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.

  • 1
    $\begingroup$ The formula in characteristic $2$ is derived in H.B.Mann's paper "On the Casus Irreducibilis", Amer. Math. Monthly 71 (1964), 288-290 (or more precisely, Mann does the parts of the derivation which aren't identical to the characteristic $0$ derivation in van der Waerden). Mann used the $A_3$-invariant $x_1^2x_2+x_2^2x_3+x_3^2x_1$, which is different from the invariant Felipe uses in his answer, although of course both values yield the same formula. $\endgroup$ Dec 18, 2022 at 13:40

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$ Nov 28, 2011 at 3:23
  • $\begingroup$ I am a bit confused. How do you solve for $\beta$? Isn't that also a nontrivial cubic equation in char 3? $\endgroup$
    – John Jiang
    Nov 28, 2011 at 9:26
  • $\begingroup$ 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. $\endgroup$ Nov 28, 2011 at 13:24
  • $\begingroup$ I know this is a long shot, but does anyone know where this paper by Dubravka Bodiroga can be found? I cannot find it anywhere. $\endgroup$ Dec 7, 2015 at 11:57

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.

  • $\begingroup$ It is perfectly legitimate to use $w$ in characteristic $2$, since $w=\wp_2^{-1}(1)$. $\endgroup$ Nov 27, 2011 at 11:36
  • $\begingroup$ Or $1^{1/3}$, for some choice of root :-). $\endgroup$ Nov 27, 2011 at 12:05
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    $\begingroup$ Characteristic 3 isn't hard. Suppose the equation is x^3+ax^2+bx+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)). $\endgroup$ Nov 27, 2011 at 12:06
  • 2
    $\begingroup$ @paul That works and $r=b/a$ works, so the only square root needed is that of $A$, in your notation. $\endgroup$ Nov 27, 2011 at 12:57
  • $\begingroup$ Just a minor addition about characteristic $2$: the initial expressions for $y$ and $y'$ determine $yy'$, namely $yy'=(x_1+x_2+x_3)^2+(x_1x_2+x_1x_3+x_2x_3)$, which is written in terms of the coefficients of the original cubic. This relation should be used to define $y'$ later on, in terms of $y$ (where the three choices of $y$ yield values $x_1$ that are the three roots of the cubic). If one instead chooses $y'$ to be an arbitrary cube root, then sometimes one will get a value of $x_1$ which is not a root of the original cubic. $\endgroup$ Dec 12, 2022 at 0:13

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