Timeline for Algebraicity of the completion of a field? Finiteness?
Current License: CC BY-SA 2.5
25 events
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| Jul 4, 2022 at 23:59 | comment | added | The Amplitwist |
The link to eom.springer.de is broken, but the article can now be found at encyclopediaofmath.org/wiki/Norm_on_a_field.
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| Jun 22, 2022 at 8:14 | history | edited | CommunityBot |
replaced http://www.math.uga.edu/~pete with http://alpha.math.uga.edu/~pete
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| Jan 29, 2010 at 21:54 | comment | added | Pete L. Clark | @Mariano: it is a theorem that if K is complete, the only topology on a finite dimensional vector space is the product topology. This does not hold in general: in my counterexample above, you can see that $\mathbb{Q}$ is dense in $\mathbb{Q}(\sqrt{2})$ in each of the two topologies I gave, so neither is the product topology. | |
| Jan 29, 2010 at 21:43 | comment | added | Mariano Suárez-Álvarez | If so, then clearly a hyperplane in $V$ is closed, for the coordinate hyperplanes in $K^n$ are closed. Since every subspace of $V$ is the intersection of the hyperplanes that contain it, we can then conclude that every such subspace is closed. This is what I had in mind above. | |
| Jan 29, 2010 at 21:24 | vote | accept | Pete L. Clark | ||
| Jan 29, 2010 at 17:31 | comment | added | Mariano Suárez-Álvarez | Given a valued field $K$ and a finite dimensional vector space $V$, the topology on $V$ is the one it gets by choosing a basis, using it to construct an $K$-linear isomorphism $V\cong K^n$, and then transporting the product topology from $K^n$ to $V$, right? | |
| Jan 29, 2010 at 16:36 | answer | added | D. Savitt | timeline score: 8 | |
| Jan 29, 2010 at 15:30 | comment | added | Gerald Edgar | The proof that finite-dimensional subspace is closed is essentially that it is complete, therefore closed. But of course completeness of the f.d. subspace requires completeness of the field. See the Q(\sqrt{2}) counterexample. In fact there are more norms than just the two listed (compatible with the usual archimedean norm on Q itself). | |
| Jan 29, 2010 at 14:48 | answer | added | S. Carnahan♦ | timeline score: 10 | |
| Jan 29, 2010 at 9:42 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Jan 29, 2010 at 9:14 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Jan 29, 2010 at 8:24 | comment | added | Pete L. Clark | @Harry: please see Exercise 1.4 of math.uga.edu/~pete/8410Chapter1.pdf. | |
| Jan 29, 2010 at 8:16 | comment | added | Harry Gindi | I read that wikipedia page fairly closely, and I'm pretty sure that the assumption is not used. The only thing it seems like they might be using is an assumption of characteristic 0?. Anyway, I know that Bourbaki has a very general treatment of topological vector spaces in volume 5, called topological vector spaces. I think there are copies on the internet floating around in French as Espaces Vectorielles Topologiques. I also saw a copy of "Topological Vector Spaces" by Grothendieck floating around. | |
| Jan 29, 2010 at 8:14 | comment | added | Pete L. Clark | OK, here's a counterexample to the automatic continuity claim: consider Q(\sqrt{2}) as a vector space over Q with the standard Archimedean norm. There are two ways to extend this norm to Q(\sqrt{2}), differing by Galois conjugation. Therefore Galois conjugation is a Q-linear map which is not continuous for either norm. More plainly, the sequence (\sqrt{2}-1)^n converges to zero in one topology but not the other. | |
| Jan 29, 2010 at 8:07 | comment | added | Pete L. Clark | But on this wikipedia page, "normed space" means normed space over the real or complex numbers, both complete fields. I'm not trying to be argumentative: I really don't know the answer myself (nor have I given myself a good chance to think about it; that's a little later in my course preparation), but I have concerns that the issue is more subtle than it appears. Can you give a reference that works with vector spaces over an arbitrary (not necessarily complete!) normed field? | |
| Jan 29, 2010 at 8:01 | comment | added | Harry Gindi | en.wikipedia.org/wiki/Discontinuous_linear_map has a proof for normed spaces, which is still enough for his claim, but another page said it was true for arbitrary TV Spaces. | |
| Jan 29, 2010 at 7:59 | comment | added | Harry Gindi | Mariano's right according to wikipedia. Linear transformations between finite dimensional TVS's are forced to be continuous. | |
| Jan 29, 2010 at 6:56 | comment | added | Pete L. Clark | I'm still unconvinced. If the ground field is complete, then finite dimensional linear maps are automatically continuous. I am worried that this is not true in general. | |
| Jan 29, 2010 at 6:53 | comment | added | Mariano Suárez-Álvarez | Sorry :) If $V$ is a subspace of a finite dimensional $K$-vector space $W$, then it is the kernel of a $K$-linear map $\phi:W\to U$ to some other vector space $U$. Since every space in sight is finite dimensional the map $\phi$ is continuous, so $V$ is closed in $W$. | |
| Jan 29, 2010 at 6:51 | comment | added | Pete L. Clark | I can't tell whether you are speculating or answering. If the latter, could you say a little more? | |
| Jan 29, 2010 at 6:46 | comment | added | Mariano Suárez-Álvarez | A finite dimensional $K$-subspace is closed, because it is the kernel of a linear map to some other space, I guess. | |
| Jan 29, 2010 at 6:42 | comment | added | Pete L. Clark | I was thinking along these lines. But unlike in the case of usual Banach spaces we don't have a complete field "at the bottom", so is it obvious that a finite-dimensional subspace is closed? | |
| Jan 29, 2010 at 6:39 | comment | added | Mariano Suárez-Álvarez | If $\hat K/K$ is finite then $K$ is a closed $K$-subvector space of $\hat K$, for it is a finite dimensional subspace, so in particular $K$ is complete, no? | |
| Jan 29, 2010 at 6:36 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Jan 29, 2010 at 6:06 | history | asked | Pete L. Clark | CC BY-SA 2.5 |