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Let $K$ be a field of characteristic $2$ ($2$ is very important in the statement -- otherwise I can do it myself :) ). On the set $K \times K$ we define the following equivalent relation: $(a, b) \equiv (a', b') $ if and only if there exists a pair $(q, \alpha) \in K^* \times K$ such that: $$ a = q^2 a' + \alpha^2 - b \alpha \quad {\rm and} \quad b = q b' $$

Can you compute explicitely the quotient set $K\times K/\equiv $? Or, if you prefere, can you give a set of representatives for the relation $\equiv$.

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  • $\begingroup$ I should say a little bit more related to the question. The quotient set $K\times K/≡$ parametrize the isomorphism class of all $2$-dimensional (unitary, associative) algebras over $K$. What is strage here is the fact that I am not able to give an explicit description of a set of representatives; since I am stupid :). If we have such a system of representative (a discution about K is mandatory -- there are two main cases: $K^2 = K$ or not) the we can list the types of isomorhism of all $2$-dimensional algebras over $K$. $\endgroup$
    – Gigel Militaru
    Commented May 4, 2013 at 8:33
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    $\begingroup$ An associative two-dimensional unitary algebra is commutative, and it is therefore the quotient of $K[X]$ by the ideal generated by a monic polynomial of degree $2$. So you want to classify polynomials of this type under the automorphism group of $K[X]$. Maybe stating it this way would make your problem more enticing :-) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 4, 2013 at 18:08
  • $\begingroup$ No! I can not do that :) You will arrive at the same problem. :) Btw the bijection between $K \times K/ \equiv$ and the set of isomorphism classes of all $2$-dimensional algebras over a field of characteristic $2$ is clear for anybody: $\overline{(a, b)} \mapsto k_{(a, b)}$, where $k_{(a, b)}$ is the algebra having $\{1, x\}$ as a basis and the multiplication is given by $x^2 = a + b x$. Now, I understand why everyone avoided the characteristic $2$ case in their classification results. $\endgroup$
    – Gigel Militaru
    Commented May 4, 2013 at 18:33
  • $\begingroup$ I did not say that you would arrive at a different problem: but the version with polynomials is expressed in terms of familiar things —polynomials and linear changes of variables— instead of weird ordered pairs under an unmotivated equivalence relation! $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 4, 2013 at 18:37

3 Answers 3

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The classification of associative algebras in low dimensions has a long history, beginning with work of Peirce. A good survey, and a classification of all nilpotent associative algebras of low dimension, over an arbitrary field, also of characteristic two, can be found in the paper of W. de Graaf, see http://arxiv.org/abs/1009.5339. I do not know if an "explicit list" of representatives is possible for the general case.

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  • $\begingroup$ Thank you very much for the link. Yes, I know the classification of Peirce etc --was done until now up to dimension 6. But this holds only for algebraically closed fields with characteristc $\neq 2$. For arbitrary fields, in particular in characteristic $2$, there are big troubles (that qoutient set above is responsible for it). The paper that you indicate is from 2012 and I have to contact the authors since over $Z_2$ is not posible to exists 52 classe of isomorphisms of algebras of dimension 2 as they claim on pag. 10. Too many :) $\endgroup$
    – Gigel Militaru
    Commented May 4, 2013 at 17:36
  • $\begingroup$ Link nitpick: When linking to arXiv, please link to the abstract, not directly to the PDF. $\endgroup$
    – Harry Altman
    Commented May 4, 2013 at 17:53
  • $\begingroup$ The 52 classes also include the non associative ones, I think. By algebra the authors mean any $K$-algebra. $\endgroup$
    – Dietrich Burde
    Commented May 4, 2013 at 17:54
  • $\begingroup$ PS: Now I got it. By 'algebras' the authors do not assume associative and unitary algebras. They just take a biliniar map -- with no other axioms. However, they do not answer my question that remains open - in characteristic $2$ they work only over the field with two elements $Z_2$. Well, in this case everyone can discribe that quotient set. The big problem with it is the case $K \neq k^2$. $\endgroup$
    – Gigel Militaru
    Commented May 4, 2013 at 18:10
  • $\begingroup$ @Gigel Militaru: you are right. I edited my answer accordingly. $\endgroup$
    – Dietrich Burde
    Commented May 4, 2013 at 18:34
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As I mentioned in a comment above, you want to classify monic polynomials of degree two over a field $K$ of characteristic two under the automorphism group of $K[X]$, which is that of maps $X\mapsto \alpha X+\beta$ with $\alpha$, $\beta\in K$ and $\alpha\neq0$.

If the polynomial is reducible, then it is easy to see that it is in the orbit of $X^2$ or of $X(X+1)$, and these two are not in the same orbit.

Suppose then that the polynomial is $X^2+aX+b$ and that it is irreducible.

If $a=0$, then the polynomial is $X^2+b$, with $b$ not a square, and two such polynomials with $b$, $b'$ nonsquares are in the same orbit iff they differ in a square.

If $a\neq0$, the change of variable $X\leadsto aX$ shows that the polynomial is in the orbit of one of the form $X^2+X+b$. Let $\wp:t\in K\mapsto t^2+t\in K$. The polynomial $X^2+X+b$ is irreducible iff $b$ is not in the image of $\wp$, and it is in the same orbit as $X^2+X+b'$ iff $b$ and $b'$ differ by an element in the image of $\wp$.

We conclude that the orbits are: (i) two special ones, (ii) one for each nonsquare of $K$ up to addition of squares and (ii) one for each nonimage of $\wp$ up to addition of elements in the image of $\wp$.

This is as far as you are going to get without being more specific about the field (because the arithmetic of the field enters as soon as you want to describe what is a square and what is not, and what is in the image of $\wp$ and what is not) This has nothing to do with the characteristic, and you would have exactly the same problem over any field: classifying irreducible polynomials of degree two is an arithmetic problem —the only difference is that in odd characteristic you need only care about the function $x\mapsto x^2$ while in even characteristic you have to worry also about the map $\wp$. You'd have a similar outcome if you try to classify cubic polynomials in characteristic three.

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  • $\begingroup$ Than you! That is probably the best thing that you can say on the problem :( On the other hand, as a said, is a little bit ... a reformulation of the equivalent relation. The first two polynomial coresspound in fact to the non-equivalent pairs $(0, 0)$ and $(0, 1)$. The third one to $(a, 0)$, where $a$ is not a square in $K$, while the last case (the one that make trouble) is the first part of the equivalent relation up to a normalization. $\endgroup$
    – Gigel Militaru
    Commented May 4, 2013 at 19:09
  • $\begingroup$ Your comment makes me think that you misunderstand what I wrote. I did noy say that the two polynomials $X^2$ and $X(X+1)$ are in the same orbit ---I said exactly the opposite. The rest of the comment, I don't understand it. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 4, 2013 at 20:18
  • $\begingroup$ My english... What I what to say is the following: dealing with polinomials or with the above relation is exactely the same thing. Is just a matter of taste. The fact that the polionomials $X^2$ and $X (X+1)$ are not in the same orbit is nothing else that a reformulation of the fact that $(0,0)$ and $(0, 1)$ are not equivalent in $K \times K$. In both cases, we arrive at the same algebras: $k<x | x^2 = 0>$ or $k<x | x^2 = 1>$. $\endgroup$
    – Gigel Militaru
    Commented May 5, 2013 at 3:32
  • $\begingroup$ The same is valid with the elements that are nonsquare in $K$. If $a$ is nonsquare, then $(a,0)$ is not equivalent to $(0,0)$ or $(0,1)$. The number of such elements is the index of the group $(K^2)^*$ in $(K^*, \cdot)$. That is exactly what you say; and the same thing follows trivialy from the relation of equivalence. $\endgroup$
    – Gigel Militaru
    Commented May 5, 2013 at 3:42
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Thank you Mariano for your stimulation. The best result that can be done is bellow. First some notations:

If $k^2 \neq k$ we denote by $R \subseteq k \setminus k^2$ a system of representative for the following relation on $k \setminus k^2$: $d \equiv_1 d'$ if and only if there exists $q \in k^{*}$ such that $d - q^{2} d' \in k^2$.

$T \subseteq k$ a system of representative for the for the following relation on $k$: $c \equiv_2 c'$ if and only if there exists $\alpha \in k$ such that $c - c' = \alpha^2 - \alpha$.

Then we have: (the numbers of types has to be computed for any field $k$ of characteristic $2$ - case by case :(:( )

  1. If $k = k^2$ then the quotient set $k \times k /\equiv $ is equal to $\{\overline{(0, 0)}, \overline{(0, 1)}\} \cup \{\overline{(c, 1)} ~|~ c \in T \} $. Thus in this case there exists $2 + |T|$ types of isomorphism of $2$-dimensional algebras namely $k_{(0, 0)}$, $k_{(0, 1)}$ and $k_{(c, 1)}$, for some $c \in T$.

  2. If $k \neq k^2$ then the quotient set $k \times k /\equiv $ is equal to $\{\overline{(0, 0)}, \overline{(0, 1)}\} \cup \{\overline{(c, 1)} ~|~ c \in T \} \, \cup \, \{\overline{(d, 0)} ~|~ d \in R \} $. Thus in this case there exists $2 + |T| + |R| $ types of isomorphism of $2$-dimensional algebras namely $k_{(0, 0)}$, $k_{(0, 1)}$, $k_{(c, 1)}$, $k_{(d, 0)}$, for some $c \in T$ and $d\in R$.

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