Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question has no justification other than a bit of fun.

We all know that the cubic is solvable by "radicals" ($\root2\of{}$ and $\root3\of{}$) in characteristics $\neq2,3$. The formula was discovered by the Italians in the 16th century (see here).

In characteristic $2$, there should be a similar formula involving $\wp_2^{-1}(\ )$ and $\root3\of{}$, and in characteristic $3$ there should be a formula involving $\root2\of{}$ and $\wp_3^{-1}(\ )$.

By $\wp_2^{-1}(a)$ and $\wp_3^{-1}(a)$ I mean a root of the polynomials $\wp_2(T)=T^2-T-a$ and $\wp_3(T)=T^3-T-a$ respectively, which give all cyclic extensions of degree $2$ and $3$ respectively.

Has somebody worked out these formulæ ?

Edit. I have accepted one of the answers --- the choice was difficult --- but I'm still curious as to whether these formulæ can be found somewhere in the literature.

share|improve this question
add comment

2 Answers

up vote 15 down vote accepted

I asked an undergraduate (Dubravka Bodiroga at Hood College) to work these results out last summer. Here is her cubic formula in characteristic 3 (paraphrasing from something she sent me):

Consder the polynomial $$ x^3 - a_1 x^2 + a_2 x - a_3, $$ where the coefficients belong to a commutative ring in which $3=0$. Assume moreover that $a_1$ is invertible. Let $b$ be a solution to $$ b^2 = -\frac{a_2^3}{a_1^6}+\frac{a_2^2}{a_1^4}-\frac{a_3}{a_1^3}, $$ and let $\beta$ be a solution to $y^3 - y - b = 0$. Then $$ x^3 - a_1 x^2 + a_2 x - a_3 = (x - (u\beta^2+v))(x-(u(\beta+1)^2+v))(x-(u(\beta+2)^2+v)), $$ where $u=-a_1$ and $v=a_1 -\frac{a_2}{a_1}$.

If one inverts the procedure, letting $$ x_i = u(\beta+i)^2 + v, $$ then ([Parson: assuming I haven't scrambled her indices]) $$ b= \frac{(x_0 - x_1)(x_1 - x_2)(x_0 - x_2)}{(x_0+x_1+x_2)^3}, $$ and $$ y= \frac{x_2 + 2x_1}{x_0+x_1+x_2} = -\frac{2\times x_2+1\times x_1+0\times x_0}{x_0+x_1+x_2}. $$

I believe she also worked out the (simpler) details for the cubic formula in characteristic $2$, but I could not find them just now. She used a heuristic method of Euler and B\'ezout to find the formulas, an exposition of which one can find in Tignol's book on Galois theory. She then solved for the auxiliary quantities $b$ and $y$ in terms of the $x_i$ to see what Lagrange would have made of her solution procedure.

share|improve this answer
    
This is certainly a nice problem for an undergraduate to work out, and I was planning to assign it to the next student with some background in Galois theory. –  Chandan Singh Dalawat Nov 28 '11 at 3:23
    
I am a bit confused. How do you solve for $\beta$? Isn't that also a nontrivial cubic equation in char 3? –  John Jiang Nov 28 '11 at 9:26
    
You take $\beta=\wp_3^{-1}(b)$, so $\beta$ is determined up to addition by $0$, $1$, $2$, just as a cube root is determined up to multiplication by a cube root of unity. –  Chandan Singh Dalawat Nov 28 '11 at 13:24
add comment

Characteristic two can be done using the standard Lagrange resolvent method. All you need is cube roots of unity.

Let $x_1,x_2,x_3$ be the roots of the cubic. Let $y=x_1+wx_2+w^2x_3,y'=x_1+w^2x_2+wx_3$, where $w$ is a primitive cube root of unity. Then $y^3,(y')^3$ are $A_3$-invariant and are roots of the quadratic with coefficients $a=y^3+(y')^3,b=y^3(y')^3$ which are $S_3$-invariants and can be computed as polynomials on the coefficients of the cubic. The roots of $x^2+ax+b$ are $a\wp_2^{-1}(b/a^2),a\wp_2^{-1}(b/a^2)+a$. Get $y,y'$ by taking cube roots of the roots of the quadratic and $x_1=y+y'+x_1+x_2+x_3$.

Characteristic three seems harder.

share|improve this answer
    
It is perfectly legitimate to use $w$ in characteristic $2$, since $w=\wp_2^{-1}(1)$. –  Chandan Singh Dalawat Nov 27 '11 at 11:36
    
Or $1^{1/3}$, for some choice of root :-). –  Felipe Voloch Nov 27 '11 at 12:05
5  
Characteristic 3 isn't hard. Suppose the equation is x^3+a*x^2+b*x+c=0. Make the b term 0 by a translation x-->x+r; this amounts to solving a quadratic equation for r. Letting y be 1/x we get an equation of the form y^3+Ay+B=0. From this (as long as A is not 0 so we have separability)we can read off P_3(y/(square root A)). –  paul Monsky Nov 27 '11 at 12:06
2  
@paul That works and $r=b/a$ works, so the only square root needed is that of $A$, in your notation. –  Felipe Voloch Nov 27 '11 at 12:57
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.