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| bio | website | |
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| visits | member for | 2 years, 7 months |
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| stats | profile views | 247 |
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May 15 |
comment |
Non-isomorphic simple extensions of the same degree of a field of positive characteristic Your approach will not work, because over the finite field $F_p$ (I assume you mean the field with $p$ elements) all irreducible polynomials are separable. However if you replace $F_p$ with a non-perfect field, then this works. It then remains to treat the case of a separably closed field $K$, that is a field that possesses only purely inseparable extensions ... |
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May 8 |
answered | Integral closure $\tilde{A}$ is flat over $A$, then $A$ is integrally closed |
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May 2 |
answered | Linear Transformations: Scaling along the line $y=x$ |
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Apr 30 |
revised |
every field of characteristic 0 has a discrete valuation ring? added 1 characters in body |
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Apr 30 |
comment |
every field of characteristic 0 has a discrete valuation ring? The answer to your question is "No". The reals do not carry discrete valuations for almost the same reason as for the complex numbers: one can take $n$-th roots of positve elements for every $n\in\mathbb{N}$. |
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Apr 29 |
revised |
every field of characteristic 0 has a discrete valuation ring? added 552 characters in body |
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Apr 29 |
comment |
every field of characteristic 0 has a discrete valuation ring? I do not agree with your statement: every field $K$ has a proper subdomain $R$ such that $K$ is the fraction field of $R$. Take a transcendence basis $T$ of $K$ over the prime field $P$ and consider the integral closure $R$ of the polynomial ring $P[T]$ in $K$. |
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Apr 29 |
comment |
$\mathbb A^n(k)$ and $\mathbb A^n(k)\setminus \{0\}$ are not homeomorphic @Rankeya: although you did not like my argument refering to spectral spaces, I restate it here: since $k^n\setminus 0$ for $n>1$ is not affine, it cannot be a spectral space, which is a purely topological property -- see en.wikipedia.org/wiki/Spectral_space I guess the problem lies in the behaviour of quasi-compact, open sets -- thus giving the opportunity to give a direct proof by contradiction. |
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Apr 28 |
revised |
every field of characteristic 0 has a discrete valuation ring? added 832 characters in body |
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Apr 28 |
comment |
every field of characteristic 0 has a discrete valuation ring? In my answer I was assuming that the DVR has fraction field equal to $\mathbb{C}$. Otherwise the statement has a trivial proof because every field of characteristic $0$ contains the rationals. |
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Apr 27 |
answered | every field of characteristic 0 has a discrete valuation ring? |
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Apr 18 |
comment |
Does every algebraically closed field contain the field of complex numbers? The cardinality of a transcendence basis of $\mathbb{C}/\mathbb{Q}$ equals the cardinality of $\mathbb{R}$. |
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Apr 18 |
answered | Does every algebraically closed field contain the field of complex numbers? |
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Apr 16 |
answered | Valuation but not Noetherian Rings |
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Apr 16 |
answered | Value range of normalization methods? min-max, z-score, decimal scaling |
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Apr 15 |
comment |
The field of Laurent series over $\mathbb{C}$ is quasi-finite I see how one can avoid general theory at various points. In particular one can specialize the proof for the fact that the Galois group is cyclic to the present particular case. However in this way one will arrive at a rather lengthy verification. And at the moment I don't see how to avoid using something like Hensel's lemma at the beginning of the whole argument ... |
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Apr 15 |
answered | The field of Laurent series over $\mathbb{C}$ is quasi-finite |
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Apr 12 |
answered | Non-trivial valuation of $\mathbb R$ |
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Apr 10 |
awarded | Custodian |
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Apr 10 |
reviewed | Reject suggested edit on Isomorphism or non-isomorphism of two specific local rings |
