Spectrum of an operator

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Can someone please elaborate on what the object K[T] is? The article mentions that a vector space, equipped with a linear operator on it, can be viewed as a module over K[T], but then subsequently only mentions the ring K[T]. If K[T] is just another name for the polynomial ring K[x], then how are the spectral properties of a particular operator T supposed to manifest in the spectrum of K[x]? Or is K[T] instead supposed to be something like "the polynomial ring mod out by the minimal polynomial of T"? — Preceding unsigned comment added by 129.97.226.227 (talk) 12:38, 10 June 2013 (UTC)Reply

You should think of K[T] as living inside GL(V). Inside GL(V), there is a field K (determined by diagonal matrices) and an element T (given to us). The subring of GL(V) generated by K and T is K[T].
This kind of notation turns up in other places. For instance, it's analogous to the notation Q(i) for the field of rational numbers together with a square root of −1 (which is a complex number—note that we've implicitly chosen a copy of C containing Q).
General facts imply that K[T] is isomorphic to K[x] modulo the minimal polynomial of T. Indeed by the universal property of K[x], there is a homomorphism K[x] → K[T] that sends x to T. The map is obviously surjective, and the kernel is generated by the minimal polynomial of T by the definition of the minimal polynomial.
By the way, on Wikipedia talk pages it's conventional to write new comments at the bottom. Ozob (talk) 14:04, 10 June 2013 (UTC)Reply

Removed example

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Header added. —Nils von Barth (nbarth) (talk) 07:48, 7 December 2009 (UTC)Reply
The functor F is co-representable by B, as over the category of algebras it is covariant. Fourier-Deligne Transgirl (talk) 03:16, 7 December 2023 (UTC)Reply

I removed the following example:

A special but quite typical case of an affine scheme is obtained as follows. Take a field K and n variables, x1,...,xn. Given m polynomials, p1,...,pm in these variables over K, there is a functor F from the category of commutative K-algebras to sets characterized by F(A)={(x1,...,xn) in An|p1=...=pm=0}. Then F is represented by Spec(B) where B is the quotient of K[x1,...,xn] by the ideal I generated by the pj.

Reasons:

  • This uses functors, but the article hasn't mentioned the functor connection yet.
  • The technical term "represented" is not explained. The functor F is not represented by Spec(B), but by B.
  • A more useful example would describe Spec(B) in detail.

AxelBoldt 16:07, 18 Jan 2004 (UTC)

IMO, this article contains too much general rubbish. Just focus on the connections to geometry. Schemes are a generalisation of this setting and reside in a seperate article. agrosquid

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See the relevant manual of style entry: External link vs. External links. It is permissible (and arguably more correct) to leave off plural in the case of a single link. - Gauge 06:59, 25 February 2006 (UTC)Reply

The value of \Gamma is a limit of objects in the category of rings? Really?

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As expressed at the moment,   is a limit of objects in the category of rings. But a limit of objects is always a product. And the product in the category of rings is just a Cartesian product of rings. Is this actually correct? There's no citation. What are the restriction maps here? --Svennik (talk) 17:14, 26 December 2023 (UTC)Reply

Should this not be a colimit of morphisms: In other words, should we not define   where each   is the localisation map? This should correspond to repeatedly localising. which is what I intuitively would've expected.--Svennik (talk) 17:23, 26 December 2023 (UTC)Reply
This is a direct limit. I have fixed notation and added a link to dirct limit. D.Lazard (talk) 18:42, 26 December 2023 (UTC)Reply
It's not a direct limit because the basis   of   doesn't form a directed set. It's likely to be the colimit I described above. I think this needs to be given in more detail. --Svennik (talk) 19:03, 26 December 2023 (UTC)Reply
In fact, this is a limit (category theory) with respect to a preordered set that is not directed. I have edited the article accordingly. D.Lazard (talk) 12:10, 27 December 2023 (UTC)Reply