Weakly quadratent rings

ABSTRACT

We completely characterize up to an isomorphism those rings whose elements satisfy the equations $x^4=x$ or $x^4=-x$. Specifically, it is proved that a ring is weakly quadratent if, and only if, it is isomorphic to either K, ${\mathbb{Z}}_3$, ${\mathbb{Z}}_7$, $K\times {\mathbb{Z}}_3$ or $K\times {\mathbb{Z}}_7$, where K is a ring which is a subring of a direct product of family of copies of the fields ${\mathbb{Z}}_2$ and ${\mathbb{F}}_4$. This achievement continues our recent joint investigation in J. Algebra (2015) where we have characterized weakly boolean rings satisfying the equations $x^2=x$ or $x^2=-x$ as well as a recent own investigation in Kragujevac J. Math. (2019) where we have characterized weakly tripotent rings satisfying the equations $x^3=x$ or $x^3=-x$.

KEYWORDS:

• Fields
• equations
• rings

2010 MATHEMATICS SUBJECT CLASSIFICATIONS:

• 16 D60
• 16 S34
• 16 U60

1. Introduction and fundamentals

Everywhere in the text of the present paper, all our rings R into consideration are assumed to be associative, containing the identity element 1 which differs from the zero element 0. Our terminology and notations are mainly in agreement with [1]. As usual, ${\mathbb{F}}_n$ denotes the finite field consisting of n elements, whereas ${\mathbb{Z}}_n\cong \mathbb{Z}/\left(n\right)=\mathbb{Z}/n\mathbb{Z}$ denotes the finite ring of n elements consisting of all integers modulo n, where $n\in \mathbb{N}$ with n>1. Clearly, for any prime number p, the equality ${\mathbb{Z}}_p={\mathbb{F}}_p$ is true.

For readers' convenience and completeness of the exposition, let us recall some classical backgrounds: Given a ring R, there is a canonical morphism of rings $\varphi :\mathbb{Z}\to R$ which maps the integer n to n. The kernel is an ideal of $\mathbb{Z}$, hence it is of the form $c\mathbb{Z}$ with $c\ge 0$ and $c\ne 1$ (this last condition is due to the assumption $1\ne 0$). This number c is the characteristic of R. The image of ϕ is the smallest subring of R, that is, the intersection of all subrings of R (it is henceforth called the prime subring of R). If now S is a (unital) subring of R, then R and S have the same prime subring, hence the same characteristic. The homomorphism ϕ induces an isomorphism between $\mathbb{Z}/c\mathbb{Z}$ and the prime subring of R. An integer $n\in \mathbb{Z}$ has the property that n=0 in R if and only if n is a multiple of the characteristic. For a prime number p, the condition p=0 in R implies that p is just the characteristic.

If now R has finite characteristic c (i.e. $c\ne 0$), any $a\in \mathbb{Z}$ prime to c is a unit in the prime field $\mathbb{Z}/c\mathbb{Z}$ (i.e. it has an inverse), whence also in R. Therefore, for an arbitrary $x\in R$, the condition ax=0 implies x=0, while the condition $ax\in S$ for a subring S of R will imply $x\in S$.

For further applicable purposes, we will need the field of four elements ${\mathbb{F}}_4$ having characteristic 2, which is constructed in a traditional manner as follows: It is well known that in the polynomial ring ${\mathbb{Z}}_2\left[x\right]$ the polynomial $1+x+x^2$ is irreducible over ${\mathbb{Z}}_2$ and hence ${\mathbb{Z}}_2\left[x\right]/(x^2+x+1)\cong {\mathbb{Z}}_2\left(\theta \right)$ is a field of 4 elements, denoting it by ${\mathbb{F}}_4$, where $\theta \notin {\mathbb{Z}}_2$ is a solution of the equation $x^3=1$. In fact, the elements in ${\mathbb{F}}_4$ are $\{0,1,\theta ,{\theta }^2\}$ taking into account that $\theta +1={\theta }^{-1}={\theta }^2$ and that $x^3-1=(x^2+x+1)(x-1)$, whence the equation $x^3=1$ has the elements $\{1={\theta }^0,\theta ,{\theta }^2\}$ as solutions. Thus ${\mathbb{F}}_4$ is the splitting field of these two polynomials.

So, we come to the following folklore fact (see, for instance, [2]):

Proposition 1.1

A ring R satisfies $x^4=x$ for all $x\in R$ i, and only i, R can be embedded as a subring of the direct product of (finitely or infinitely many) copies of the fields ${\mathbb{Z}}_2$ and ${\mathbb{F}}_4$.

Proof.

Letting firstly R to be a domain, we detect that $x^3=1$ for all nonzero $x\in R$. So, R must be a field because any $0\ne x\in R$ is invertible with inverse $x^2$. Therefore, since the polynomial $x^3-1$ has at most three roots in the domain (i.e. in the field) R, either $R\cong {\mathbb{Z}}_2$ or $R\cong {\mathbb{F}}_4$. Indeed, either $R=\{0,1|2=0\}$ or $R=\{0,1,x,x^2|1+x+x^2=0;2=0\}$.

Furthermore, we claim in general that 2=0. To show this, we observe that $2^4=2$ leads to 14=0, while $3^4=3$ leads to 78=0. Thus 8=78−14.5=0, whence 0=14−2.8=−2 and finally 2=0, as claimed. Since $J\left(R\right)=\left\{0\right\}$ according to the already established above fact that R is a field (see, e.g. [1]), we consequently deduce that R can be embedded in the direct product of domains D of characteristic 2, and hence by what we have just shown above, these D are isomorphic to either ${\mathbb{Z}}_2={\mathbb{F}}_2$ or ${\mathbb{F}}_4$, concluding the assertion after all.

We shall hereafter call quadratent such a ring R for which $x^4=x$ for all $x\in R$.

The following notion is our crucial tool.

Definition 1.2

A ring R is called weakly quadratent if, for every $x\in R$, we have that $x^4=x$ or $x^4=-x$.

The aim of the present paper is to describe up to an isomorphism these weakly quadratent rings.

A brief history of some principally known results concerning rings which satisfy the equations $x^n=x$ or $x^n=-x$ for some fixed natural $n\ge 2$ are these: For n=2 we obtain the well-known weakly boolean rings and the case is settled in [3]. For n=3 we get the so-termed weakly tripotent rings and the case is exhausted in [4]. As aforementioned, we will examine here the isomorphic structure of such rings when n=4. The general situation for an arbitrary natural n is unknown yet, and so it is stated as an open question at the end of the article.

2. Main results

We first begin with the following technicality.

Lemma 2.1

If R is a ring of characteristic ab, where a and b are relatively prime integers $\ge 2,$ then R is isomorphic to a direct product of rings $R_1\times R_2,$ where $R_1$ has characteristic a and $R_2$ has characteristic b.

In particular, if R is weakly quadratent, then the two factors $R_1,R_2$ are weakly quadratent, whereas if $R_1$ is quadratent and $R_2$ is weakly quadratent, then $R_1\times R_2\cong R$ is also weakly quadratent.

Proof.

The canonical map $R\to \left(R/aR\right)\times \left(R/bR\right)$ is injective since $aR\cap bR=\left\{0\right\}$. It is also surjective by the usage of Chinese Remainder Theorem, namely, if au+bv=1, then $(x_1,x_2)$ is the image of $au{x}_2+bv{x}_1$. And finally, R/aR has characteristic a, because its prime field is $\left(\mathbb{Z}/ab\mathbb{Z}\right)/\left(a\mathbb{Z}/ab\mathbb{Z}\right)\cong \mathbb{Z}/a\mathbb{Z}$, whereas by the same token R/bR has characteristic b.

The second part-half is self-evident.

We continue with the next useful observation.

Lemma 2.2

In a weakly quadratent ring the equality ${2.3}^2.7=0$ holds. In addition, the characteristic of a weakly quadratent ring is either 2, 3, 6, 7 or 14.

Proof.

Write $2^4=2$ or $2^4=-2$. Then 2.7=0 or ${2.3}^2=0$, and thus the first assertion follows.

Next, by what we just have shown, the characteristic divides ${2.3}^2.7$. Further, one proves that a quadratent ring cannot contain ${\mathbb{Z}}_9$ nor be a product of rings, where one of them is ${\mathbb{Z}}_9$, since in ${\mathbb{Z}}_9$ one has the equality $3^2=0$ which is not compatible with $3^4=\pm 3$.

If now the characteristic is 21, then R contains ${\mathbb{Z}}_{21}$ which is manifestly not weakly quadratent (as a subring of a weakly quadratent ring is obviously again weakly quadratent), while if the characteristic is 42, then R contains ${\mathbb{Z}}_{42}$ which is also manifestly not weakly quadratent.

We are now ready to establish our central result.

Theorem 2.3

A ring R is weakly quadratent i, and only i, R is isomorphic to either K, ${\mathbb{Z}}_3,$ ${\mathbb{Z}}_7,$ $K\times {\mathbb{Z}}_3$ or $K\times {\mathbb{Z}}_7,$ where K is a quadratent ring, that is, a ring which is a subring of a direct product of family of copies of the fields ${\mathbb{Z}}_2$ and ${\mathbb{F}}_4$.

Proof.

Utilizing a combination of Lemmas 2.1 and 2.2, one writes that $R\cong K\times L\times M$ for some weakly quadratent rings K,L,M such that $K=\left\{0\right\}$ or char $\left(K\right)=2$, $L=\left\{0\right\}$ or char $\left(L\right)=3$ and $M=\left\{0\right\}$ or char $\left(M\right)=7$.

We foremost consider a weakly quadratent ring of characteristic 2, and as above let us denote it by K. Thus K is necessarily quadratent, and its structure has been settled in Proposition 1.1.

Let us prove that the only weakly quadratent ring of characteristic 3 is ${\mathbb{Z}}_3$. Denote it as above by L. We claim that $L\cong {\mathbb{Z}}_3$. To that aim, for any $x\in L$, we have $(x+1{)}^4=x+1$ or $(x+1{)}^4=-(x+1)$. In the first case, we deduce in combination with $x^4=x$ or $x^4=-x$ that $x^3=-x$ or $x^3=x$, respectively, whence in both cases multiplying by x we get that $x^2=-x$. In the second case, we detect by combining with $x^4=x$ that $x^3=1$, whereas in combination with $x^4=-x$ we derive that $x^3=-x+1$ and again multiplying both sides with x we infer that $x^2=-x$ holds at once. We thus finally obtain that $x^2=-x$ or $x^3=1$. It now directly follows from these two equalities that L does not have nontrivial nilpotents. However, $x^3=1$ amounts to $(x-1{)}^3=0$ because 3=0, and hence x=1 giving up that $x^2=x$. We therefore may apply [3] to conclude after all that $L\cong {\mathbb{Z}}_3$, as expected.

Let us prove that the only weakly quadratent ring of characteristic 7 is ${\mathbb{Z}}_7$. Denote it as above by M. We assert that $M\cong {\mathbb{Z}}_7$. In order to show this, suppose P is the subring of M, generated by its identity element. Hence, it is elementarily seen that $P\cong {\mathbb{Z}}_7$ as 7=0. We intend to prove that M=P. To that goal, assume in a way of contradiction that there exists an element $b\in M\setminus P$. But we may also assume with no harm of generality that $b^4=-b$ since $b^4=b$ implies that $(3b{)}^4=-(3b)$ as 7=0 and $3b\notin P$ (for otherwise $3b\in P$ yields that $6b=-b\in P$ which is false). It is easily checked also that $(2b{)}^4=-(2b)$, because by assumption $b^4=-b$, where $2b\notin P$ too (if not, $8b=b\in P$, which is false). Let us now consider the case $(1+b{)}^4=1+b$. Hence $4b^3+6b^2+2b=0$. We now claim that $(1+3b{)}^4\ne \pm (1+3b)$, however. Firstly, in the case of sign “+”, we get that $3b^3+5b^2-2b=0$. Summing the two equations, we infer that $11b^2=0$ as 7=0. Thus $b^2=22b^2=0$ and $b=0\in P$ since $b^4=-b$, a contradiction. Secondly, in the case of sign “−”, we get that $3b^3+5b^2+4b+2=0$ which, in conjunction with $4b^3+6b^2+2b=0$, ensures that $4b^2-b+2=0$ taking into account once again that the characteristic of M is exactly 7. Thus, multiplying the lastly obtained equality by 2, one infers that $b^2=2b+3$ and substituting this in the equation $3b^3+5b^2+4b+2=0$, one derives that $4b^3=-4$, and hence $8b^3=-8$, that is, $b^3=-1$, because as already used above 7=0 in M. Now, we have two possibilities: $(1-b{)}^4=1-b$ or $(1-b{)}^4=-(1-b)$. The first equality leads to $6b^2-4b+4=0$ and since we already have $6b^2+2b-4=0$, these two equations imply 6b=8, that is, $b=-1=6\in P$ – false. The second equality leads to $3b^2-3b+3=0$ and since we already have $3b^2+b-2=0$, these two equations yield 4b=5, i.e. $b=3\in P$ – wrong.

That is why, it must be that $(1+b{)}^4=-(1+b)$ accomplished with $b^4=-b$. Consequently, $4b^3+6b^2+4b+2=0$. Here, again $(1-b{)}^4=1-b$ or $(1-b{)}^4=-(1-b)=b-1$. Thus, in the first case, we arrive at $-4b^3+6b^2-4b=0$ and the comparison of the existing two equations enables us that $b^2=1$. Hence $1=b^4=-b$ gives that $b=-1=6\in P$ which is impossible. In the second case, we receive $4b^3-6b^2-b-2=0$ and the comparison of the two valid equations allows us to get that $b^3=-3b$. So, $3b^2=b$ and the substitution these two new relations in the initial equation $4b^3+6b^2+4b+2=0$ assures that 6b=2 giving up that $b=-2=5\in P$ which is impossible, too.

Finally, after all considerations, one concludes that such an element b from M does not exist outside P, and therefore the identity M=P is really true, as promised.

Remark 2.4

The rings ${\mathbb{Z}}_3\times {\mathbb{Z}}_7\cong {\mathbb{Z}}_{21}$ and ${\mathbb{Z}}_2\times {\mathbb{Z}}_3\times {\mathbb{Z}}_7\cong {\mathbb{Z}}_{42}$ are not weakly quadratent. In fact, the 4th power of the element $(1,1,3)$, lying in the latter ring, is equal to $(1,1,-3)$ but it is totally different from $\pm (1,1,3)$.

However, a direct check shows that ${\mathbb{Z}}_2\times {\mathbb{Z}}_3\cong {\mathbb{Z}}_6$ as well as ${\mathbb{Z}}_2\times {\mathbb{Z}}_7\cong {\mathbb{Z}}_{14}$ are weakly quadratent.

We end our work with the following:

Problem 2.5

Describe the isomorphic structure of those rings whose elements satisfy the polynomial identities $x^n=x$ or $x^n=-x$, where n is an arbitrary fixed natural number.

It is worthwhile noticing that the structure of rings satisfying the polynomial identity $x^n=x$ is already known (see, e.g. [2]).

The next problem is somewhat worth-standing: The assumption $x^4=x$ or $x^4=-x$ yields that $x^8=x^2$. The converse is not true, since, for instance, in the ring ${\mathbb{Z}}_3\times {\mathbb{Z}}_3$, the later condition holds but not the former one. However, it might be of interest to study the rings satisfying the polynomial equation $x^8=x^2$ and, more generally, given two integers n and m with $n>m\ge 1$, the condition $x^n=x^m$. It turns out that the domains satisfying this polynomial equation are easy to characterize: these are the finite fields ${\mathbb{F}}_q$ with q elements (q being a power of a prime) such that q−1 divides n−m. For our concrete example with n=8 and m=2, the condition q−1 divides 6 forces that q=2,3,4 or 7.

Indeed, on the one hand, since the multiplicative group of a finite field with q elements has order q−1, any nonzero element satisfies $x^{q-1}=1$, and hence if q−1 divides n−m then any nonzero element satisfies also $x^{n-m}=1$ and, therefore, any element (including also 0) satisfies $x^n=x^m$.

In the other direction, if R is a domain in which any element satisfies $x^n=x^m$, then R is of necessity finite as the polynomial $x^n-x^m$ has at most n roots in a domain, whence it is a finite field because a finite domain is always a field, say with q elements, and the multiplicative group of this field is a cyclic group with q−1 elements. Letting x be its generator, we deduce from $x^{n-m}=1$ that q−1 divides n−m, as required. This substantiates our claim after all.

Disclosure statement

No potential conflict of interest was reported by the authors.