View source for Talk:Field (mathematics)

{{GA|14:21, 5 June 2017 (UTC)|oldid=783918528|topic=Mathematics and mathematicians|page=1}}
{{WikiProject banner shell|class=GA|vital=yes|1=
{{WikiProject Mathematics|importance=Top}}
}}

{{User:MiszaBot/config
| algo = old(365d)
| archive  = Talk:Field (mathematics)/Archive %(counter)d
| counter  = 1
| maxarchivesize  = 150K
| archiveheader  = {{Automatic archive navigator}}
| minthreadstoarchive = 1
| minthreadsleft = 10
}}
{{Archive box |search=yes |bot=Lowercase sigmabot III |age=12 |units=months |auto=yes }}

==Is vector field a field?==
At the end of the article, in the {{section link|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.  [[User:Micha7a|Micha7a]] ([[User talk: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? [[User:Jakob.scholbach|Jakob.scholbach]] ([[User talk: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? [[User:Micha7a|Micha7a]] ([[User talk: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. [[User:Jakob.scholbach|Jakob.scholbach]] ([[User talk: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! 😁😁😁 [[Special:Contributions/67.198.37.16|67.198.37.16]] ([[User talk: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. [[User:D.Lazard|D.Lazard]] ([[User talk:D.Lazard|talk]]) 08:09, 6 October 2020 (UTC)

== On a condition of field morphisms ==

It is written that :

"[[Field homomorphism]]s are maps {{math|''f'': ''E'' → ''F''}} between two fields such that {{math|1=''f''(''e''<sub>1</sub> + ''e''<sub>2</sub>) = ''f''(''e''<sub>1</sub>) + ''f''(''e''<sub>2</sub>)}}, {{math|1=''f''(''e''<sub>1</sub>''e''<sub>2</sub>) = ''f''(''e''<sub>1</sub>)''f''(''e''<sub>2</sub>)}}, and '''{{math|1=''f''(1<sub>E</sub>) = 1<sub>F</sub>}},''' where {{math|''e''<sub>1</sub>}} and {{math|''e''<sub>2</sub>}} are arbitrary elements of {{math|''E''}}. "

Why such a condition is given to be a field homomorphism ? It is a consequence of : {{math|1=''f''(''e''<sub>1</sub>  ''e''<sub>2</sub>) = ''f''(''e''<sub>1</sub>) ''f''(''e''<sub>2</sub>)}} for every {{math|''e''<sub>1</sub>}} and {{math|''e''<sub>2</sub>}} in {{math|''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? [[Special:Contributions/93.173.65.88|93.173.65.88]] ([[User talk: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 <math>f(x)= \frac{x}{x} \ne 1 </math>, 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. [[User:TMM53|TMM53]] ([[User talk:TMM53|talk]]) 00:51, 25 October 2022 (UTC)[[User:TMM53|TMM53]] ([[User talk: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. --[[User:Trovatore|Trovatore]] ([[User talk: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). [[Special:Contributions/2A01:CB08:8607:CC00:7C0A:14D4:51F9:DC0A|2A01:CB08:8607:CC00:7C0A:14D4:51F9:DC0A]] ([[User talk: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. [[User:JBW|JBW]] ([[User talk: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. [[Special:Contributions/2A01:CB08:8607:CC00:15BC:2EE1:4CB8:F98A|2A01:CB08:8607:CC00:15BC:2EE1:4CB8:F98A]] ([[User talk:2A01:CB08:8607:CC00:15BC:2EE1:4CB8:F98A|talk]]) 11:41, 11 August 2023 (UTC)
:::Suppose there are two nonzero elements, {{math|''a''}} and {{math|''b''}}, such that {{math|''ab''}} = 0. Then {{math|''a''<sup>−1</sup>}}{{math|''a''}}{{math|''b''}} = ({{math|''a''<sup>−1</sup>}}{{math|''a''}}){{math|''b''}} = 1{{math|''b''}} = {{math|''b''}}, but also {{math|''a''<sup>−1</sup>}}{{math|''a''}}{{math|''b''}} = {{math|''a''<sup>−1</sup>}}({{math|''a''}}{{math|''b''}}) = {{math|''a''<sup>−1</sup>}}0 = 0, which is a contradiction, as b is nonzero. That makes use of the fact that {{math|''x''}}0 = 0 for any {{math|''x''}}, but that follows immediately from {{math|''x''}}0 = {{math|''x''}}(0 + 0) = {{math|''x''}}0 + {{math|''x''}}0. [[User:JBW|JBW]] ([[User talk: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.". [[Special:Contributions/2A01:CB08:8607:CC00:5AEE:AA1D:1B8F:205B|2A01:CB08:8607:CC00:5AEE:AA1D:1B8F:205B]] ([[User talk: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 ? [[Special:Contributions/2A01:CB08:8607:CC00:8DB5:AA47:A0B7:45B|2A01:CB08:8607:CC00:8DB5:AA47:A0B7:45B]] ([[User talk: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 <math>F</math> to represent an arbitrary field, except for this one specific section, [[Field (mathematics)#Geometry: field of functions|§&nbsp;Geometry: field of functions]], which uses lower-case <math>k</math> instead (but without mentioning the difference). This seems unnecessarily confusing. Is there some reason it can't also use <math>F</math> or at least capital <math>K</math>? –[[user:jacobolus|jacobolus]] [[User_talk:jacobolus|(t)]] 19:46, 27 November 2023 (UTC)
: None that I can see, so I have replaced <math>k</math> by <math>F</math>. [[User:JBW|JBW]] ([[User talk:JBW|talk]]) 22:37, 27 November 2023 (UTC)

== What does the tag mean? ==

{{ping|Beland}} I have tried for 5 minutes to understand the meaning of the [https://en.wikipedia.org/w/index.php?title=Field_(mathematics)&diff=prev&oldid=1198867509 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?  --[[User:JayBeeEll|JBL]] ([[User_talk:JayBeeEll|talk]]) 20:47, 30 January 2024 (UTC)
:{{reply|JayBeeEll}} This article uses Unicode character {{unichar|2216|SET MINUS}} in {{tl|math}} blocks. According to [[MOS:MATHSPECIAL]], those should be converted to {{tag|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 {{tag|math}} blocks already in the article, so maybe that's our answer.) The rest of the {{tl|math}} blocks containing this character also need to be converted to avoid complaints from other editors who think (and I agree) nesting {{tag|math}} blocks inside {{tl|math}} is too messy and hard to untangle. In doing these conversions, I find [[Help:Displaying a formula]] and general LaTeX syntax guides helpful. -- [[User:Beland|Beland]] ([[User talk:Beland|talk]]) 23:14, 30 January 2024 (UTC)
::{{ping|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.  --[[User:JayBeeEll|JBL]] ([[User_talk:JayBeeEll|talk]]) 17:15, 1 February 2024 (UTC)
::{{reply|JayBeeEll}} Looks good; thanks for your help! -- [[User:Beland|Beland]] ([[User talk: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 [[User:Farkle Griffen|Farkle Griffen]] ([[User talk: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. [[User:D.Lazard|D.Lazard]] ([[User talk: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 <sup>-1</sup> notation reveals ab<sup>-1</sup>(cfd<sup>-1</sup>f<sup>-1</sup>+edf<sup>-1</sup>d<sup>-1</sup>)=ab<sup>-1</sup>((cf+ed)d<sup>-1</sup>f<sup>-1</sup>), 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. [[User:OiT42|OiT42]] ([[User talk: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. [[Special:Contributions/35.139.154.158|35.139.154.158]] ([[User talk: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. [[User:D.Lazard|D.Lazard]] ([[User talk: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. –[[user:jacobolus|jacobolus]] [[user_talk:jacobolus|(t)]] 18:08, 26 November 2024 (UTC)

== Field versus [[Division ring]] ==

Shouldn't the article mention to older English convention where a field is not assumed to be commutative, and mention [[Division ring]] in a hatnot or See also? -- [[User:Chatul|Shmuel (Seymour J.) Metz Username:Chatul]] ([[User talk:Chatul|talk]]) 20:42, 27 May 2025 (UTC)

:There is already a section {{alink|Division rings}}. [[User:D.Lazard|D.Lazard]] ([[User talk:D.Lazard|talk]]) 21:12, 27 May 2025 (UTC)

:: Yes, but that doesn't clearly address the point, which {{u|Chatul}} has raised, that in older works "field" had a broader meaning than it has now. There is a note about this, but it was shown only in a footnote, along with numerous other notes and references, where few people would be likely to see it. I have put the note into the text of the section on division rings. [[User:JBW|JBW]] ([[User talk:JBW|talk]]) 10:28, 28 May 2025 (UTC)