tricky stuff; its still fine i guess. no integration in first course, which is weird
analysis-2 was super weird, it was mostly metric spaces and then multivariable with a little bit of riemann integration done badly squished between.
In some sense, though, you can think of it as a machine that takes something geometric/topological, that is unwieldy, and converting it to a convenient algebraic form that one can actually do things with (and is invariant under whatever you want). I'd still say it's algebraization of homotopy theory in some sense.
If $k$ is an integer, and $X$ is a finite CW-complex, $y\in K_R(X)$, then there exists a non-negative integer $e$ such that $k^e(\Psi^k-1)y$ maps to zero in $J(X)$.
