Talk:Algebra over a field

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This article has comments.

It's nice to see this separated out into its own thing as it should be. The problem is that there's really no good name for it. "algebra over a field" isn't quite right, since algebras can be defined over any commutative ring with no change in the concept, and one does study these things with the same terminology and notation in (for example) algebraic geometry. But "algebra over a commutative ring" isn't too hot either, being less familiar and (with "commutative") rather unwieldy.

One idea is "algebra over a base", since the commutative ring in question is called the base ring (or base field). One does hear phrases such as "Let A be an algebra over the base ring K.", for example. But I can easily imagine that this means something else too. Another idea is "algebra over a base ring", which has some of the problems of the version with "commutative" above, but is not so enormous.

Ideas?

-- Toby 06:39 Mar 7, 2003 (UTC)

Well, while editing this page, I decided that it's OK as it is. The fact the "vector space" and "module" are completely different terms is a much bigger problem (and much bigger than Wikipedia) that already prevents discussing the general case in the same breath as the more familiar. So let it be. -- Toby 07:46 Mar 7, 2003 (UTC)

Alternative algebra bug?[edit]

This page states that the only finite dimensional alternative algebras over the reals are R, C, the quaternions and the octonions. This doesn't seem correct, since any matrix ring is a finite dimensional alternative algebra over the reals, or any real vector space with xy := 0 as multiplication. I don't know how to correct this -- maybe any finite dimensional alternative division algebra? --Sven 217.9.27.15 00:36, 20 Apr 2005 (UTC)

Beats me. Maybe the statement is "any irreducible finite dimensional alternative algebra that is not an associative algebra"? (since a matrix ring is associative, and a real vector space with xy=0 but xx!=0 seems to be completely reducible, to me).linas 01:38, 20 Apr 2005 (UTC)
I see no one fixed this in 15 months. What was probably meant was the Bruck-Kleinfeld theorem specialized to the field of real numbers: the only finite dimensional real alternative division algebras are R, C, the quaternions and octonions. The original author might also have meant Hurwitz' theorem: the only finite dimensional real normed alternative algebras are R, C, the quaternions and octonions. But I'm guessing the Bruck-Kleinfeld result is what was meant. --Michael Kinyon 12:19, 31 July 2006 (UTC)

index free notation ???[edit]

I have big doubts about the validity of the section "index free notation".

  • first, say why is it called like this
  • second, should "k-algebra" be corrected to "K-algebra" everywhere ? (simultaneous use of K and k is very unfortunate ; in addition the prefix "k-" often means something else not related to any element "k" or set "K")
  • third, according to the introduction, we do not suppose all algebras to be associative, thus they are not rings in general
  • fourth, they should even less be supposed to be unital, and I think that eta(1) must be a unit if not eta=o

So I strongly suggest to move that paragraph (and the following, based upon it) away from the main page (e.g. into the Talk page) until these issues are clarified. — MFH:Talk 15:02, 2 June 2006 (UTC)

  • fifth, the range of η must lie in the centre of A to really make it a k-algebra. -- EJ (talk) 16:03, 25 February 2008 (UTC)

Simplification[edit]

I was wondering if the definition at the beginning be simplified for non-math majors. I found a very nice and simple definition that explained the definition of an algebra perfectly: A set together with a pair of binary operations defined on the set. Usually, the set and the operations include an identity element, and the operations are commutative or associative. Nice, plain English. Start talking about the Ks and As in the next section, not the first definition. It's quite intimidating. ForestAngel (talk) 01:43, 24 July 2008 (UTC)

That's not right. You have to have not only the ring structure (a set with two binary operations obeying certain algebraic laws) but also the vector-space structure (where the addition in the vector space is the same as the addition in the ring, but you've also got multiplication by scalars). So it's really three operations:
(1) addition,
(2) multiplication of two members of the algebra, and
(3) multiplication of a member of the algebra by a scalar.
Michael Hardy (talk) 20:05, 8 December 2009 (UTC)
I think I've addressed this now. Mlm42 (talk) 02:43, 13 November 2008 (UTC)

Missing examples[edit]

I'm not an algebraist, but when I try to think of examples of this concept, the first things that come to my mind are (1) the algebra of polynomials over a field, and (2) the algebra of n×n matrices over a field. Yet I don't see those mentioned in the section titled "examples". Should they be there, or am I missing something? Michael Hardy (talk) 20:02, 8 December 2009 (UTC)

...oh, I see there are more parts of the article with examples than the part I was looking at.....
to be continued..... Michael Hardy (talk) 20:13, 8 December 2009 (UTC)

Classification is incomplete[edit]

This article claims that there exist only two two-dimensional unital associative algebras over the complex numbers up to isomorphism, but the classification given does not include quaternions (the case aa=-1). It also misses the case 1≈[[1,0][0,1]] (the 2-2 identity matrix) and a≈[[1,0][0,0]] which would have aa=a. Both these have two complex dimensions and are associative and unital. Clearly the classification is inaccurate, but I don't have access to the source given. The source looks legit, so someone should double check it. —Preceding unsigned comment added by 98.149.109.69 (talk) 00:52, 17 February 2010 (UTC)

The result is correct (and would still be correct without the word "associative"). Note that quaternion multiplication isn't bilinear over C (though it is bilinear over R), so the quaternions do not form an algebra over the complex numbers. --Zundark (talk) 09:09, 17 February 2010 (UTC)
Reference to Eduard Study for the classification of algebras over C looks way back. The subject of classification of algebras stimulated the work of many mathematicians back then; see hypercomplex number for more context and results.Rgdboer (talk) 21:01, 6 December 2011 (UTC)
Working from first principles makes this clearer:
  • The case of aa=−1 does not generate the quaternions. Putting b=ia we get bb=1, so it is isomorphic to the case with aa=1.
  • The case of aa=a: Putting b=2a−1 we get bb=1, so it is also isomorphic to the case with aa=1. Quondumtalkcontr 09:41, 7 December 2011 (UTC)

Improved, but.... (distributive ≠ bilinear)[edit]

The latest edit is a very neat, understandable nutshell, and thus I'm loathe to change it. However, it has a significant shortcoming: it is no longer perfectly correct, because it replaces bilinearity with distributivity, and the two are not equivalent. One can thus find examples that fit the new definition (distributivity) that do not satisfy the other axiom of (bi-)linearity: homogeneity (referred to in the article as compatibility with scalars). Since the standard definition does require homogeneity, can we find a wording that includes this requirement without destroying the elegance of the new wording? — Quondum 04:33, 16 October 2012 (UTC)

Why not: "In mathematics, an algebra over a field is a vector space equipped with a bilinear product that distributes over vector addition"? D.Lazard (talk) 07:57, 16 October 2012 (UTC)
Thanks, I have incorporated this suggestion; it is nicely compact. — Quondum 09:07, 16 October 2012 (UTC)
In an edit which was in conflict with the preceding one, I have crossed out "that distributes over vector addition" in my preceding post and added in comment:
In fact, bilinearity implies distributivity and mention of the distributivity is not needed here. By the way, the lead of bilinear map is rather unclear and deserve to be rewritten in the spirit of that bilinear form. Also, the section "Definition and motivation" gives only examples of unital associative algebras. For not being confusing, it should contain a less restricted example, like a 3-dimensional vector space equipped with the cross product.
I'll be bold and will remove the mention of distributivity in the lead. Please revert if I have missed something --D.Lazard (talk) 09:33, 16 October 2012 (UTC)
Yup, nice. I feel a bit silly for missing something that obvious. — Quondum 14:45, 16 October 2012 (UTC)
I like the new version, but I felt something really had to be said about the common usage of "algebra" for "associative algebra". I hope my attempt doesn't interfere too much with the lead. Rschwieb (talk) 13:06, 16 October 2012 (UTC)
"Because", "ubiquity", "importance" seem WP:OR. Moreover "algebra" does not appear only in textbooks, as suggested in the last paragraph of the lead. I suggest: 'In many texts, "algebra" is used for "unital associative algebra". It is almost always the case when "algebra" is used without further specification.' --D.Lazard (talk) 14:21, 16 October 2012 (UTC)
Or perhaps (forgive my nitpicking; feel free to ignore it): 'In many texts, "algebra" is used to mean "unital associative algebra". It is typically the case when "algebra" is used without further qualification.'? Underlining is purely to highlight changes. — Quondum 15:03, 16 October 2012 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── I agree D.Lazard... let's take out "importance". "Texts" for textbooks is fine too. However, I don't know what to do about "ubiquity". How else do you describe the predominance of associative algebras over nonassociative ones in print? There is probably a better word. Q: Perhaps my wording is not the best for all English speakers... if there is a more straightforward version, have at it :) Rschwieb (talk) 15:13, 16 October 2012 (UTC)

To Rschwieb: the "predominance" is a matter of opinion that has not its place in this lead. Is there really a predominance of associative algebras over Lie algebras? The use of "almost always" or "typically" says essentially the same in a less controversial way.
To Quondum: I have used "specification" to include the case of the texts, like textbooks, that give a definition of "algebra". By "without specification", I meant "without definition nor qualification". The latter wording may be better. About "almost always" vs. "typically": we have to think to the reader who have read the word of "algebra" and want to know its meaning. "Almost always" means clearly that he has to read it as "unital associative algebra". My level of English is not enough to be sure that "typically" is as clear. --D.Lazard (talk) 15:47, 16 October 2012 (UTC)

I would like to propose that we create associative algebra over a field, either renaming this article or whatever else. As for the convention, the simplest solution is to skip the discussion on it altogether. I agree with R that "algebra" "often" refers to associative ones (and almost never say Lie or Jordan ones), but the point is that Wikipedia doesn't have to concern with such issue. -- Taku (talk) 17:28, 16 October 2012 (UTC)

I'm not entirely comfortable with the statement of "almost always" or "typically", as it actually seems to invite challenge (on the rest I tend to go along with D.Lazard). But is it the case that WP can adopt, without comment, a convention of using the term "algebra" in a specific sense (e.g. the most general mentioned here, namely not necessarily either unital or associative, alternatively the most restricted sense, being unital and associative)? — Quondum 17:44, 16 October 2012 (UTC)
IMO, the point is not WP conventions, there are fine as they are. The point is that, when someone consult WP because he has read "algebra" without explanation, he has to be informed that this probably means "unital associative algebra". If providing such an information breaks some WP rule, this is a case of WP:Ignore all rules. --D.Lazard (talk) 18:16, 16 October 2012 (UTC)
Thus I propose:
In many texts, "algebra" is used to mean "unital associative algebra". In particular, if "algebra" 
is used without any definition nor qualification, it means very probably "unital associative algebra".
--D.Lazard (talk) 18:30, 16 October 2012 (UTC)
This seems to repeat itself, somewhat unnecessarily. A reader will get the idea from the first of the two sentences. Indicating that the term can also be used in the general sense may be of value:
In many texts, "algebra" is used to mean "unital associative algebra". A more general algebra may be indicated by a definition or qualification.
Quondum 19:35, 16 October 2012 (UTC)

Bilinear[edit]

There is no reason to put the bilinear condition on the algebra product. This restriction is not given in references: Hazewinkle has (p 3): An algebra over a field k (or k-algebra) is a set A which is both a ring and a vector space over k in such a manner that the additive group structures are the same and the axiom (λa)b = a(λ b) = λ(ab) is satisfied for all λ in k and a,b in A.

Bahturin (1993) has (p 29): A ring R which also is a vector space over a field F is called an algebra over F, or F-algebra, if for any λ in F and any x,y in R

λ(xy)=(λ x) y = x(λ y)

In Basic Structures of Modern Algebra, Kluwer Academic, ISBN 0-7923-2459-5.

The definition of Steven Kantz (2000) is wide enough to include Lie algebras: A vector space (over a field) on which is also defined an operation of multiplication. In Dictionary of Algebra, Arithmetic and Trigonometry, CRC Press, ISBN 978-1-4200-3802-6.

The term "bilinear" is commonly used for bilinear form, but not as an axiom for algebras. Usage here provides an unnecessary complication for someone trying to determine the exact requirements for this mathematical structure.Rgdboer (talk) 22:32, 24 October 2012 (UTC)

Your quotes from Hazewinkle and Bahturin (defining what I would call a unital associative algebra) give three separate requirements:
  • It is a vector space over a field K.
  • It is a ring.
  • The multiplication of the ring is homogeneous of degree 1 over K.
This implies bilinearity, even if one avoids the term.
What is defined first in the lead is the more general case (that is not required to be unital or associative), which may be given with the following axioms:
  • It is a vector space over a field K.
  • It has a multiplication operation that is closed on the vector space
  • The multiplication operation distributes over the vector space addition.
  • The multiplication operation is homogeneous of degree 1 over K.
The last two axioms are encapsulated in "bilinear", so the term cannot be removed without replacing it with something equivalent, such as the explicit axioms.
The definition by Steven Kantz would normally be regarded as deficient: it allows the class being defined to include cases that do not satisfy the axiom of homogeneity, and in particular includes objects that are neither associative algebras nor non-associative algebras. — Quondum 08:43, 25 October 2012 (UTC)
Since the mentioned axiom is precisely bilinearity, I wouldn't say it's misleading but maybe it does put another layer between the reader and understanding. In this case, Krantz's definition appears to be vague to the point of incorrectness. If you exclude it, then you are omitting the fact that the copy of the scalar field is central in the algebra. In some texts, that is how algebra structure is defined: for a ring (or Lie ring) A, and a commutative ring R, A has an R algebra structure if there is a ring homomorphism (or Lie ring homomorphism) from R to the center of A. The axiom relating the module structure and the multiplication appears in all texts I have ever read on the topic in some form. Rschwieb (talk) 14:44, 25 October 2012 (UTC)
Actually while thinking about it now, I'm not sure the "central homomorphism" description of Lie algebras works that way. But I am sure it works that way for associative algebras. Rschwieb (talk) 14:46, 25 October 2012 (UTC)

Very valuable responses. Thank you. Continuing to look at reliable sources, there is JDH Smith and AB Romanowska in Post-modern Algebra (1999) who confirm use of bilinear, but in a slightly broader context. They write on page 158:

If K is a unital commutative ring, then a K-algebra (AK, · , 1) is a unital K-module AK and a monoid (A, · , 1) such that the monoid multiplication A × A → A; (a,b) → ab is bilinear.

This terse text with little history or commentary includes the opinion:

The rather grandiose name [algebra] probably arose because early twentieth-century algebra was almost exclusively devoted to the study of "algebras" in this sense.

Algebras have considerable structure. Perhaps inclusion of bilinearity as a defining property of the product helps prepare students for tensor products of algebras, where the bilinear property comes to the fore.Rgdboer (talk) 20:48, 25 October 2012 (UTC)

I do not agree with Smith-Romanovska opinion for the origin of the name of "algebra". IMO, this name arose before than algebras began to be studied for themselves. At the beginning, an algebra was simply a framework where one can operate like in usual algebra, now called elementary algebra. Similar shifts from the name of an area of mathematics to the name of a structure among others arose also for geometry (non Euclidean geometries), arithmetic (floating point arithmetic, Presburger arithmetic) and even for (mathematical) logic. D.Lazard (talk) 07:41, 26 October 2012 (UTC)

@Rgdboer I agree with your sentiment in the third to last paragraph about bilinearity, but I want to emphasize that I think a lot of results for both associative and Lie algebras use the fact that the scalars commute with the multiplication operation (although I can't think of a specific description atm :( ).

Now I'm curious to know if sesquilinearity is ever substituted for the axiom we are talking about. I haven't seen it before, but it seems like it could be possible for some sort of weird complex algebra-like object. Rschwieb (talk) 13:08, 26 October 2012 (UTC)

Sesquilinearity seems to occur frequently in representations in physics as a sesquilinear form (i.e. onto complex scalars) rather than a sesquilinear mapping onto the whole algebra space. Nevertheless, it should not be weird: surely any sesquilinear mapping on a complex vector space can always be expressed as a real bilinear mapping on a real vector space, yielding a normal ℝ-algebra? — Quondum 14:46, 26 October 2012 (UTC)

Hans Zassenhaus addressed the definition of an algebra in his Theory of Groups (1936). First he identifies a semi-module as an additive abelian semi-group.(p 92). Then he identifies a semi-ring as a semi-module with a multiplication that is both left and right distributive (p 93). Then he writes "A quasi-ring is a semi-ring which under addition is a module. In other words, the axioms of a quasi-ring are obtained from the ring axioms by omitting the associative law of multiplication"(p 95). On page 103 the definition of an algebra is given:

A quasi-ring over a field as coefficient ring is called an algebra. An algebra always has a basis over its coefficient field. An associative algebra is an algebra satisfying the associative law of multiplication.

An advantage of Zassenhaus terminology is the integration of Lie rings and Lie algebras in the treatment (instead of making them exceptional in nomenclature as when the definition of algebra is framed too narrowly).Rgdboer (talk) 01:36, 10 November 2012 (UTC)

Hitsuji Santos (talk) 04:03, 6 November 2013 (UTC) I wish every mathematician could be so clear as you. Congratulations.