Mal'cev condition for variety of rings generated by finite fields to be arithmetical This is an exercise of Burris & Sankappanavar (Universal Algebra), Chapter II, section 12.
It asks to prove that, if $V$ is a variety of rings generated by finitely many finite fields, then $V$ is arithmetical.
Since rings are congruence permutable (with Mal'cev term $p(x,y,z)=x-y+z$), it remains to show that this variety is congruence-distributive.
It seems to me that $V$ is precisely the variety of commutative rings with $1$ that satisfy $nx \approx 0$, where $n$ is the product of the distinct prime numbers which are characteristics of each of the fields that generate the variety.
So I tried to prove it using each of the following strategies:


*

*Define a ternary majority term, that is a term $M(x,y,z)$ in the language of the variety which satisfies $M(x,x,y) \approx M(x,y,x) \approx M(y,x,x) \approx x$. (This is a sufficient, but not necessary, condition for the variety to be congruence-distributive.)

*Define a term that characterizes arithmetic varieties, that is $m(x,y,z)$ such that $m(x,y,x) \approx m(x,y,y) \approx m(y,y,x) \approx x$.

*Define a list of (Jónsson) terms which characterize congruence-distributive varieties, that is, $d_0(x,y,z), \ldots, d_n(x,y,z)$ such that $d_0(x,y,z) \approx x$, $d_n(x,y,z) \approx z$, $d_i(x,y,x) \approx x$, and $d_i(x,x,y) \approx d_{i+1}(x,x,y)$, for $i$ even, and $d_i(x,y,y) \approx d_{i+1}(x,y,y)$, for $i$ odd.


As I said, the majority term is only a sufficient condition, but for arithmetical varieties, there is always such term, because it can be obtained from the term that characterizes arithmetical varieties making $M(x,y,z) = m(x,m(x,y,z),z)$.
Anyway, I was able to find a ternary majority term if the variety is generated by fields of characteristic $2$.
We know that $\mathbb{F}_{2^k} \models x^{2^k} \approx x$.
So, consider $M(x,y,z) = (xy+yz+zx)^{2^{k-1}}$; the expression in the base evaluates, if two of the variables coincide, to the square of that variable, so that
$$M(x,x,y) \approx M(x,y,x) \approx M(y,x,x) \approx (x^2)^{2^{k-1}} \approx x^{2^k} \approx x.$$
Unfortunately, I was not able to generalize to fields of characteristic $3,5,\ldots$, and even if I had, I'm not very confident I could somehow merge these results into a general one.
A possible hint on how to make such merging is that, if a ring as a term $p(x)$ such that $p(0) = 0$ and $r=r \cdot p(r)$, for $r \neq 0$ (on a field this just means that $p(r)=1$, if $r\neq 0$, that is $p(x) = 1 - \delta_{0x}$, where $\delta_{xy}$ is the Kronecker symbol), then
$m(x,y,z) = x + (z-y) \cdot p(z-x)$ satisfies the identities in item 2 on the list above.
Now, if we only have a finite field of characteristic $p$ (or equivalently, several with the same characteristic), say, $\mathbb{F}_{p^k}$, then $x^{p^{k-1}} \approx 1$, except for $0$, so we could take the above term function $p(x)$ to be $x^{p^{k-1}}$.   
Following the other two approaches didn't prove any useful at all, since I wasn't able to get any partial result.
Thanks in advance

EDIT:

Of course I can give a majority ternary term for the variety generated by $\mathbb{F}_{p^k}$: as I pointed, $m(x,y,z) = x + (z -y) \cdot (z-x)^{p^{k-1}}$ satisfies the conditions that characterize arithmetic varieties, and thus, $M(x,y,z) = m(x,m(x,y,z),z)$ is one such majority term (but I don't know whether or not it coincides with the one I gave in the case $p=2$).


EDIT:
Well, that first edit doesn't follow.
The equational property is $x^{p^k} \approx x$, not $x^{p^{k-1}} \approx 1$, if $x \neq 0$, and $0$ otherwise; this last one doesn't survive the journey to direct products.
So actually, I don't have a majority term for varieties generated by rings of characteristic other than $2$. 
 A: Ok, so I finally got it and now I'm answering my own question.
It turns out that that first edit had the right clue, and the problem I pose in the second one, doesn't actually arise, even if what I wrote there is right.
If $p$ is a prime number, and $k$ a positive integer, then a majority ternary term for the variety of rings generated by $\mathbb{F}_{p^k}$, is given by
$$M(x,y,z) = z-(z-y)(z-x)^{p^k-1}.$$
Here, we use the fact that $\mathbb{F}_{p^k} \models x^{p^k} \approx x$, and it will easily follow that 
$$M(x,x,y) \approx M(x,y,x) \approx M(y,x,x) \approx x.$$
Now, if $x^n \approx x$, then an induction argument on $m$ yields $x^{n+m(n-1)} \approx x$.
Thus, $$x^{(n-1)m+1} \approx x^{n+(m-1)(n-1)} \approx x.$$
So if we have the fields $\mathbf{F}_1 = \mathbb{F}_{p_1^{k_1}}, \ldots, \mathbf{F}_n = \mathbb{F}_{p_n^{k_n}}$, 

where we may suppose that $p_1, \ldots, p_n$ are distinct prime numbers because, if we have $p^n$ and $p^m$ among the orders of the fields, we may replace both with $p^r$, where $r=\mathrm{lcm}(n,m)$, since then both $\mathbb{F}_{p^n}$ and $\mathbb{F}_{p^m}$ are subfields of $\mathbb{F}_r$ and $\mathbb{F}_r$ is the least such field.

Then, in $\mathbf{F}_i$, we have $x^{n_i} \approx x$, where $n_i = p_i^{k_i}$, and by the above argument, if we make
$$m= \prod_{i=1}^n (n_i-1),$$
then $x^{m+1} \approx x$ is valid in $F_1 \cup \cdots \cup F_n$.
Take $M(x,y,z) = z - (z-y)(z-x)^m$. Then,
\begin{align}
M(x,x,z) &= z - (z-x)(z-x)^m = z - (z-x)^{m+1} = z - (z-x) = x,\\
M(x,y,x) &= x - (x-y)(x-x)^m = x - (x-y)\cdot 0 = x,\\
M(y,x,x) &= x - (x-x)(x-y)^m = x - 0 \cdot (x-y)^m = x.
\end{align}
So we have a majority ternary term for all the generating rings of the variety, and thus, it is congruence-distributive.
Since all rings are congruence-permutable, we get that the variety is arithmetical.
