Talk:Field (mathematics)

What element does not have a multiplicative inverse?
The use of 0 and 1 for multiplicative and additive identities is distracting and redirects the reader down the wrong path of thinking. This should be expressed more abstractly. The reason is because Im still not sure what element does not have a multiplicative inverse or why. It is said that 0 does not have a multiplicative inverse but Im trying to understand why, what is so unique about 0, that makes this statement true under arbitrary binary operations and arbitrary sets. 0 is the additive identity. Is it the case that the additive identity doesnt have a multiplicative inverse in any field, or just in this particular field? Or perhaps multiplicative invertibility has nothing to do with the additive identity at all, and that is just coincidental? Perhaps there is something else unique about 0 under multiplication that prohibits its invertibility? There is an unfortunate and I think false tie being drawn between what element is not invertible under multiplication and what element is the additive identity. 50.251.241.179 (talk) 01:44, 12 August 2018 (UTC)
 * The definition, as written, is (a) correct and (b) standard. We cannot change it to something unsourced.
 * Surely, there is something unique about 0 under multiplication that prohibits its invertibility. Here is it. Assume existence of $$a$$ such that $$0\cdot a=1.$$ Then $$1=0\cdot a=(0+0)\cdot a=0\cdot a+0\cdot a=1+1,$$ whence $$0=1$$ and therefore $$x=x\cdot1=x\cdot0=0$$ for all $$x.$$ Boris Tsirelson (talk) 05:10, 12 August 2018 (UTC)
 * It is by definition that every element but zero has to be invertible; otherwise all you have is a commutative ring. See division by zero.--Jasper Deng (talk) 07:36, 12 August 2018 (UTC)
 * Adding some explicitness to Boris Tsirelson: the multiplication and the addition are not freely moving around their respective domains, decreeing arbitrarily their unities, but they are both confined to almost the same domain by the distributive laws, which also take care of the minus × minus is plus. Purgy (talk) 08:06, 12 August 2018 (UTC)
 * Probably you mean $$(-1)\times(-1)=1;$$ generally a field is not ordered, and then, what are "plus" and "minus"? Boris Tsirelson (talk) 08:10, 12 August 2018 (UTC)
 * Yes, I did; I apologize for having been carried away to using just popular, but never the less inappropriate lingo :). Thanks. Purgy (talk) 08:28, 12 August 2018 (UTC)


 * Tell me if Im wrong, but what I have come to understand is that 0 is an absorption element under multiplication. Thus for all x in the Field, 0*x = x*0 = 0. So if 0*x = 0 and 0*y = 0 then there is no unique inverse of 0, as inv(0)*0 must be x and y.  Thus either the inverse is not unique or its application to 0 under the binary operation does not produce a unique result, and is not a function.  This is why 0 is not invertible under multiplication and why we remove it from consideration of invertible elements.  Thus a Field is any commutative ring with unity (F,+,*), and Z is a set of absorption elements in the monoid (F,*) and (F\Z,*) is a group. And it would seem then that there is no immediately obvious tie between the identity element 0 under + and the absorption element 0 under *. The nice thing is that I dont have to invoke distributivity or the addition operator to make this argument. As a side note, notice how concise my definition of a Field is. Something else nice about this definition is that I dont to name specific elements 0,1, etc., nor do I risk confusing the reader into thinking numbers specifically.  I also avoid the risk of drawing a false parallel between the additive identity and the element we reject for multiplicative inverses.  Tell me what you think. I dont know anything about "freely moving" elements. 50.35.97.238 (talk) 16:04, 8 November 2018 (UTC)
 * To be correct, your definition of a field should be rewritten as "a field is a commutative ring such that the elements different from the additive identity form a group under multiplication". This is certainly concise, but not more than the definition given in the article, if one adds to it the definition of the linked terms. Also, giving this definition in the article would be useful only for people who know commutative rings and do not know fields. As such people are very rare, there is no reason for giving this definition in the article. D.Lazard (talk) 16:37, 8 November 2018 (UTC)
 * I believe youve missed the entire point of this discussion. Different from the additive identity? Why? This whole conversation was about precisely that point. That the additive identity is irrelevant.  The definition in the article has like 20 different bullet point criterion to be met.  Far more verbose and convoluted. Why would anyone be studying fields without knowing rings first, since after all most of the required properties of the field precisely define it as a ring. It would be far easier for the layman to learn what a ring is first, dont you think? 50.35.97.238 (talk) 01:56, 9 November 2018 (UTC)
 * No, the point is that you are not satisfied by the standard definition of a field (this is your right), and you want changing the definition that appears in all algebra textbooks. There is no hope that you will succeed. D.Lazard (talk) 02:10, 9 November 2018 (UTC)


 * @50.35.97.238, I think the comment above by Boris Tsirelson gives a connection between the identity of addition and the non-invertible/absorbing of multiplication. This is also the point I tried to hint to with my informal (I beg pardon) attribute of "freely moving".
 * I share your opinion about defining a field via more elaborate algebraic structures (group, ring) being more clearly laid out, but I oppose to estimating this as more accessible to non-experts.
 * Finally, you have no chance of inserting something into WP that is so much off a well known mainstream, and furthermore, not prominently (second) sourced, not even as a viable alternative. Purgy (talk) 07:48, 9 November 2018 (UTC)

Edits on April 13-14, 2019
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 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 $f(e_{1} + e_{2}) = f(e_{1}) + f(e_{2})$, $f(e_{1}e_{2}) = f(e_{1})f(e_{2})$, and $f(1_{E}) = 1_{F}$, where $e_{1}$ and $e_{2}$ are arbitrary elements of $E$. "

Why such a condition is given to be a field homomorphism ? It is a consequence of : $f(e_{1} e_{2}) = f(e_{1}) f(e_{2})$ for every $e_{1}$ and $e_{2}$ 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 $$f(x)= \frac{x}{x} \ne 1 $$, 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, 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^{−1}$$a$$b$ = ($a^{−1}$$a$)$b$ = 1$b$ = $b$, but also $a^{−1}$$a$$b$ = $a^{−1}$($a$$b$) = $a^{−1}$0 = 0, which is a contradiction, as b is nonzero. That makes use of the fact that $x$0 = 0 for any $x$, but that follows immediately from $x$0 = $x$(0 + 0) = $x$0 + $x$0. 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?
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)
 * This article uses Unicode character in math blocks. According to MOS:MATHSPECIAL, those should be converted to  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  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  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)
 * 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)
 * 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)