現代数学の系譜11 ガロア理論を読む25 [無断転載禁止]©2ch.net

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631: 現代数学の系譜11 ガロア理論を読む [sage] 2016/12/03(土) 14:03:09.06 ID:6Rgz8i9T

>>628 ついで

http://math.stackexchange.com/questions/86762/finding-a-basis-of-an-infinite-dimensional-vector-space
Finding a basis of an infinite-dimensional vector space?  asked Nov 29 '11 at 16:30 InterestedGuest

2 Answers  answered Jan 20 '12 at 19:25 Qiaochu Yuan

For many infinite-dimensional vector spaces of interest we don't care about describing a basis anyway; they often come with a topology and we can therefore get a lot out of studying dense subspaces, some of which, again, have easily describable bases.
In Hilbert spaces, for example, we care more about orthonormal bases (which are not Hamel bases in the infinite-dimensional case); these span dense subspaces in a particularly nice way.

4.  answered Jan 20 '12 at 19:09 David Wheeler
The "hard case" is essentially equivalent to this one:

Find a basis for the real numbers R over the field of the rational numbers Q.

The reals are obviously an extension field of the rationals, so they form a vector space over Q. It should be clear that such a basis has to be uncountable (for if it were countable, the reals would likewise also be countable).

It should also be clear that such a basis is a subset of {1}∪R?Q. The trouble is, that the power set of the reals is "so big" that it's not even clear how to name the sets we need to apply the axiom of choice TO. Linearly independent subsets however, DO satisfy the requirements for Zorn's Lemma, a form of the Axiom of Choice.

A relatively easy-to-follow proof of the existence of a basis for any vector space using Zorn's Lemma can be found here: http://planetmath.org/encyclopedia/EveryVectorSpaceHasABasis.html

http://rio2016.5ch.net/test/read.cgi/math/1477804000/631

632: 現代数学の系譜11 ガロア理論を読む [sage] 2016/12/03(土) 14:20:00.87 ID:6Rgz8i9T

>>631 ついで

http://math.stackexchange.com/questions/271536/ring-of-formal-power-series-finitely-generated-as-algebra
Ring of formal power series finitely generated as algebra?  asked Jan 6 '13 at 13:44 user55354

I'm asked if the ring of formal power series is finitely generated as a K-algebra. Intuition says no, but I don't know where to start. Any hint or suggestion?

2 Answers

Let A be a non-trivial commutative ring. Then A[[x]] is not finitely generated as a A-algebra.

Indeed, observe that A must have a maximal ideal m, so we have a field k=A/m, and if k[[x]] is not finitely-generated as a k-algebra, then A[[x]] cannot be finitely-generated as an A-algebra. So it suffices to prove that k[[x]] is not finitely generated.
Now, it is a straightforward matter to show that the polynomial ring k[x1,…,xn] has a countably infinite basis as a k-vector space, so any finitely-generated k-algebra must have an at most countable basis as a k -vector space.

However, k[[x]] has an uncountable basis as a k-vector space. Observe that k[[x]] is obviously isomorphic to kN, the space of all N-indexed sequences of elements of k, as k-vector spaces. But it is well-known that kN is of uncountable dimension: see here, for example.

http://rio2016.5ch.net/test/read.cgi/math/1477804000/632

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