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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)
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)
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)
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(1E) = 1F, 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.
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 , 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)
"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)
This whole article uses capital to represent an arbitrary field, except for this one specific section, § Geometry: field of functions, which uses lower-case instead (but without mentioning the difference). This seems unnecessarily confusing. Is there some reason it can't also use or at least capital ? – jacobolus (t) 19:46, 27 November 2023 (UTC)
@ 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)
<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)
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)
![]() | Field (mathematics) has been listed as one of the
Mathematics good articles under the
good article criteria. If you can improve it further,
please do so. If it no longer meets these criteria, you can
reassess it. Review: June 5, 2017. ( Reviewed version). |
![]() | This ![]() It is of interest to the following WikiProjects: | ||||||||||
|
This page has archives. Sections older than 365 days may be automatically archived by Lowercase sigmabot III when more than 10 sections are present. |
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)
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)
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)
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(1E) = 1F, 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.
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 , 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)
"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)
This whole article uses capital to represent an arbitrary field, except for this one specific section, § Geometry: field of functions, which uses lower-case instead (but without mentioning the difference). This seems unnecessarily confusing. Is there some reason it can't also use or at least capital ? – jacobolus (t) 19:46, 27 November 2023 (UTC)
@ 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)
<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)
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)