Talk:Finite field

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I would like to suggest using the GF(q) notation in the article, as it's easy to write GF(pn), but the equivalent with Fq notation cannot be written very satisfactarily in HTML (witness the penultimate paragraph of the article). Would anyone object strongly if I changed the article to use GF(q) notation throughout? --Zundark, 2001 Dec 10

Having written that, I see that Fpm does display correctly in Netscape. But in IE4 it looks the same as Fpm, which is not good. --Zundark, 2001 Dec 10

In IE 5.5, it displays ok but looks somewhat crappy. It seems to be safer to use the GF notation. --AxelBoldt

It seems to be time to use Fpm as it should be supported by modern browsers. -- 14:07, 5 August 2006 (UTC)'s changes

Maybe someone who knows about the subject wouldn't mind reviewing the changes done by to this article. She seems to have subtly vandalised Fascism and started the bogus entry Marxianism. Thanks. --snoyes 15:15 Feb 19, 2003 (UTC)

It looks as if she changed "T^3 + T^2 + T + 1 is irreducible" to "T^3 + T^2 + T - 1 is irreducible". Though my algebra is rusty, this seems to be a good and necessary change as (T^3 + T^2 + T + 1) = (T^2+1)(T+1) whereas there are no factors (mod 3) of (T^3 + T^2 + T - 1).

Thanks. Admittedly I didn't bother trying to understand the problem (it is of course fairly easy as you have demonstrated). I just saw a sign change in the tex source and "saw red" ;-) --snoyes 15:59 Feb 19, 2003 (UTC)

how finite field may be implemented in AES cryptanalysis can be elaborated with in this article —Preceding unsigned comment added by (talk) 11:31, 5 September 2008 (UTC)

classification section

A few things in the "classification" section bothered me, so I rewrote it. First, no effort had been made to justify the existence of an irreducible polynomial of the appropriate degree, which is by no means a priori obvious; in fact I'm pretty sure you need to prove existence of the finite field before you can assert such a polynomial exists. (Please someone correct me if this is wrong.) Secondly, after I worked on filling in more details of the existence proof, it seemed worthwhile to put the main facts of the classification upfront, and then put the derivation afterwards. Dmharvey 13:31, 4 April 2006 (UTC)

  • The second part is definively too long. You simply forget that in a ring finite or not finite the additive law is abelian (a consequence of distributivity ), this fact should be set clearly at the beginning of the article to avoid some confusions related to commutativity.

irreducible poly equation?

It seems from your examples that to actually get the multiplication table of a finite field we will need the appropriate irreductible polynomial. 1) Is there a general formula for given p and n? reference? 2) Are there simple formulas for some large class of p and n? reference? 3) If the answer to 1 is no then I think this should be mentioned--preferably with a reference.

~ interested

more on irreducible poly

I think someone who understands these things should write a paragraph or two on the construction of the needed irreducible poly. It might start out with something like

"Note that without the needed irreducible polynomial that GF(p^n) is a purely theoretical object and somewhat nebulous. There seems to be no general foumula. There are varous tables of polynomials for various p and n. Such a polynomial always exists and there are various algorithms for constructing it."

The knowledgable writer could then talk about the efficiency of such algorithms, etc. It need not be formal--just enough to impress upon the reader that you have got to have such a poly to do anything with the field


In fact, I believe that there is no polynomial-time construction of any finite field. Conjectures of the Geometric Langlands Program will give one, though. Eulerianpath (talk) 22:43, 5 January 2008 (UTC)

Finite division rings vs finite fields

I believe this removal is incorrect. Finite fields are the same as finite division rings, per Wedderburn's theorem. As such, even if the Japanese Wikipedia is different from the English Wikipedia as far what is a field/division ring, the Japanese article removed by the above edit is the perfect counterpart of our article at finite field and the link must come back. Comments? Oleg Alexandrov (talk) 18:53, 6 April 2006 (UTC)

As clarification, here is the corresponding Japanese article which DYLAN thinks should not be the interwiki counterpart of our finite field article. Oleg Alexandrov (talk) 18:58, 6 April 2006 (UTC)

I'm not saying that finite division ring is not finite field. I'm just saying that Wedderburn's theorem is about the property of finite division ring. Happening to be a finite field is just the consequence of the theorem. Therefore, it shouldn't be listed as it now is.

Same thing to Japanese interwiki DYLAN LENNON 19:15, 6 April 2006 (UTC)

Disagree with DYLAN. The result is closely enough related to finite fields to warrant a mention on this page. By the way DYLAN, thanks for expanding on your actions on this talk page; it's greatly appreciated. Please continue to do that. Dmharvey 19:45, 6 April 2006 (UTC)

But the primary thing I want to get across is not so much the removal by WAREL of the paragraph

Division rings are algebraic structures more general than fields, as they are not assumed to be necessarily commutative. Wedderburn's (little) theorem states that all finite division rings are commutative, hence finite fields.

That may be a matter of taste. My point is that since a finite field is the same as a finite division ring, the interwiki link from this finite field article must go to the Japanese "finite division ring or whatever" article, as they are talking about the same thing, even if starting with different assumptions about commutativity. Does the Japanese Wikipedia have separate entries for "finite division ring" and "finite field"? I hightly doubt. Oleg Alexandrov (talk) 20:25, 6 April 2006 (UTC)

If there is indeed only one article on wikpedia.ja that deals with this topic, I have no idea how WAREL could convince even him/herself that it improves the article to remove the link. But perhaps that wasn't really the issue. Elroch 22:30, 6 April 2006 (UTC)

Since it's a math article, we should be strictly as we can about what we are talking. 有限体 means finite division ring. Even if every finite division ring is indeed finite field, there is a difference. We won't link a artcle of "natural number" to that of "integer" even every natural number is integer, do we? DYLAN LENNON 23:50, 6 April 2006 (UTC)

Yes, actually we do. Splendid example. Dmharvey 00:23, 7 April 2006 (UTC)
DYLAN, what I am saying is that our English article about finite fields is the direct analog of the Japanese article about finite division rings. Things start with different assumptions, but the material in both articles is essentially the same (and you don't need to know Japanese to understand that). As such, the interwiki link to the Japanese article is appropriate. Oleg Alexandrov (talk) 01:13, 7 April 2006 (UTC)
Also, your example of integers is invalid. Every natural number is an integer, but not every integer is natural number. However, every finite division ring is a finite field, and every finite field is a finite division ring. Oleg Alexandrov (talk) 01:15, 7 April 2006 (UTC)

Since every finite field is a finite division ring, I am pursuaded that the link is valid. However, I think Wedderburn's theorem should not be listed here but on the article of "division ring". DYLAN LENNON 12:15, 7 April 2006 (UTC)

DYLAN, I find it amazing that you have not yet inferred that if you repeatedly make an edit with which everyone disagrees, you will get repeatedly reverted back. This has happened to you numerous times already. You will not get your way by repetition alone. Rather, you will get blocked. The only way you will get your way is if you write down an argument that convinces other editors. It is not good enough to write down an argument that convinces yourself. I find it puzzling that you haven't realised this yet, because you are obviously a reasonably intelligent person. Can you explain why you continue to change the article, when all of the evidence should make it clear that someone will revert it back straight away? Dmharvey 13:06, 7 April 2006 (UTC)


isn't this article sloppy in discriminating between things that are equal and things that are just isomorphic? --Lunni Fring 21:11, 23 September 2006 (UTC)

I can see one place where it says 'the', and it might say 'a'. Why do you ask? Charles Matthews 21:32, 23 September 2006 (UTC)
Well one example is "Then GF(q) = GF(p)[T] / <f(T)>." But GF(q) is really isomorphic to the RHS. -- 22:37, 23 September 2006 (UTC)

Mathematicians who contribute to WP are sloppy about most everything, especially with the idea that people who aren't mathematicians might want to learn a bit about higher mathematics, but most Wikipedia articles covering any area of mathematics are as heavy on terseness as articles in a particular specialty, but in snippets and paragraphs yanked out of a context the WP writer knows so well they assume everyone else does. And the notations, some non-standard usage, rarely defined..I could go on. Brevity and clarity rarely coincide. This is why introductory books are so much better for the self taught. The forces of comepetition give the author an incentive to actually be clear and concise, and lay the ideas out in an orderly fashion. Books of this type don't interject advanced results in the middle of an introductory paragraph on finite fields. Unless mathematics on Wikipedia is for mathematicians only, I don't think everyone just looking up the basic idea of finite fields is also looking to worry about whether they should have to chase down the meaning of Morita equivalence classes just to get the gist of something, especially in the first paragraph.

Lead Section Inconsistency

The definition "GF is a field that contains only finitely many elements." is inconsistent with the fact that a field of 6 elements has a finite number of elements, yet it is not a GF. John 20:12, 13 March 2007 (UTC)

There is no field of six elements, so there is no inconsistency. --Zundark 22:36, 13 March 2007 (UTC)
It seems to me that is true only if you define a field as a GF, and that makes it a circular definition. Why is there no field of 6 elements? Do we need to clarify or define field so that is true? Otherwise why should a reader who is reading this to find out what a GF is assume that? John 05:47, 15 March 2007 (UTC)
This article links to the field (mathematics) article, which provides the definition of field. The non-existence of fields of 6 elements follows from the definition by elementary arguments (which are already given in the article). --Zundark 09:45, 15 March 2007 (UTC)
You are right, I agree, I was under the mistaken impression that field was not defined or linked. I see now. Thanks. John 16:54, 15 March 2007 (UTC)

Sum of all field elements

Please see Talk:Field (mathematics)#Sum of all field elements. —The preceding unsigned comment was added by (talk) 20:27, 15 April 2007 (UTC).

Naming of "Galois Fields"

I have been told that referring to finite fields as "Galois fields" is a little odd, with the reason being that Galois himself never worked with finite fields. This may be worth mentioning if anyone can confirm it. 19:11, 15 May 2007 (UTC)

The EoM entry on Galois fields says they were first considered by Galois, and gives a reference. --Zundark 20:05, 15 May 2007 (UTC)
Galois not only worked with finite fields, it's quite probable that he was the first one to come up with the concept. See Évariste Galois, "Sur la théorie des nombres". Bulletin des Sciences mathématiques XIII: p. 428ff (1830). It's also part of the collection of Galois' mathematical works published by Joseph Liouville in 1846, and the Bibliothéque Nationale de France helpfully makes this available here (use the interface to go to page 381. It's in French, of course, but those are the breaks). --Stormwyrm (talk) 04:45, 19 February 2009 (UTC)

Proof that a Finite Field has Order [tex]p^n[/tex]

The first proof claims that a finite field is a vector space over Z/pZ, where p is the characteristic, and then goes on to conclude that the order of the field must be [tex]p^n[/tex]. But it seems proving that it is a vector space over Z/pZ is the heart of the proof; why leave it out? —Preceding unsigned comment added by (talk) 07:08, 25 February 2008 (UTC)

  • I agree, this is the heart of the proof. Yet in this writeup it is simply stated as fact. I think a few words or a sentence or two explaining this could make the proof infinitely better. The Raven (talk) 21:51, 19 August 2010 (UTC)

simpler explanation

Something I think would be useful is to be told that the additive group of a finite field is primitive abelian, and its multiplicative group is cyclic. This seems to me to explain in simple terms exactly how every finite field works. But instead I see stuff like "(Z/2Z)[T]/(T2+T+1)", which is completely opaque to me, and I suspect far more complicated than necessary. Maproom (talk) 08:56, 29 May 2008 (UTC)

The article already mentions that the multiplicative group is cyclic. What does primitive abelian mean? In any case, information about the additive and multiplicative groups does not "explain exactly how every finite field works", since it is the relationship between them that matters. Algebraist 16:03, 11 February 2009 (UTC)
A "primitive Abelian" group is the direct product of a number of cyclic groups of prime order. Maproom (talk) 16:08, 11 February 2009 (UTC)
That's mentioned in the article too. Do you think it should be more prominent? Algebraist 16:12, 11 February 2009 (UTC)
The additive group of a finite field is an elementary abelian group. There may be a translation problem; there is exactly one hit from mathscinet for "primitive abelian group", and such a group is torsion-free, so never finite. A primitive permutation group that is abelian is cyclic of prime order, and this appears to account for all other usages on google.
At any rate, to agree with both Maproom and Algebraist, viewing a finite field as an especially nice subring of the full matrix ring M_n(Z/pZ) is a very good way to understand finite fields that explicitly describes the additive group, the multiplicative group, and the action of the multiplicative group on the additive group. A reasonable choice of subring is the polynomials in the companion matrix of an irreducible polynomial of degree n over Z/pZ. This also explains many applications such as LFSRs. JackSchmidt (talk) 16:48, 11 February 2009 (UTC)

I think a simpler explanation is definitely needed. An example from the article's intro, "Each finite field of size q is the splitting field of the polynomial xq - x, and thus the fixed field of the Frobenius endomorphism which takes x to xq." This type of jargon is extremely opaque to those unfamiliar with the subject (read, me). The subject matter should be discussed in terms of simpler mathematical structures. The reason to read this article is to learn what a finite field is, so it should be assumed that "splitting fields" and "Frobenius endomorphisms" are also foreign concepts. --Pesto (talk) 06:45, 11 September 2009 (UTC)

Could someone explain how the notation in "(Z/2Z)[T]/(T2+T+1)" works? It is completely new to me and I can't find any articles on it. (talk) 05:08, 14 August 2010 (UTC)Nathan
Z/2Z is the ring of integers modulo 2. (Z/2Z)[T] is the polynomial ring in the variable T over Z/2Z. T2 + T + 1 is an element in the polynomial ring (Z/2Z)[T] (i.e. it is a polynomial with coefficients in Z/2Z). (T2 + T + 1) is the principal ideal of (Z/2Z)[T] generated by T2 + T + 1, and (Z/2Z)[T]/(T2 + T + 1) is the quotient of the ring (Z/2Z)[T] by the ideal (T2 + T + 1). RobHar (talk) 05:33, 14 August 2010 (UTC)

Reference suggestion

I suggest that reference be made to "Arithmetic over finite fields" in Collected Algorithms of ACM (?467) by Lam & McKay. It uses Zech logarithms, is short and efficient. John McKay132.205.221.161 (talk) 15:53, 12 February 2009 (UTC)

the finite field is not completely known

I tried to delete the following passage "The finite fields are completely known", first of all because it is a false statement, secondly, since when is something "completely known"? There is no reason to emphasize that a theorem completely holds, and thirdly and most important the quoted sentence imply that there is nothing new to discover in the area of finite fields and that research in that area is pointless. Whatever you (the nice editor that reverted my delete) are trying to say it comes off wrong.

I think the wording should change to include a context, or be deleted altogether, please express your concerns if you have them or I'll delete it again. —Preceding unsigned comment added by Sonoluminesence (talkcontribs) 09:03, 28 March 2009 (UTC)

I (the reverter) am not trying to say anything. An unexplained deletion of a large percentage of the introduction is a standard thing to revert (indistinguishable from random vandalism). An WP:edit summary is normally sufficient to explain the deletion, but in this case it is better to try improve the introduction.
We can guess at what the original editor meant, and try to improve it. A purpose of the WP:lead section, amongst others, is to summarize the article, and at least half the article is describing the classification of finite fields. Presumably the original author was trying to summarize this. I changed the wording to be more direct, "The finite fields are classified by order." JackSchmidt (talk) 05:58, 29 March 2009 (UTC)

Hello JackSchmidt, thank you for your response and change, I think the wording is now exponentially better. Sonoluminesence (talk) 07:48, 29 March 2009 (UTC)

"The finite fields are fully known" makes sense to me, and tells me something useful about finite fields. "The finite fields are classified by order" merely tells me how people have chosen to classify them. The former statement (if true, which I think it is) seems far more useful. Maproom (talk) 22:24, 29 March 2009 (UTC)
Unfortunately there are many very basic questions about Z/pZ that are not known, so the statement is either meaningless (my opinion) or false (probably Sono's opinion). A silly example of great practical importance is implicit in Primitive root modulo n#Order of magnitude of primitive roots; without the GRH estimates on the smallest generator are fairly poor, so until GRH is proven, our knowledge on even a very basic question is extremely incomplete (and even then, I don't know that the estimates are sharp enough for more serious computational applications). At any rate, if you can find WP:reliable sources to support the claim, then feel free to expand the article (not the lead) with the new material. This will make a good excuse to fix the current introduction which neglects to summarize over half the article. JackSchmidt (talk) 22:43, 29 March 2009 (UTC)
Also, to make it clear research is on-going in finite fields (which might make a nice "Finite fields are completely classified, but are still actively researched." sort of comparison), there is the short section Kakeya conjecture#Kakeya sets in vector spaces over finite fields. There are lots of problems over uncountable fields that are not resolved over finite fields (and lots of the reverse). Basically finite fields are a healthy, vibrant area of research mathematics. JackSchmidt (talk) 22:56, 29 March 2009 (UTC)
I'll be happy with "Finite fields are completely classified, but are still actively researched." "Completely classified" is the kind of thing I can understand, and want to know. Maproom (talk) 08:49, 30 March 2009 (UTC)
Why do you insist on putting the word "complete"? It doesn't appear with that context in the article. If you think that it is complete, please explain in what way are the classifications complete? The wording you suggest doesn't contribute anything - not about the finite field nor its classifications, the current wording actually says something new which is explained right on the next paragraph. I vote to keep the current wording. Sonoluminesence (talk) 10:55, 31 March 2009 (UTC)
When I used the word "complete" I was quoting a phrase which has been removed from the article. It seems appropriate - all finite fields are know, there is one for each prime power and no others (unless I have misunderstood something). But you write "not about the finite field nor its classifications", which puzzles me – there is more than one finite field – so we may be talking about different things. Maproom (talk) 15:06, 31 March 2009 (UTC)
First of all, thank you for participating, second of all, there is a finite field for every p^k where k is an integer, third of all, there are finite fields for integers other than p^k, they just can't be used in the same context (although they share similar properties), 4th of all, even if you know there is a finite field for every p^k you haven't constructed every single one, and finally, I think it is better to say the actual meaning of a mathematical theorem rather than inflating its importance. Sonoluminesence (talk) 08:34, 1 April 2009 (UTC)
You write "there are finite fields for integers other than p^k", but the article states "We give two proofs that a finite field has prime-power order." Can you explain this contradiction, and maybe correct the article? Maproom (talk) 09:30, 1 April 2009 (UTC)
I can't do that without violating WP:COI and WP:NOR, but there is a meaning to objects which are identical to the finite field but lack the p^k property. For example, one can take the multiplication table modulo 6 and see similarities with a finite field although 6 is not p^k. But that is not the point, my argument was only that the sentence "the finite fields are completely known" is false. Are we in agreement on this? Sonoluminesence (talk) 07:26, 2 April 2009 (UTC)
You write of objects "which are identical to the finite field but lack the p^k property", and give the integers modulo 6 as an example. So what do you think is 3/2 in that "field"?
Anyway, I am not in agreement. That statement was true and useful. To justify its removal, you first used "finite field" in the singular as if there were only one such field, then you claimed that there are finite fields not of prime power order. My point is that it is fully known what finite fields exist, and it would be helpful to have this stated in the lead-in. I accept that "the finite fields are completely known" may mislead people into thinking that everything about them is known, which is false. So can we find a way of making it clear, in the lead-in, that it is known what finite fields exist? In my view "The finite fields are classified by their size" does not do this. Maproom (talk) 09:16, 2 April 2009 (UTC)
As I said, the object is used in different context which exclude it from being a field. If a theorem existed that included all the p^k finite fields, and it included objects for any integer k, that are the same for some attributes needed for the theorem but are not "fields" in the traditional sense (call it a semi-field…) – wouldn't you call your understanding of finite fields incomplete? Besides how do you know the finite fields doesn't have connections to other mathematical areas which we are unaware of, for example to the Riemann hypothesis, or to the prime number theorem? I still think the term "completely known" is completely false. Sonoluminesence (talk) 13:25, 2 April 2009 (UTC)
Perhaps this would be reasonable:
The finite fields are classified by size; there is exactly one finite field up to isomorphism of size pk for each prime p and positive integer k. Each finite field of size q is the splitting field of the polynomial xq - x, and thus the fixed field of the Frobenius endomorphism which takes x to xq. Similarly, the multiplicative group of the field is a cyclic group. Wedderburn's little theorem states that the Brauer group of a finite field is trivial, so that every finite division ring is a finite field. Finite fields have applications in many areas of mathematics and computer science, including coding theory, LFSRs, modular representation theory, and the groups of Lie type. Finite fields are an active area of research, including recent results on the Kakeya conjecture and open problems on the size of the smallest primitive root.
One can add the applications and open problems to the end of the article fairly easily. JackSchmidt (talk) 12:52, 2 April 2009 (UTC)
It is more than reasonable it is great, good job! Sonoluminesence (talk) 13:25, 2 April 2009 (UTC)
I too am very happy with JackSchmidt's proposed wording. Maproom (talk) 14:40, 2 April 2009 (UTC)

Some small finite fields example

The example given for F4 is:

 + | 0 1 A B       × | 0 1 A B
 --+--------       --+--------
 0 | 0 1 A B       0 | 0 0 0 0
 1 | 1 0 B A       1 | 0 1 A B
 A | A B 0 1       A | 0 A B 1
 B | B A 1 0       B | 0 B 1 A

Shouldn't that be like this:

 + | 0 1 2 3       × | 0 1 2 3
 --+--------       --+--------
 0 | 0 1 2 3       0 | 0 0 0 0
 1 | 1 2 3 0       1 | 0 1 2 3
 2 | 2 3 0 1       2 | 0 2 0 2
 3 | 3 0 1 0       3 | 0 3 2 1


So the translation should go like this:

 + | 0 1 A B       × | 0 1 A B
 --+--------       --+--------
 0 | 0 1 A B       0 | 0 0 0 0
 1 | 1 A B 0       1 | 0 1 A B
 A | A B 0 1       A | 0 A 0 A
 B | B 0 1 0       B | 0 B A 1

Am I wrong? —Preceding unsigned comment added by Sonoluminesence (talkcontribs) 11:19, 3 April 2009 (UTC)

Yes, you are. This is the field of four elements, not the ring of integers mod 4. Algebraist 12:53, 7 April 2009 (UTC)

for gods sake lets have some EXAMPLES of the galois field phenom without notation or explanation. just plain number patterns. some of us are dyslexic & respond best to number patterns. isnt that what math is all about anyway. were not all writing phd theses. EXAMPLES PLEASE. THE MORE THE BETTER. dieutoutpuissant —Preceding unsigned comment added by Dieutoutpuissant (talkcontribs) 20:08, 27 May 2009 (UTC)

I thought maybe some one would like to see some more examples of small
finite fields.  James

Field of 8 elements represented as matrices
integers are modulo 2

element (0)     element (1)     element (2)     element (3)     

  0  0  0         1  0  0         0  1  0         0  0  1       
  0  0  0         0  1  0         0  0  1         1  1  0       
  0  0  0         0  0  1         1  1  0         0  1  1       

element (4)     element (5)     element (6)     element (7)     

  1  1  0         0  1  1         1  1  1         1  0  1       
  0  1  1         1  1  1         1  0  1         1  0  0       
  1  1  1         1  0  1         1  0  0         0  1  0       

+/  (0) (1) (2) (3) (4) (5) (6) (7) 
(0)  0   1   2   3   4   5   6   7  
(1)  1   0   4   7   2   6   5   3  
(2)  2   4   0   5   1   3   7   6  
(3)  3   7   5   0   6   2   4   1  
(4)  4   2   1   6   0   7   3   5  
(5)  5   6   3   2   7   0   1   4  
(6)  6   5   7   4   3   1   0   2  
(7)  7   3   6   1   5   4   2   0  

x/  (0) (1) (2) (3) (4) (5) (6) (7) 
(0)  0   0   0   0   0   0   0   0  
(1)  0   1   2   3   4   5   6   7  
(2)  0   2   3   4   5   6   7   1  
(3)  0   3   4   5   6   7   1   2  
(4)  0   4   5   6   7   1   2   3  
(5)  0   5   6   7   1   2   3   4  
(6)  0   6   7   1   2   3   4   5  
(7)  0   7   1   2   3   4   5   6  


Field of 9 elements represented as matrices
integers are modulo 3

element (0)     element (1)     element (2)     

  0  0            1  0            0  1          
  0  0            0  1            1  1          

element (3)     element (4)     element (5)     

  1  1            1  2            2  0          
  1  2            2  0            0  2          

element (6)     element (7)     element (8)     

  0  2            2  2            2  1          
  2  2            2  1            1  0          

+/  (0) (1) (2) (3) (4) (5) (6) (7) (8) 
(0)  0   1   2   3   4   5   6   7   8  
(1)  1   5   3   8   7   0   4   6   2  
(2)  2   3   6   4   1   8   0   5   7  
(3)  3   8   4   7   5   2   1   0   6  
(4)  4   7   1   5   8   6   3   2   0  
(5)  5   0   8   2   6   1   7   4   3  
(6)  6   4   0   1   3   7   2   8   5  
(7)  7   6   5   0   2   4   8   3   1  
(8)  8   2   7   6   0   3   5   1   4  

x/  (0) (1) (2) (3) (4) (5) (6) (7) (8) 
(0)  0   0   0   0   0   0   0   0   0  
(1)  0   1   2   3   4   5   6   7   8  
(2)  0   2   3   4   5   6   7   8   1  
(3)  0   3   4   5   6   7   8   1   2  
(4)  0   4   5   6   7   8   1   2   3  
(5)  0   5   6   7   8   1   2   3   4  
(6)  0   6   7   8   1   2   3   4   5  
(7)  0   7   8   1   2   3   4   5   6  
(8)  0   8   1   2   3   4   5   6   7

Section on irreducibility

While I am extremely impressed by this article as a whole (and think the creators have done an excellent job), I still think that it lacks a section purely on irreducible polynomials. There are many aspects of the theory of irreducible polynomials over finite fields, most notably the Berlekamp algorithm, that deserve mention in this article. For instance, it may be worthwhile to mention the following formula for the number N of monic irreducible polynomials over GF(q) (where GF(q) is, of course, the (unique) finite field with q elements, where q is an arbitrary prime power):


where s runs over all divisors of n that are products of distinct primes (allowing ) and where μ is the Möbius function. PST 13:46, 16 April 2010 (UTC)

Error in the examples section

I don't think that (Z/3Z)[T]/(T^2+1) creates a field of size 9

Attempting to generate the field with T as a primitive element results in:


T^2=-1=2 mod 3

Using this equality, the elements of the field would be:







So the root of T^2+1 has order 4, and not 8 as required to generate a field of size 9.

A valid example for the field of size 9 is (Z/3Z)[T]/(T^2+2T+2) or (Z/3Z)[T]/(T^2+T+2) Jellenburg (talk) 01:48, 6 May 2010 (UTC)

For any irreducible polynomial f(T) over Z/3Z of degree two, the field (Z/3Z)[T]/(f(T)) is a degree two extension of Z/3Z, i.e. it has dimension two as a Z/3Z vector space. Hence, it has size 9. What is confusing you is the notion of a primitive element of the field extension (Z/3Z)[T]/(T2+1) versus the notion of a primitive element of the finite field (Z/3Z)[T]/(T2+1). The element T in your example is a primitive element in the first sense, which means that every element of the order 9 field (Z/3Z)[T]/(T2+1) is a polynomial in T (as opposed to simply a power of T). RobHar (talk) 02:52, 6 May 2010 (UTC)

Unicity of automorphisms

D.Lazard has expanded the statement "Any two finite fields with the same number of elements are isomorphic", with "and the isomorphism between two such fields is unique." Quite likely I have misunderstood something, but it seems to me that this is false. Finite fields of prime-power order, such as F4, have non-trivial automorphisms. Between two fields of order 4 there are two distinct isomorphisms. The unicity applies only to fields of prime order. Maproom (talk) 09:22, 10 November 2013 (UTC)

OOOPS. D.Lazard (talk) 22:17, 10 November 2013 (UTC)