Definition talk:Algebraic Number Field

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They're only equivalent without the restriction to finite extensions in the first! --Linus44 17:42, 27 March 2011 (CDT)

That went so far over my head I didn't hear it whistle ... I think I need to do some reading. --prime mover 00:23, 28 March 2011 (CDT)

I understand now what I did wrong when I added the second definition for a number field - I quoted an unnecessarily loose sentence in 1969: C.R.J. Clapham: Introduction to Abstract Algebra.

I believe what needs to be done is that the second sentence should be extracted into a proof that such a number field is an element of the set of all number fields in that it has $\Q$ as a subfield - but this can not be used as a definition. --prime mover 00:40, 18 August 2011 (CDT)


I think this section is a bit misleading as it stands.

I think the premise for the note is that $\C$ is some universally fixed field; and so in general $\Q[X]/(X^2 - 2) \nsubseteq \C$ as sets.

But this depends on the construction of $\C$; there's loads of ways of constructing $\C$, and none of them is "standard"; so a priori it's not clear how to decide if $\Q[X]/(X^2 - 2)$ is contained in $\C$ or not.

In any given theorem using the complex numbers we assume fixed some field that's isomorphic to $\C$: it's unique up to unique isomorphism; but we still don't know how to decide if a given field is contained in it.

Hence the point of view that field extensions are monomorphisms from one field to another: with this definition we do have $\Q[X]/(X^2 - 2) \hookrightarrow \C$.

It's misleading because in constructing algebraic numbers like $\sqrt{2}$, it's preferable to do this without working in some larger field $\C$; and explicitly construct the appropriate extension of $\Q$, so in many field theory texts $\Q[\sqrt{2}]$ is defined to be $\Q[X]/(X^2 - 2)$, and there's no notational difference to distinguish this from the definition of $\Q[\sqrt{2}]$ as the subfield of $\C$ generated by $\sqrt{2}$ over $\Q$.

On the other hand, I'm mostly writing this because I realised I can't be bothered to write out Taylor's theorem properly, so settled on bringing up notational pedantry instead. Feel free to ignore. --Linus44 (talk) 19:44, 17 October 2012 (UTC)

It's in the "unique up to unique isomorphism" clause that we are allowed, or agree, to ignore the specific definition. $\Q[\sqrt2]$ as embedded in $\C$ is a perfectly correct instance of $\Q[X]/(X^2-2)$.
But this is exactly what I'm saying: $\Q[\sqrt2]$ as embedded in $\C$ is a perfectly correct instance of $\Q[X]/(X^2-2)$; therefore one would invariably consider $\Q[X]/(X^2-2)$ as a subfield of $\C$, since it is as good an instance of $\Q[\sqrt2]$ as any other. The note suggests that we don't have this relationship. --Linus44 (talk) 20:16, 17 October 2012 (UTC)
Hm, but what about adding more than one indeterminate, e.g. $\Q[X,Y]/(XY-1)$? In general $\C$ will be too small for such things, not? --Lord_Farin (talk) 20:20, 17 October 2012 (UTC)
Oh, that hasn't got finite degree / $\Q$. Probably it works out, but that's a nontrivial theorem, then. --Lord_Farin (talk) 20:20, 17 October 2012 (UTC)
If $K/\Q$ is finite it has to work out, since $\C$ contains an algebraic closure of $\Q$, and a finite extension is algebraic; hence contained in $\C$. Thinking about it, this should be given as a proposition to show the equivalence.
I wondered about the non-finitely generated case once; but I don't know much about transcendental number theory: we have $\Q[X,Y]/(XY-1) \simeq \Q[X,X^{-1}] \simeq \Q[e,e^{-1}]$ since $e := \exp(1)$ is transcendental, but adding more variables it's not clear. For instance can $\Q[X_i]_{i \in \N}$ be embedded in $\C$? I don't know of a way of constructing infinitely (or uncountably) many algebraically independent transcendental numbers. --Linus44 (talk) 20:36, 17 October 2012 (UTC)
That's true; I accept the finite case, then. For the infinite case, I find it hard to believe $\Q[X_z: z \in \C]$ could be embedded in $\C$. Maybe Krull dimension can help in proving such impossibilities. --Lord_Farin (talk) 20:40, 17 October 2012 (UTC)
If it helps, consider $\Q[X]$ as the adjoint of the forgetful functor $\mathbf{Rng_*} \to \mathbf{Rng}$ (forgetting base point) to establish it's unique up to uq. iso-ness. --Lord_Farin (talk) 20:07, 17 October 2012 (UTC)

Regarding the infinite (uncountable) case: I've done some reading, it seems it is possible.

We have the relation $|\C| = \max\{|\Q|, \text{transcendence degree}_\Q\C \}$, as described here. This shows that the basis for the algebraically independent elements of $\C$, transcendental over $\Q$ is uncountable. Let $S$ be a such a basis. Then $\Q(S) \subseteq \C$ since $\Q(S)$ is the smallest field containing $S \subseteq \C$ and $\Q$ (since by the definitions $\C$ is algebraic over $\Q(S)$, $\C$ is an algebraic closure of this field).

In particular choosing a bijection $\C \simeq S$, we can embed $\Q[X_z: z \in \C] \hookrightarrow \Q(S) \hookrightarrow \C$. Interesting. --Linus44 (talk) 21:59, 18 October 2012 (UTC)

Thanks for pointing me to that interesting fact. Of course, we can supersede $\C$ and consider $\mathcal P (\C)$ to make this fail; more indeterminates than elements of $\C$ will surely prohibit an injection. Fascinating stuff; however, I'd better get back to the double pointed topologies, much work to be done there. --Lord_Farin (talk) 22:05, 18 October 2012 (UTC)

Suggested rename to Algebraic Number Field

... why? --prime mover (talk) 21:13, 18 October 2012 (UTC)

I'm currently following a course on these; it's plainly called number field there as well. --Lord_Farin (talk) 21:25, 18 October 2012 (UTC)
Because "number field" makes "standard number field" sound like a special case of this. That's really it. --abcxyz (talk) 22:44, 18 October 2012 (UTC)
I argue a) not, and b) so what? --prime mover (talk) 05:24, 19 October 2012 (UTC)
Seems like you prefer the way it is, so let's just leave it like that then. --abcxyz (talk) 05:32, 19 October 2012 (UTC)
You have a point there; I suggest to at least add a note about possible confusion and your new proposed term. --Lord_Farin (talk) 08:15, 19 October 2012 (UTC)
Unhelpful linguistic note: in english, almost invariably nouns of the form "<adjective> noun" are a special case of the noun. Counter-examples are unusually frequent in mathematics, for example skew field. It's still quite rare. --Linus44 (talk) 10:00, 19 October 2012 (UTC)

The difference between Definition:Number Field and Definition:Standard Number Field is indeed because the latter are a special case of a Definition:Number Field which includes things like $\Q (\sqrt 2)$, i.e. numbers of the form $a + b \sqrt 2$ where $a, b \in \Q$. There's an "also known as" which covers the "Algebraic Number Field" bit, so we're done.

I'm against just renaming a page because a particular book (or school) has it under a different name. There are options one way or another with a lot of these things, and we pick one, and by default that one is the de facto standard on this site. We would rename if a) it turns out that the one chosen by ProofWiki is a seriously minority definition (not sure that's the case here) or b) there are other related concepts which are often given the same name, and it is necessary for us to pick a naming scheme which has least ambiguity (the recent renaming of "neighborhood" to "open ball" in metric spaces is a case in point). Renaming is a nuisance and tedious, so it's only worth doing if there is an overwhelming need for it. In all other cases, an "also see" and a redirect page should cover it.

Also in this case, "algebraic" in this field of activity usually stands in counterpoint to "transcendental". Is the definition given on the Definition:Transcendental page an instance of this? If so, then maybe there is a case for "algebraic", but if it happens to be a coincidence of terminology, then certainly not. --prime mover (talk) 18:58, 19 October 2012 (UTC)

It's not quite a special case: a number field as defined must have finite degree over $\Q$. There was an example (I think on this page) that $[\R:\Q] = \infty$, so $\R$ is not a number field.
There is a case for algebraic number field, for the reason that you stated. The fact that $\R$ contains transcendental numbers is only possible because $[\R:\Q] = \infty$ (though this condition isn't sufficient to guarantee transcendental elements). There's a result (Finite Field Extension is Algebraic) that a finite extension of $\Q$ cannot contain transcendental elements.
I think it's not too important: "algebraic number fields" is also common terminology; but no-ones going to be totally thrown by either. In my experience, the use of "standard number field" for $\Q,\R,\C$ is less common. --Linus44 (talk) 19:07, 19 October 2012 (UTC)
"Standard number field" was me, so as to have a way to identify these three entities simply and more-or-less obviously. --prime mover (talk) 19:52, 19 October 2012 (UTC)
Got it. This is where we've gone wrong. Linus44 put this page up Mar 2011 as a finite extension of $\Q$. Prime Mover then posted up that a number field is a subfield of the complex numbers (that is, $\Q,\R,\C$ and e.g. $\Q \sqrt 2$) from the book he was studying. It was then pointed out that the two defs are not equivalent.
But a "number field" is, by intuition, a "field of numbers". So $\R$ ought to be a number field.
Bugger it, more domestic trivia to attend to. I swear I'm gonna have to kill myself. --prime mover (talk) 19:57, 19 October 2012 (UTC)
Okay I concede. This is one of the ambiguity cases. Call an "algebraic number field" the finite extension, and a "number field" a subfield (finite extension of $\Q$ or not) of $\C$. Standard number fields are a specialisation of number fields, as are algebraic number fields.
See, it's all in the way you explain what you mean. --prime mover (talk) 22:40, 19 October 2012 (UTC)