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I know that we cannot use the cubic formula to find the solution of the cubic equation in the field of characteristic $3$. If I have an equation say, $y^3 - y - a = 0$, is there a way to find the solution of this cubic equation if the field has characteristic $3$?

Anton Vrdoljak
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A. Sharma
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  • The roots of $Y^3-Y+a$ are $x,x+1,x+2$ for any root $x$. On the other hand, $Y^3-Y+a$ is irreducible in $\mathbb{F}_3(a)$ so you need to cook up a degree 3 extension. See also this MO question. – user10354138 Oct 06 '23 at 20:45
  • I believe this depends on the field. For k((t)) where k is of char 3, I think we can write down a solution explicitly. – Cranium Clamp Oct 06 '23 at 20:48
  • @CraniumClamp Often we can. There is the interesting special case of $a=1/t$, when the solution looks like $$y=t^{-1/3}+t^{-1/9}+t^{-1/27}+\cdots$$ I guess you can call that explicit. IIRC we need such series as well as the Puiseux series to get an algebraic closure. – Jyrki Lahtonen Oct 08 '23 at 10:17
  • A local incarnation of the basic lemma of Artin-Schreier theory in characteristic $p$. See the comments (by others) under my answer for more extensive versions. – Jyrki Lahtonen Oct 08 '23 at 10:20

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It's unclear what you mean by "find" here. The solutions are just what they are and there's not really any easier way to refer to them. In the particular case of the polynomial you write down, we can say the following about the solutions. Since $\beta^3$ is the Frobenius homomorphism in characteristic $3$, it satisfies $(\beta + 1)^3 = \beta^3 + 1$, from which it follows that if $\beta^3 - \beta = a$ then the same is true of $\beta + 1$ and $\beta + 2$. So the three roots have the form $\beta, \beta + 1, \beta + 2$ and the polynomial is irreducible over the base field $F$ iff $a$ is not of the form $b^3 - b$ for $b \in F$.

$\beta$ generates what is called an Artin-Schreier extension. These are a substitute for Kummer extensions of degree $p$ in characteristic $p$.

Qiaochu Yuan
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A very special case where we can explicitly write down the solutions is that of the underlying field $k$ being finite, say $k=\Bbb{F}_{3^m}$, and we look for solutions in $k$ also.

The additive version of Hilbert 90 says that $$ y^3-y=a, $$ $a\in k$, has solutions in the field $k$ if and only if the trace of $a$ vanishes: $$ tr(a)=\sum_{i=0}^{m-1}a^{3^i}=0. $$ Assuming that is the case, and that we write the elements of $k$ using a normal basis $\{\beta,\beta^3,\beta^9,\beta^{3^{m-1}}\}$, where the element $\beta$ is chosen carefully, we can say the following.

If $$a=\sum_{i=0}^{m-1}a_i\beta^{3^i},$$ $a_i\in\Bbb{F}_3$ for all $i$, then one solution $y$ is gotten by the recipe $$y=\sum_{i=0}^{m-2}y_i\beta^{3^i}$$ with $y_{m-2}=a_{m-1}$, $y_{m-3}=a_{m-1}+a_{m-2}$, and then recursively downwards $y_j=y_{j+1}+a_{j+1}=\sum_{\ell=j+1}^{m-1}a_{\ell}$. The coordinate vectors for $y^3$ and $y$ w.r.t. the normal basis are thus $$ y^3=(\ldots,a_{m-3}+a_{m-2}+a_{m-1},a_{m-2}+a_{m-1},a_{m-1}) $$ and $$ y=(\ldots,a_{m-2}+a_{m-1},a_{m-1},0) $$ respectively. We see a mass cancellation in the difference $y^3-y=(a_0,a_1,\ldots,a_{m-1})$. The first coordinates will match because the trace condition implies that $\sum_{i=0}^{m-1}a_i=0$.

As explained by Qiaochu and the commenters, the other solutions are then $y+1$ and $y+2$. This reflects the fact that the mass cancellation above goes the same way, if we add a constant from the prime field $\Bbb{F}_3$ to all the coordinates of $y$.

Jyrki Lahtonen
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  • In a sense I am taking advantage of the fact that cubing is linear over the prime field. Hence, using a suitable basis, the question turns into one of linear algebra in the case, where no field extension is required. – Jyrki Lahtonen Oct 08 '23 at 11:17