Talk:Field (mathematics)

Edits on April 13-14, 2019

HOTmag has recently done two edits in section "Classic definition", and has redone them after being reverted. By WP:BRD they must wait for a consensus here before redo them again. IMO, these edits do not improve the article and cannot be accepted for the following reasons. Both edits concern the heading of field axioms.

The first edit insert the word "division", with a link to a disambiguation page. This is wrong, as division is a binary operation, while the axiom is about multiplicative inversion, which is a unary operation.

The second edit replaces "Additive inverses" by "Additive invertibility". This breaks the symmetry with the other existential axioms, which are all implicitly preceded by "Existence of". On the other hand, "invertibility" introduces a term that is not defined in the linked article. Moreover "invertibility" does not apply to a binary operation; one may talk only of the invertibility of the addition by a fixed element. So the use of "invertibility" may be confusing for some readers. D.Lazard (talk) 10:46, 14 April 2019 (UTC)

By WP:BRD they must wait for a consensus here before redo them again.

• Must? Indeed, BRD "does not encourage reverting" (ibid.). However, the process of BRD "is not mandated by Wikipedia policy" (ibid.).
• Additionally, I'm not reverting now, but rather editing the section anew - according to a fair compromise, for keeping all main interests of both parties.

these edits do not improve the article.

• Indeed, you can claim my original version contains some defects your version does not, but you can't claim my original version improves nothing - while it actually removes some defects the version supported by you does contain; unless you haven't read my old edit summery, where I explained why my original version is an improvement.

The first edit insert the word "division", with a link to a disambiguation page.

• This technical problem can easily be fixed, by linking to division.

This is wrong, as division is a binary operation, while the axiom is about multiplicative inversion, which is a unary operation.

• You ignore the last comment in my old edit summery, about division ring, whose special axiom (besides the other axioms of regular rings) is about multiplicative inversion - which is a unary operation - while division is a binary operation...
• Actually, the reason for using the term "division" in division ring, instead of the term "multiplicative inverse ring", is because the latter is wrong, because: Just as the title of the property "additive inverse" (in a ring) - means that every element has an additive inverse, so the title of the property "multiplicative inverse" (in a ring) - should mean that every element has a multiplicative inverse. Hence the name of such a ring, shouldn't be "multiplicative inverse ring", but rather a division ring. The same is true for defining the field: As long as the title of the property "additive inverses" means - that every element has an additive inverse, also the title of the property "multiplicative inverses" should mean - that every element has a multiplicative inverse. Hence that property shouldn't be titled: "multiplicative inverses".

The second edit replaces "Additive inverses" by "Additive invertibility". This breaks the symmetry with the other existential axioms, which are all implicitly preceded by "Existence of" .

• Please notice, that also the definition of division ring leans on an existential axiom about - the existence of a multiplicative inverse - for almost every element unless it's zero. However, the name of this ring breaks the symmetry you're talking about, so that ring - is not called "multiplicative inverse ring" - but rather division ring.
• However, I definitely understand your (legitimate) interest, of titling the multiplicative property by a title that reflects the existential quantifier; but you, too, should understand my (legitimate) interest, of titling this property by a title that reflects the difference between - this title about the existence of "multiplicative inverses" which is not a universal phenomenon (for every element) - as opposed to the title "additives inverses" about a universal phenomenon (for every element).
• If we want to keep the main interests of both parties, we can agree on the following compromise: The additive property should be titled: "additive inverses universally" (thus reflecting also the universal quantifier that precedes the existential one), while the multiplicative property should be titled: "multiplicative inverses almost-universally", enabling "division". Too long? I know that, but your short one is defective. To avoid lengthiness, I'd written "division" in one word, as it appears also in the term "division ring" (instead of a longer name: "multiplicative-inverse-of-every-nonzero-element ring")...

On the other hand, "invertibility" introduces a term that is not defined in the linked article.

• It seems you ignore many mathematical articles in Wikipedia (e.g. Outline of algebraic structures), that use the term "invertibility" while linking to inverse element.

Moreover "invertibility" does not apply to a binary operation; one may talk only of the invertibility of the addition by a fixed element. So the use of "invertibility" may be confusing for some readers.

• This technical problem can easily be fixed, by adding "universally" after "additive inverses", and by replacing my original phrase "Division (i.e multiplicative invertibility of every nonzero element)", by the phrase: "multiplicative inverses almost-universally" enabling "division". Too long? I know that. See above. This what sometimes happens with compromises.

HOTmag (talk) 14:55, 14 April 2019 (UTC)

I think WP:COMMONNAME covers this situation. This attempt to replace common terminology by undocumented "proper terminology" is blatant OR. --Bill Cherowitzo (talk) 21:20, 14 April 2019 (UTC)

Is vector field a field?

At the end of the article, in the Field (mathematics) § Related notions section, there is a figure with an example of vector field, and I think it's out of place, because a) vector field is not mentioned anywhere else in the article b) vector field is not a field in this sense. I could fix it, but I'm not sure if it's better to remove it or to explain the relation between vector field and field. Micha7a (talk) 11:37, 17 September 2019 (UTC)

No, a vector field is not a field and not related to the notion of a field. (In other languages the two meanings of field use different words, for example corps (=field in algebra) and champs (=field in the sense of vector field) in French.) I don't think it is necessary and / or a good idea to remove the illustration; the fact that a vector field is not a field is clearly discernible from the linked subpage, don't you think? Jakob.scholbach (talk) 12:34, 17 September 2019 (UTC)

I personally found it confusing, that's why I wrote here. For me it wasn't clear that they are different things until I read the definitions, and I was able to do it quickly, but for someone new to algebra this might be an exercise to make sure that those are different things. I'm not saying it is wrong to have the illustration, I just don't really see what is the point in the first place. If the point is to show that there exist a different concept also called "field" then why not state it explicitly? Micha7a (talk) 12:58, 17 September 2019 (UTC)

No, the point of the picture is to illustrate the hairy ball theorem, which is not a theorem about fields, but rather is meant to convey some geometric ideas about division algebras.

I think the article has to be meaningful for a variety of readers. In the beginning, everything is explained very slowly, but as the article progresses, it is (IMO rightly so) the case that some concepts are merely mentioned. (E.g. in the section on topological fields.) From this point of view I find it natural that the article in the end merely mentions (i.e., does not explain) vector fields.

If we were to include an explanation that there are two different senses of the (mathematical) word "field" I think it should be included much further up, but I am not sure we have to do this here. Jakob.scholbach (talk) 13:19, 17 September 2019 (UTC)

At least three senses. Grad students in physics are regularly told they must take a class in "field theory", and cautioned to not accidentally sign up for the one the math dept gives! 😁😁😁 67.198.37.16 (talk) 20:38, 5 October 2020 (UTC)

There are much more meanings, even in mathematics only. This is the reason for the hatnote at the top of the article.

I have edited the hatnote for making clearer that the two meanings are unrelated, and do not limiting the "other uses" to mathematics. D.Lazard (talk) 08:09, 6 October 2020 (UTC)

On a condition of field morphisms

It is written that :

"Field homomorphisms are maps f: E → F between two fields such that f(e₁ + e₂) = f(e₁) + f(e₂), f(e₁e₂) = f(e₁)f(e₂), and f(1_E) = 1_F, where e₁ and e₂ are arbitrary elements of E. "

Why such a condition is given to be a field homomorphism ? It is a consequence of : f(e₁ e₂) = f(e₁) f(e₂) for every e₁ and e₂ in E.

Parenthesis in the distribution axiom

Can we throw the Parenthesis in the right side of the phrase, becuase we can rely on the fact that multiplication has priority over addition. Or because this is a "pure" definition of a field , we first have to use the parenthesis in the distribution, and only then talk about a convention that in a field, multiplication has priority over addition? 93.173.65.88 (talk) 09:30, 2 August 2022 (UTC)

Field of rational functions?

This article states, "as fields of rational functions." However, this reference from nLab states "Rational functions on a field do not form a field. This comes from the fact that the reciprocal function is not a multiplicative inverse of the identity function $f(x)={\frac {x}{x}}\neq 1$, due to the fact that f(x) is undefined at 0, and thus has a different domain than the constant function 1, which is the multiplicative unit." The rational function link is actually to rational fractions. I recommend changing "rational function" to "rational fraction." Any comments or suggestions? https://ncatlab.org/nlab/show/rational+function#:~:text=Properties-,Rational%20functions%20on%20a%20field%20do%20not%20form%20a%20field,which%20is%20the%20multiplicative%20unit. TMM53 (talk) 00:51, 25 October 2022 (UTC)TMM53 (talk) 01:36, 25 October 2022 (UTC)

Your source seems to be assuming that the multiplication is pointwise. That's not necessarily the case. If you follow the link to field of fractions#Examples, you'll see (after a little unpacking) that we're talking about formal quotients of polynomials, with the multiplication defined in terms of the product of the numerators by the product of the denominators, up to a certain equivalence relation. If you evaluate the result at a point, it will agree with the pointwise product, provided all subterms are defined, but the definition per se is not pointwise, and is not refutable by the argument your link gives. --Trovatore (talk) 02:07, 25 October 2022 (UTC)

I don't agree with the summarized definition

"This may be summarized by saying: a field has two commutative operations, called addition and multiplication; it is a group under addition with 0 as the additive identity; the nonzero elements are a group under multiplication with 1 as the multiplicative identity; and multiplication distributes over addition."

If the nonzero elements are a group under multiplication with 1 as the multiplicative identity then one has a field : I agree.

But I think that the converse needs explanation. Because it is not clear that the set of nonzero elements is stable under multiplication (that is : it is not clear that the product of two nonzero elements is a nonzero element). 2A01:CB08:8607:CC00:7C0A:14D4:51F9:DC0A (talk) 07:09, 11 August 2023 (UTC)

Yes, you are right. It is very easy to derive from the given axioms that the product of nonzero elements is nonzero, but it isn't obvious from just reading the axioms. This problem could be addressed by adding a proof, but it seems to me preferable to avoid putting proofs in the definition section. That being so, I see two possible solutions: add another axiom to cover this, or reword the statement, so that instead of stating that the second version is merely a summary of the first definition, it describes it as an alternative and equivalent definition. I prefer the latter, and propose to make that change. JBW (talk) 09:30, 11 August 2023 (UTC)

Thank you for your quick answer. I am interested by a reference for the proof that the product of two nonzero elements is nonzero. 2A01:CB08:8607:CC00:15BC:2EE1:4CB8:F98A (talk) 11:41, 11 August 2023 (UTC)

Suppose there are two nonzero elements, a and b, such that ab = 0. Then a⁻¹ab = (a⁻¹a)b = 1b = b, but also a⁻¹ab = a⁻¹(ab) = a⁻¹0 = 0, which is a contradiction, as b is nonzero. That makes use of the fact that x0 = 0 for any x, but that follows immediately from x0 = x(0 + 0) = x0 + x0. JBW (talk) 21:19, 11 August 2023 (UTC)

I just realize that the first part of your proof lies some lines later in the article: every field is an integral domain. So, adding here the second part ( That makes use of the fact that x0 = 0 for any x, but that follows immediately from x0 = x(0 + 0) = x0 + x0. ) it seems to me that the four last lines of the section "Classical definition" have their natural place in the section "Consequences of the definition" :

"An equivalent, and more succinct, definition is: a field has two commutative operations, called addition and multiplication; it is a group under addition with 0 as the additive identity; the nonzero elements are a group under multiplication with 1 as the multiplicative identity; and multiplication distributes over addition.

Even more succinct: a field is a commutative ring where and all nonzero elements are invertible under multiplication.". 2A01:CB08:8607:CC00:5AEE:AA1D:1B8F:205B (talk) 06:00, 12 August 2023 (UTC)

The proof above shows that for any field F, (F/{0},.) is a multiplicative group. Therefore, by Gödel's completeness theorem, it is a theorem of field Theory. It seems strange to me that a so simply formulated theorem needs a so sophisticated proof! Do you know a simpler one ? 2A01:CB08:8607:CC00:8DB5:AA47:A0B7:45B (talk) 15:19, 15 August 2023 (UTC)

Can field-of-functions section use F for its field?

This whole article uses capital $F$ to represent an arbitrary field, except for this one specific section, § Geometry: field of functions, which uses lower-case $k$ instead (but without mentioning the difference). This seems unnecessarily confusing. Is there some reason it can't also use $F$ or at least capital $K$? –jacobolus (t) 19:46, 27 November 2023 (UTC)

None that I can see, so I have replaced $k$ by $F$. JBW (talk) 22:37, 27 November 2023 (UTC)

What does the tag mean?

@Beland: I have tried for 5 minutes to understand the meaning of the tag you recently placed on this article, but failed. Following the link to MOS:MATHSPECIAL was not illuminating. What's the problem supposed to be? --JBL (talk) 20:47, 30 January 2024 (UTC)

@JayBeeEll: This article uses Unicode character U+2216 ∖ SET MINUS in {{math}} blocks. According to MOS:MATHSPECIAL, those should be converted to <math>...</math> markup using \setminus or \smallsetminus for this character. I was not sure which is best, which is why I left the conversion to a math expert. (Looking back now, I do see \setminus in <math>...</math> blocks already in the article, so maybe that's our answer.) The rest of the {{math}} blocks containing this character also need to be converted to avoid complaints from other editors who think (and I agree) nesting <math>...</math> blocks inside {{math}} is too messy and hard to untangle. In doing these conversions, I find Help:Displaying a formula and general LaTeX syntax guides helpful. -- Beland (talk) 23:14, 30 January 2024 (UTC)

@Beland: Thanks. There is no distinction of meaning between \setminus and \smallsetminus, it's just a matter of authorial preference how to write it (similar to the various forms of lowercase epsilon or phi). Personally I like smallsetminus. I believe I've got every copy of the unicode character, but perhaps you could double check that I didn't miss any. --JBL (talk) 17:15, 1 February 2024 (UTC)

@JayBeeEll: Looks good; thanks for your help! -- Beland (talk) 19:18, 1 February 2024 (UTC)

0 ≠ 1 discrepancy

In the Classic Definition, there is no restriction forbidding a field over the 0 ring.

However, in the commutative ring definition just below, it stipulates that 0≠1.

Either this stipulation should be removed from the latter definition, or added to the former Farkle Griffen (talk) 04:40, 28 June 2024 (UTC)

In the first definition, the third axiom says explicitely that 0 and 1 are distinct. In the second one the nonzero elements are supposed to form a group with 1 as identity; that is 1 is nonzero. If you remove the condition 0≠1, you state that the zero ring is a field, which contradicts all textbooks that define fields. D.Lazard (talk) 07:52, 28 June 2024 (UTC)

Law of Distributivity Proof Mistake

I notice that the proof of the distributive property actually uses the distributive property in the proof, it is simply disguised. Step three states that (a/b)*((cf/df)+(ed/fd))=(a/b)*((cf+ed)/df). While true, rearranging this to use ^-1 notation reveals ab^-1(cfd^-1f^-1+edf^-1d^-1)=ab^-1((cf+ed)d^-1f^-1), its simply applying the distributive property to inverses. If the intent is to prove from axioms, it cannot use this and would need to take distributivity as an axiom itself. I'm a physics/math major who has taken several proof based mathematics classes and this is not sufficiently rigorous. OiT42 (talk) 15:02, 26 November 2024 (UTC)

Looks fine to me. A rational number is simply an equivalence class of pairs of integers (the article doesn't really emphasize this, but that's probably okay). It invokes the distributive property for integers in order to prove the property for rationals...again, this isn't really emphasized, but what it's doing is still valid. 35.139.154.158 (talk) 15:22, 26 November 2024 (UTC)

More explicitly: The step three that you quote is simply the rule for adding fractions with the same denominator. All but one other steps involve only the basic rules for manipulating fractions and integers. The only exception is in the last-but-one line, where distributivity of integers is involved (this is not a surprise for proving distributivity of fractions). On the other hand, your rearrangement is not convenient, since it involves operations with multiplicative inverses of integers, which are not well defined when one has not yet proved that the fractions form a field. D.Lazard (talk) 16:10, 26 November 2024 (UTC)

Often it is taken as a definition of addition of rational numbers that (c/d + e/f) = (cf + ed)/df. We expanded this 1 step into 2, instead only taking addition of fractions with like denominators for granted, like (a/d + b/d) = (a + b)/d. You have to have some kind of definition like this if you want to prove the distributive property, or else, as you notice, you could alternately take the distributive property as true and then prove the rule for addition of fractions based on it. –jacobolus (t) 18:08, 26 November 2024 (UTC)