I am a Part III student in Cambridge, focusing on Algebraic Number Theory.
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May
15 |
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accepted | Category of adjunctions inducing a particular monad |
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May
15 |
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asked | Category of adjunctions inducing a particular monad |
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May
6 |
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awarded | Caucus |
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Apr
17 |
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comment |
Alternating sum of multiple zetas equals always 1? The proof doesn't generalize to non-integer arguments since $z_m = x^{m-1}y$ only makes sense if $m$ is a natural number. But maybe the result for natural $m$ implies the result for non-integer $m$ by the identity theorem. If $m \mapsto 1-A_m$ is holomorphic on a connected open subset $D$ of the Riemann sphere containing $\mathbb N$ and the point at infinity, then since $\mathbb N$ has the point at infinity as limit point, the identity theorem forces $1-A_m$ to vanish on all of $D$. |
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Apr
17 |
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revised |
Alternating sum of multiple zetas equals always 1? fixed an error |
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Apr
17 |
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answered | Alternating sum of multiple zetas equals always 1? |
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Apr
16 |
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answered | field extension is Galois extension if and only if this extension are both normal extension and separable extension |
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Apr
7 |
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comment |
$f \in K[X]$ of degree $n$ with Galois group $S_n$, why are there no non-trivial intermediate fields of $K \subset K(a)$ with $a$ a root of $f$? You're right, that case needs to be taken care of separately. See the edit in my answer. |
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Apr
7 |
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revised |
$f \in K[X]$ of degree $n$ with Galois group $S_n$, why are there no non-trivial intermediate fields of $K \subset K(a)$ with $a$ a root of $f$? fixed the proof |
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Apr
7 |
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answered | $f \in K[X]$ of degree $n$ with Galois group $S_n$, why are there no non-trivial intermediate fields of $K \subset K(a)$ with $a$ a root of $f$? |
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Mar
28 |
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revised |
Proof on p. $16 \;$ of Lang's Algebraic Number Theory \mathfrak B should be \mathfrak P |
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Mar
28 |
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answered | Proof on p. $16 \;$ of Lang's Algebraic Number Theory |
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Mar
19 |
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comment |
Inverting formal power series wrt. composition Actually, I couldn't find a reference for that variant of Hensel's lemma. However, lemma 10.5 here (people.ds.cam.ac.uk/eb525/ec-notes.pdf) comes very close and I think the proof generalizes to the variant given above. Alternatively, look at lemma 11.6. This is exactly the statement you want to prove, and the proof actually goes along the lines of Hensel's lemma, by constructing successive approximations of the inverse power series. Integrality of $A$ is only needed for the uniqueness in Hensel's lemma, but since we didn't use that, it can probably be dropped. |
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Mar
19 |
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revised |
Inverting formal power series wrt. composition fixed typo |
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Mar
18 |
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answered | Inverting formal power series wrt. composition |
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Mar
18 |
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answered | Holomorphic function $f$ such that $f'(z_0) \neq 0$ |
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Mar
18 |
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answered | Kernel of $p$-adic logarithm. |
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Mar
18 |
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awarded | abstract-algebra |
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Mar
15 |
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answered | number of distinct prime Ideal of a ring |
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Mar
11 |
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comment |
Is all algebraic commutative operation always associative? See also here: math.stackexchange.com/questions/160945/… |
