Talk:Field (mathematics)

It is written that :

"Field homomorphisms are maps {{math|f: E → F}} between two fields such that {{math|1=f(e₁ + e₂) = f(e₁) + f(e₂)}}, {{math|1=f(e₁e₂) = f(e₁)f(e₂)}}, and {{math|1=f(1_E) = 1_F}}, where {{math|e₁}} and {{math|e₂}} 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(e₁ e₂) = f(e₁) f(e₂)}} for every {{math|e₁}} and {{math|e₂}} in {{math|E}}.

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)

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 $f(x)=\; \backslash frac\{x\}\{x\}\; \backslash ne\; 1$, 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#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)

"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, {{math|a}} and {{math|b}}, such that {{math|ab}} = 0. Then {{math|a⁻¹}}{{math|a}}{{math|b}} = ({{math|a⁻¹}}{{math|a}}){{math|b}} = 1{{math|b}} = {{math|b}}, but also {{math|a⁻¹}}{{math|a}}{{math|b}} = {{math|a⁻¹}}({{math|a}}{{math|b}}) = {{math|a⁻¹}}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. 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)

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)

{{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? --JBL (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. -- 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. --JBL (talk) 17:15, 1 February 2024 (UTC)

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

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)

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)

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

:There is already a section {{alink|Division rings}}. 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. JBW (talk) 10:28, 28 May 2025 (UTC)

The definition explicitly mentions negative elements. It could just be a convention to denote additive inverses as negative but I think it's misguided. Do others here feel the same?

To be clear, the simplest example to illustrate this is that of a finite field that doesn't have any 'negative' elements, and yet has additive inverses.

It could be expressed equally well by saying the additive inverse of a is b such that a+b=0. CallumMScott (talk) 13:12, 28 June 2025 (UTC)

:I do not understand your complaint; the word "negative" does not appear in the article. Certainly, the notation {{tmath|-x}} appears in the article, but this does not mean that it represets something negative. Even in the case of numbers, {{tmath|-x}} does not represent a negative number if {{tmath|x<0}}. In any case, the additive inverse of {{tmath|a}} is defined exactly as you suggest, except that it is denoted {{tmath|-a}} instead of {{tmath|b}}. D.Lazard (talk) 13:35, 28 June 2025 (UTC)

:{{pb}}{{ec}} Denoting the additive inverse of $a$ by $-a$ is completely standard (which I'm assuming is the actual complaint). In the field with 3 elements (say, 0, 1, and 2), it just so happens that $-1\; =\; 2$. This is all fine. 35.139.154.158 (talk) 13:38, 28 June 2025 (UTC)

::And indeed, using the symbols {{tmath|\{-1, 0, 1\} }} for the elements of this field is often quite a bit clearer than {{tmath|\{0, 1, 2\} }} in my opinion. –jacobolus (t) 19:56, 28 June 2025 (UTC)

:This seems like approximately the same confusion as the one that leads students to become confused by the sentence "if x is a negative real number then the absolute value of x is negative x". The first use of the word "negative" is about a trichotomy, the second use of the word "negative" is synonymous with "additive inverse". The first use (trichotomy) doesn't appear in general fields, but second (additive inverse) does and is completely standard, along with the associated notation $-x$ for the additive inverse of $x$, and the consequent validity of the notation $x\; -\; y$. (In fact these are all natural and standard in any abelian group when represented additively.) --JBL (talk) 17:41, 28 June 2025 (UTC)