Bounds for the number of idempotents in finite rings

Abstract

We find bounds for the number of idempotents in a ring of matrices over an arbitrary finite commutative ring. Using these results, we find a bound on the number of idempotents in an arbitrary finite ring, whereby we improve upon the currently known bounds for the number of idempotents.

Keywords:

• Finite ring
• idempotent
• matrix

2020 Mathematics Subject Classification:

• 16P10
• 16N40
• 16U99

1. Introduction

The study of idempotents has always been an important tool in ring theory and in particular, it has proved to be useful in studying Artinian and finite rings. Idempotents dominate the structure theory of rings. In 1966, Steger [14] checked the conditions of diagnosability of idempotent matrices over commutative rings with identity. Much of the work has been done to examine the properties of idempotents in connection with the structure of rings, see for example [3, 8, 13]. It is well-known that idempotents induce direct sum decompositions of rings (Peirce decompositions) which in turn determine the structure of rings, provided that the rings have enough idempotents. For some recent results in this area see [2] and [1]. Much research has also been devoted to the properties of rings with either small or large number of idempotents - see [4, 6]. Since the exact number of idempotents in a ring can be difficult to obtain in general, some authors have tried to establish the number of idempotents in some special classes of rings ([9, 12, 15]), while others have tried to find bounds for the number of idempotents in a ring. The first result in this direction was [10], where in 1982 MacHale found an upper bound for the number of idempotents in a finite ring. This bound has recently been improved upon in [5].

Throughout this paper, we shall assume that R is a finite ring with a multiplicative identity. We say that an element $e\in R$ is called an idempotent if $e^2=e.$ We shall denote the ring of n by n matrices over ring R by ${\mathrm{M}}_n(R).$ We say that a matrix is a 0/1 matrix if all its elements are equal to either 0 or 1. We will denote the cardinality of an arbitrary set X by $\left|X\right|.$ Furthermore, the cardinality of the group of units of ${\mathrm{M}}_n(R)$ will be denoted by $U_n(R),$ while I(R) will denote the set of all idempotents in ring R. We will assume that $U_0(R)=1.$ Finally, the Jacobson radical of a ring R will be denoted by J(R). For all the necessary theory of finite rings, we refer the reader to [11].

In this paper, we investigate bounds for the number of idempotents. In Section 2, we examine the rings of matrices over finite local commutative rings, and their direct products. In Theorem 2.5, we find lower and upper bounds for the number of idempotents in the setting of matrices over an arbitrary finite local commutative ring and Theorem 2.6 generalizes this result to the ring of matrices over arbitrary finite commutative rings. Using this result, we improve upon the result from [5, Lemma 3.3] which states that for a field F of cardinality q, we have $|I({\mathrm{M}}_n(F))|\le 2q^{n^2-1}.$ In Corollary 2.7, we namely prove that $|I({\mathrm{M}}_n(F))|\le q^{\frac{n(n+1)}{2}}.$

For an arbitrary finite ring, it has been proven in [10, Theorem 1] that if p is the smallest prime dividing $\left|R\right|,$ then $|I(R)|\le \frac{2}{p}\left|R\right|,$ while Corollary 3.6 in [5] improves upon this by stating that $|I(R)|\le {(\frac{2}{p})}^t\left|R\right|,$ where t denotes the number of distinct primes dividing $\left|R\right|.$ We improve upon both of these statements in Section 3. Specifically, Theorem 3.2 proves that $|I(R)|\le \frac{2^{t-k}}{p^t}\left|R\right|,$ where t is the number of simple rings in the decomposition of the semisimple ring $R/J(R)$ as a direct sum of simple rings, and k the number of those simple rings that are not fields.

2. Idempotents in rings of matrices

In this section, we shall investigate idempotents in ${\mathrm{M}}_n(R).$ We shall make use of the following Theorem.

Theorem 2.1.

Let R be a finite local commutative ring and $A\in {\mathrm{M}}_n(R)$ an idempotent. Then there exists an invertible matrix $P\in {\mathrm{M}}_n(R)$ and a 0/1 diagonal matrix $B\in M_n(R)$ such that $A=PB{P}^{-1}.$

Proof.

By [14, Theorem 9] every idempotent matrix in ${\mathrm{M}}_n(R)$ is diagonalizable. Thus, there exists an invertible matrix $P\in {\mathrm{M}}_n(R)$ and a diagonal matrix $B\in M_n(R)$ such that $A=PB{P}^{-1}.$ Clearly, B is an idempotent. By [11, Theorem VII.7], R contains only trivial idempotents, so B has to be a 0/1 matrix. □

This now enables us to find the number of idempotents in ${\mathrm{M}}_n(R)$ for a finite local commutative ring R, thereby extending Theorem 3.1(ii) from [5] from fields to local rings.

Lemma 2.2.

Let R be a finite local commutative ring and $n\ge 1$ an integer. Then $|I({\mathrm{M}}_n(R))|=U_n(R){\sum }_{i=0}^n\frac{1}{U_i(R)U_{n-i}(R)}.$

Proof.

Choose an idempotent $A\in {\mathrm{M}}_n(R).$ By Theorem 2.1, there exists an invertible matrix $P\in {\mathrm{M}}_n(R)$ and a 0/1 diagonal matrix $B\in M_n(R)$ such that $A=PB{P}^{-1}.$ Suppose B₁ and B₂ are two diagonal 0/1 matrices. If B₁ and B₂ have the same number of elements equal to 1, there exists a permutation matrix P such that $B_1=P{B}_2{P}^{-1}.$ On the other hand, if B₁ and B₂ do not have the same number of elements equal to 1, they are not similar matrices, since their projections to $M_n(R/J(R))$ do not have the same rank and are thus not similar matrices. This argument proves that we can use the exact same proof as the proof of Theorem 3.1(ii) from [5]. □

In order to find the bounds for the number of idempotents, we shall need the following lemma on the number of units in ${\mathrm{M}}_n(R).$

Lemma 2.3.

Let R be a finite local commutative ring with p^kr elements and the factor field $R/J(R)$ with $q=p^r$ elements for a prime number p and integers r and k. Then for every integer $n\ge 1,$ $U_n(R)=q^{\frac{n}{2}(2nk-n-1)}(q-1)\dots (q^n-1).$

Proof.

Since R/J is a field with q elements, a well-known theorem from linear algebra (see for example [7, p. 23]) states that $|{\mathrm{M}}_n(R/J(R))|=q^{(\begin{array}{c}n\\ 2\end{array})}(q-1)\dots (q^n-1).$ Observe that $J({\mathrm{M}}_n(R))={\mathrm{M}}_n(J(R))$ and thus ${\mathrm{M}}_n(R)/J({\mathrm{M}}_n(R))\simeq {\mathrm{M}}_n(R/J(R)).$ Since $J({\mathrm{M}}_n(R))$ is a nilpotent ideal, every $A\in {\mathrm{M}}_n(R)$ is a unit if and only if $A+J({\mathrm{M}}_n(R))\in {\mathrm{M}}_n(R/J)$ is a unit. The fact that $|J(R)|=q^{k-1}$ implies $|J({\mathrm{M}}_n(R))|=q^{n^2(k-1)},$ and since $(\begin{array}{c}n\\ 2\end{array})+n^2(k-1)=\frac{n}{2}(2nk-n-1),$ the lemma is now proven. □

Let us examine the cardinalities of the sets of idempotents in some “small” examples.

Example 2.4.

Using Lemmas 2.2 and 2.3 we can easily calculate the cardinalities of sets of idempotents in ${\mathrm{M}}_n({\mathbb{Z}}_2)$ for small integers n. The following Table 1 summarizes the results. We shall need these numbers in the proof of Corollary 2.7.

We can now find the upper and lower bound for $|I({\mathrm{M}}_n(R))|$ for a finite local commutative ring R. The following theorem extends the upper bound from Lemma 3.3 in [5] to the case of matrices over an arbitrary finite local commutative ring, as well as establishing a lower bound for the number of idempotents.

Theorem 2.5.

Let R be a finite local commutative ring with p^kr elements and the factor field $R/J(R)$ with $q=p^r$ elements for a prime number p and integers r and k. Then for every $n\ge 1,$ $$2+{\left((q-1)q^{2\lceil \frac{n}{2}\rceil k-1}\right)}^{n-\lceil \frac{n}{2}\rceil }\le |I({\mathrm{M}}_n(R))|\le \frac{q^{\frac{n(nk+1)}{2}}}{{(q-1)}^2}\left(1+\frac{1}{{(q-1)}^{\lfloor \frac{n}{2}\rfloor -2}}\right).$$

Proof.

Lemmas 2.3 and 2.2 yield $$|I({\mathrm{M}}_n(R))|=2+\sum _{i=1}^{n-1}\frac{q^{\frac{n}{2}(2nk-n-1)}(q-1)\dots (q^n-1)}{q^{\frac{i}{2}(2ik-i-1)}(q-1)\dots (q^i-1)q^{\frac{n-i}{2}(2(n-i)k-(n-i)-1)}(q-1)\dots (q^{n-i}-1)}.$$

Since a short calculation shows that $$\frac{n}{2}\left(2nk-n-1\right)-\frac{i}{2}\left(2ik-i-1\right)-\frac{n-i}{2}\left(2(n-i)k-(n-i)-1\right)=i(2(n-i)k-n+i),$$

we arrive at (1) $$|I({\mathrm{M}}_n(R))|=2+\sum _{i=1}^{n-1}{q}^{i(2(n-i)k-n+i)}\frac{(q^{i+1}-1)\dots (q^n-1)}{(q-1)\dots (q^{n-i}-1)}.$$(1)

Let us first examine the upper bound. If n = 1, the statement is clear. Obviously, $\frac{q-1}{q}{q}^i\le q^i-1<q^i$ for every $i\ge 1,$ therefore $$|I({\mathrm{M}}_n(R))|\le 2+\sum _{i=1}^{n-1}\left(q^{i(2(n-i)k-(n-i))}{q}^{i(n-i)}{\left(\frac{q}{q-1}\right)}^{n-i}\right)=2+\sum _{i=1}^{n-1}{\left(\frac{1}{q-1}{q}^{2ik+1}\right)}^{n-i}.$$

Define $j=\lfloor \frac{n}{2}\rfloor$ and choose any $j\ne i\in \{1,\dots ,n-1\}.$ We now prove that the inequality (2) $$(2ik+1)(n-i)<(2jk+1)(n-j)$$(2)

holds. Observe that (2)(2) $$(2ik+1)(n-i)<(2jk+1)(n-j)$$(2) is equivalent to the inequality (3) $$(i-j)(2k(n-(i+j))-1)<0.$$(3)

If i > j then $i+j\ge n,$ so $2k(n-(i+j))-1<0$ and (3)(3) $$(i-j)(2k(n-(i+j))-1)<0.$$(3) holds. If i < j then $i+j<n$ and therefore $2k(n-(i+j))-1>0$ since $k\ge 1$ and thus (3)(3) $$(i-j)(2k(n-(i+j))-1)<0.$$(3) holds in this case as well.

We now have $$|I({\mathrm{M}}_n(R))|\le 2+{\left(\frac{1}{q-1}{q}^{2jk+1}\right)}^{n-j}+\sum _{i=1,i\ne j}^{n-1}{\left(\frac{1}{q-1}{q}^{2ik+1}\right)}^{n-i}.$$

We notice that $(i-j)(2k(n-(i+j))-1)=0$ if and only if i = j, which implies that the function $i\mapsto (2ik+1)(n-i)$ is injective on the set $\{1,\dots ,n-1\}.$ Also observe that $2ik+1\ge 3$ for every $i\ge 1$ and $2=\frac{1}{q-1}2(q-1)\le \frac{1}{q-1}{q}^2.$ The above remarks, together with inequality (2)(2) $$(2ik+1)(n-i)<(2jk+1)(n-j)$$(2) yield $$2+\sum _{i=1,i\ne j}^{n-1}{\left(\frac{1}{q-1}{q}^{2ik+1}\right)}^{n-i}\le \frac{1}{q-1}{q}^2+\frac{1}{q-1}\sum _{i=1,i\ne j}^{n-1}{q}^{(2ik+1)(n-i)}\le \frac{1}{q-1}\sum _{i=1}^{(2jk+1)(n-j)-1}{q}^i.$$

Furthermore, (4) $$(2jk+1)(n-j)\le \frac{n(nk+1)}{2},$$(4) so $$\sum _{i=1}^{(2jk+1)(n-j)-1}{q}^i\le \sum _{i=1}^{\frac{n(nk+1)}{2}-1}{q}^i\le \frac{q^{\frac{n(nk+1)}{2}}}{q-1}.$$

Now this, together with (4)(4) $$(2jk+1)(n-j)\le \frac{n(nk+1)}{2},$$(4) finally gives us $$|I({\mathrm{M}}_n(R))|\le \frac{q^{\frac{n(nk+1)}{2}}}{{(q-1)}^{\lfloor \frac{n}{2}\rfloor }}+\frac{q^{\frac{n(nk+1)}{2}}}{{(q-1)}^2}=\frac{q^{\frac{n(nk+1)}{2}}}{{(q-1)}^2}\left(1+\frac{1}{{(q-1)}^{\lfloor \frac{n}{2}\rfloor -2}}\right).$$

We now proceed to find a lower bound. We start again with Equation (1)(1) $$|I({\mathrm{M}}_n(R))|=2+\sum _{i=1}^{n-1}{q}^{i(2(n-i)k-n+i)}\frac{(q^{i+1}-1)\dots (q^n-1)}{(q-1)\dots (q^{n-i}-1)}.$$(1) and considering that $\frac{q-1}{q}{q}^i\le q^i-1<q^i$ for every $i\ge 1,$ we arrive with a similar calculation as above at $$|I({\mathrm{M}}_n(R))|\ge 2+\sum _{i=1}^{n-1}{\left((q-1)q^{2ik-1}\right)}^{n-i},$$

so obviously $$|I({\mathrm{M}}_n(R))|\ge 2+{\left((q-1)q^{2\lceil \frac{n}{2}\rceil k-1}\right)}^{n-\lceil \frac{n}{2}\rceil }.$$ □

We can now obtain the following estimate on the number of idempotents in an arbitrary matrix ring over a finite commutative ring, which is one of the main results of this paper.

Theorem 2.6.

Let R be a finite commutative ring. Then there exist an integer t, integers $k_1,\dots ,k_t,r_1,\dots ,r_t$ and primes $p_1,\dots ,p_t$ such that ${\mathrm{M}}_n(R)\simeq {\mathrm{M}}_n(R_1)\times \cdots \times {\mathrm{M}}_n(R_t)$, where for every i, R_i is a local ring with $p_i^{k_i{r}_i}$ elements with the factor field $R_i/J(R_i)$ having $q_i=p_i^{r_i}$ elements and $$\prod _{i=1}^t\left(2+{\left((q_i-1)q_i^{2\lceil \frac{n}{2}\rceil k_i-1}\right)}^{n-\lceil \frac{n}{2}\rceil }\right)\le |I({\mathrm{M}}_n(R))|\le \prod _{i=1}^t\frac{q_i^{\frac{n(n{k}_i+1)}{2}}}{{(q_i-1)}^2}\left(1+\frac{1}{{(q_i-1)}^{\lfloor \frac{n}{2}\rfloor -2}}\right).$$

Proof.

Since R is a finite commutative ring, it is isomorphic to a direct product of local rings (see for example [11, Theorem VI.2]). Thus there exist an integer t, integers $k_1,\dots ,k_t,r_1,\dots ,r_t$ and primes $p_1,\dots ,p_t$ such that $R\simeq R_1\times \cdots \times R_t,$ where for every i, R_i is a local ring with $p_i^{k_i{r}_i}$ elements and the factor field $R_i/J(R_i)$ with $q_i=p_i^{r_i}$ elements. Obviously, ${\mathrm{M}}_n(R)\simeq {\mathrm{M}}_n(R_1)\times \cdots \times {\mathrm{M}}_n(R_t)$ and thus $|I({\mathrm{M}}_n(R))|={\prod }_{i=1}^t|I({\mathrm{M}}_n(R_i))|,$ so we can use Theorem 2.5 to conclude the proof. □

Also, we have the following corollary estimating the number of idempotents in a direct product of matrix rings over fields. This corollary improves upon the result from [5, Lemma 3.3].

Corollary 2.7.

Let t be an integer and $F_1,\dots ,F_t$ fields, where for every i, $q_i=\left|F_i\right|=p_i^{r_i}$ for a prime p_i and an integer r_i. For integers $n_1,\dots ,n_t$, denote $R={\mathrm{M}}_{n_1}(F_1)\times \cdots \times {\mathrm{M}}_{n_t}(F_t)$. Then $$|I(R)|\le \prod _{i=1}^t{q}_i^{\frac{n_i(n_i+1)}{2}},$$ where equality holds if and only if R is a direct product of rings isomorphic to ${\mathrm{M}}_2({\mathbb{Z}}_2)$ or ${\mathbb{Z}}_2.$

Proof.

It is enough to prove that $|I({\mathrm{M}}_{n_i}(F_i))|\le q_i^{\frac{n_i(n_i+1)}{2}}$ for every $i\in \{1,\dots ,t\}.$ If n_i = 1, then the fact that $\left|F_i\right|\ge 2$ implies $2=|I(F_i)|\le \left|F_i\right|=q_i^{\frac{n_i(n_i+1)}{2}},$ whilst noting also that we have the equality here if and only if $\left|F_i\right|=2.$ Assume now that $n_i\ge 2.$ If $q_i\ge 3,$ we have by Theorem 2.5 $$|I({\mathrm{M}}_{n_i}(F_i))|\le \frac{q_i^{\frac{n_i(n_i+1)}{2}}}{{(q_i-1)}^2}\left(1+\frac{1}{{(q_i-1)}^{\lfloor \frac{n_i}{2}\rfloor -2}}\right)\le \frac{q_i^{\frac{n_i(n_i+1)}{2}+1}}{{(q_i-1)}^2}.$$

To prove our statement, we have to check that $\frac{q_i^{\frac{n_i(n_i+1)}{2}+1}}{{(q_i-1)}^2}<q_i^{\frac{n_i(n_i+1)}{2}},$ which is equivalent to $q_i<{(q_i-1)}^2,$ so the statement holds. Finally, assume that q_i = 2. If $n_i\le 4,$ then Example 2.4 shows that the Corollary holds. So, we only have to examine the case $n_i\ge 5.$ As in Equation (1)(1) $$|I({\mathrm{M}}_n(R))|=2+\sum _{i=1}^{n-1}{q}^{i(2(n-i)k-n+i)}\frac{(q^{i+1}-1)\dots (q^n-1)}{(q-1)\dots (q^{n-i}-1)}.$$(1) in the proof of Theorem 2.5, we have $$|I({\mathrm{M}}_{n_i}(F_i))|=2+\sum _{l=1}^{n_i-1}{2}^{l(n_i-l)}\frac{(2^{l+1}-1)\dots (2^{n_i}-1)}{(2-1)\dots (2^{n_i-l}-1)},$$

so by noting that $2^j-1<2^j$ for every j, we get (5) $$|I({\mathrm{M}}_{n_i}(F_i))|<2+\sum _{l=1}^{n_i-1}{2}^{l(n_i-l)}\frac{2^{(l+1)(n_i-l)}{2}^{\frac{(n_i-l-1)(n_i-l)}{2}}}{(2-1)\dots (2^{n_i-l}-1)}.$$(5)

We further observe that $u^2\ge 3u-1$ for all $u\ge \frac{3+\sqrt{5}}{2},$ so by choosing $u=2^j,$ we arrive at the inequality $(2^j-1)(2^{j+1}-1)\ge 2^{2j},$ which holds for all $j\ge 2.$ If $n_i-l$ is odd, then $(2-1)\dots (2^{n_i-l}-1)\ge 2^4·2^8\cdots 2^{2(n_i-l-1)}=2^{4(1+2+\dots +\frac{n_i-l-1}{2})}=2^{\frac{(n_i-l-1)(n_i-l+1)}{2}}.$ Therefore, whenever $n_i-l$ is odd, the summands on the right side of Inequality (5)(5) $$|I({\mathrm{M}}_{n_i}(F_i))|<2+\sum _{l=1}^{n_i-1}{2}^{l(n_i-l)}\frac{2^{(l+1)(n_i-l)}{2}^{\frac{(n_i-l-1)(n_i-l)}{2}}}{(2-1)\dots (2^{n_i-l}-1)}.$$(5) are bounded above by $2^{l(n_i-l)}\frac{2^{(l+1)(n_i-l)}{2}^{\frac{(n_i-l-1)(n_i-l)}{2}}}{2^{\frac{(n_i-l-1)(n_i-l+1)}{2}}}=2^{(n_i-l)(2l+\frac{1}{2})+\frac{1}{2}}.$

On the other hand, if $n_i-l$ is even, then $n_i-l+1$ is odd, so $(2-1)\dots (2^{n_i-l}-1)\ge \frac{2^{\frac{(n_i-l)(n_i-l+2)}{2}}}{2^{n_i-l+1}-1}$ by the above. In this case, the summands on the right side of Inequality (5)(5) $$|I({\mathrm{M}}_{n_i}(F_i))|<2+\sum _{l=1}^{n_i-1}{2}^{l(n_i-l)}\frac{2^{(l+1)(n_i-l)}{2}^{\frac{(n_i-l-1)(n_i-l)}{2}}}{(2-1)\dots (2^{n_i-l}-1)}.$$(5) are bounded above by $2^{l(n_i-l)}\frac{2^{(l+1)(n_i-l)}{2}^{\frac{(n_i-l-1)(n_i-l)}{2}}{2}^{n_i-l+1}}{2^{\frac{(n_i-l)(n_i-l+2)}{2}}}=2^{(n_i-l)(2l+\frac{1}{2})+1}.$

Similarly, as in the proof of Theorem 2.5, we observe that the function $$l\mapsto \{\begin{array}{cc}(2l+\frac{1}{2})(n_i-l)+1,\hfill & \mathrm{ }\text{for}\mathrm{ }{n}_i-l\mathrm{ }\text{even}\hfill \\ (2l+\frac{1}{2})(n_i-l)+\frac{1}{2},\hfill & \mathrm{ }\text{for}\mathrm{ }{n}_i-l\mathrm{ }\text{odd}\hfill \end{array}=\lfloor (2l+\frac{1}{2})(n_i-l)+1\rfloor$$ is injective on the set $\{1,\dots ,n_i-1\}$ and that for every $l\in \{1,\dots ,n_i-1\}$ we have $(2l+\frac{1}{2})(n_i-l)\le (2j+\frac{1}{2})(n_i-j)$ where $j=\lfloor \frac{n_i}{2}\rfloor .$ Inequality (5)(5) $$|I({\mathrm{M}}_{n_i}(F_i))|<2+\sum _{l=1}^{n_i-1}{2}^{l(n_i-l)}\frac{2^{(l+1)(n_i-l)}{2}^{\frac{(n_i-l-1)(n_i-l)}{2}}}{(2-1)\dots (2^{n_i-l}-1)}.$$(5) now transforms into $$|I({\mathrm{M}}_{n_i}(F_i))|<2+\sum _{l=1}^{n_i-1}{2}^{\lfloor (2l+\frac{1}{2})(n_i-l)+1\rfloor }=2+2^{\lfloor (2j+\frac{1}{2})(n_i-j)+1\rfloor }+\sum _{l=1,l\ne j}^{n_i-1}{2}^{\lfloor (2l+\frac{1}{2})(n_i-l)+1\rfloor }.$$

Together with the fact that $(2l+\frac{1}{2})(n_i-l)\ge 2$ for all l, we get $$|I({\mathrm{M}}_{n_i}(F_i))|<2^{\lfloor (2j+\frac{1}{2})(n_i-j)+1\rfloor }+\sum _{l=1}^{\lfloor (2j+\frac{1}{2})(n_i-j)+1\rfloor -1}{2}^l\le 2^{\lfloor (2j+\frac{1}{2})(n_i-j)+2\rfloor }.$$

Now, we only have to prove that $\lfloor (2j+\frac{1}{2})(n_i-j)+2\rfloor \le \frac{n_i(n_i+1)}{2}.$ Again, we have to examine the cases of n_i being odd and even separately. Suppose first that $n_i=2m\ge 6$ is even. Then j = m, so $(2j+\frac{1}{2})(n_i-j)+2=2m^2+\frac{m}{2}+2.$ We need to prove that $\lfloor 2m^2+\frac{m}{2}+2\rfloor \le 2m^2+m=\frac{n_i(n_i+1)}{2}$ which is equivalent to $\lfloor \frac{m}{2}+2\rfloor \le m.$ This obviously holds, since $m\ge 3.$ In the case $n_i=2m+1\ge 5$ is odd, we have j = m and $(2j+\frac{1}{2})(n_i-j)+2=2m^2+\frac{5m}{2}+\frac{5}{2}.$ In this case, we need to prove that $\lfloor 2m^2+\frac{5m}{2}+\frac{5}{2}\rfloor \le 2m^2+3m+1=\frac{n_i(n_i+1)}{2},$ or equivalently $\lfloor \frac{m}{2}+\frac{3}{2}\rfloor \le m,$ which again holds for all $m\ge 2.$ □

3. Idempotents in an arbitrary finite ring

In this section, we find the bounds for the number of idempotents in an arbitrary finite ring. First, we prove the following lemma.

Lemma 3.1.

Let R be a finite ring. Then $$|I(R/J(R))|\le |I(R)|\le |J(R)||I(R/J(R))|.$$

Proof.

Obviously, for every idempotent $e\in R,e+J(R)$ is an idempotent in $R/J(R),$ so $|I(R)|\le |J(R)||I(R/J(R))|.$ On the other hand, every idempotent in $R/J(R)$ can be lifted (see [11, Theorem VII.11]) to an idempotent in R, therefore $|I(R)|\ge |I(R/J(R))|.$ □

If p is the smallest prime dividing $\left|R\right|,$ then Theorem 1 in [10] states that $|I(R)|\le \frac{2}{p}\left|R\right|,$ while Corollary 3.6 in [5] states that $|I(R)|\le {(\frac{2}{p})}^t\left|R\right|,$ where t denotes the number of distinct primes dividing $\left|R\right|.$ We can now improve upon both of these statements in the following theorem.

Theorem 3.2.

Let R be a finite ring, p be the smallest prime dividing $\left|R\right|$ and $R/J(R)\simeq {\mathrm{M}}_{n_1}(F_1)\times \cdots \times {\mathrm{M}}_{n_t}(F_t)$ for some integers $t,n_1,\dots ,n_t$ and fields $F_1,\dots ,F_t$. Let $k=\left|\right\{i\in \{1,\dots ,t\};n_i>1\left\}\right|$. Then (6) $$|I(R)|\le \frac{2^{t-k}}{p^t}\left|R\right|.$$(6)

Furthermore, equality in (6)(6) $$|I(R)|\le \frac{2^{t-k}}{p^t}\left|R\right|.$$(6) holds if and only if either R is a direct product of $t\ge 1$ rings isomorphic to ${\mathrm{M}}_2({\mathbb{Z}}_2)$ or ${\mathbb{Z}}_2$, or R is a direct product of $t\ge 1$ fields isomorphic to the field ${\mathbb{Z}}_p.$

Proof.

We can assume without loss of generality that $n_1,\dots ,n_k>1.$ For every i, F_i is a field with $q_i=p_i^{r_i}$ elements for some prime p_i and integer r_i. By Lemma 3.1, we have $$\frac{|I(R)|}{\left|R\right|}\le \frac{|I(R/J(R))|}{|R/J(R)|}.$$

By Corollary 2.7, we have $$\frac{|I({\mathrm{M}}_{n_1}(F_1)\times \cdots \times {\mathrm{M}}_{n_k}(F_k))|}{|{\mathrm{M}}_{n_1}(F_1)\times \cdots \times {\mathrm{M}}_{n_k}(F_k)|}\le \prod _{i=1}^k{q}_i^{\frac{n_i(n_i+1)}{2}-n_i^2}=\prod _{i=1}^k\frac{1}{q_i^{\frac{n_i^2-n_i}{2}}}\le \prod _{i=1}^k\frac{1}{p_i}\le {\left(\frac{1}{p}\right)}^k.$$

We also have $$\frac{|I({\mathrm{M}}_{n_{k+1}}(F_{k+1})\times \cdots \times {\mathrm{M}}_{n_t}(F_t))|}{|{\mathrm{M}}_{n_{k+1}}(F_{k+1})\times \cdots \times {\mathrm{M}}_{n_t}(F_t)|}=\prod _{i=k+1}^t\frac{2}{\left|F_i\right|}\le \prod _{i=k+1}^t\frac{2}{p_i}\le {\left(\frac{2}{p}\right)}^{t-k}.$$

Together, this gives us $$\frac{|I(R)|}{\left|R\right|}\le \frac{2^{t-k}}{p^t}.$$

If R is a direct product of $t\ge 1$ rings isomorphic to ${\mathrm{M}}_2({\mathbb{Z}}_2)$ or ${\mathbb{Z}}_2,$ or R is a direct product of $t\ge 1$ fields isomorphic to ${\mathbb{Z}}_p,$ then we can easily verify that we have an equality in (6)(6) $$|I(R)|\le \frac{2^{t-k}}{p^t}\left|R\right|.$$(6) .

Conversely, suppose that we have equality in (6)(6) $$|I(R)|\le \frac{2^{t-k}}{p^t}\left|R\right|.$$(6) . Assume that $J(R)\ne 0$ and choose a nonzero $x\in J(R).$ The fact that we have $|I(R)|=|J(R)|·|I(R/J(R))|$ implies that for every $e\in R$ such that $e+J(R)$ is an idempotent in $R/J(R),$ e + j has to be an idempotent in R for all $j\in J(R).$ Since $1+J(R)$ is an idempotent in $R/J(R),1+x$ has to be an idempotent in R. However, J(R) is nilpotent, so $1+x$ is a unit in R, which implies that x = 0, a contradiction. Thus J(R) = 0 and R is a semisimple ring. If $k\ge 1$ in the proof above, then for every $1\le i\le k$ we have $p_i=2=q_i$ and n_i = 2, whilst for every i > k we have $\left|F_i\right|=p_i=2,$ so R is a direct product of $k\ge 1$ rings isomorphic to ${\mathrm{M}}_2({\mathbb{Z}}_2)$ and t – k rings isomorphic to ${\mathbb{Z}}_2.$ On the other hand, if k = 0, we have $\left|F_i\right|=p_i=p$ for every $i\in \{1,\dots ,t\},$ so R is a direct product of $t\ge 1$ fields isomorphic to ${\mathbb{Z}}_p.$ □

Table 1 Cardinalities of different sets in ${\mathrm{M}}_n({\mathbb{Z}}_2).$

n	$|{\mathrm{M}}_n({\mathbb{Z}}_2)|$	$|U_n({\mathbb{Z}}_2)|$	$|I({\mathrm{M}}_n({\mathbb{Z}}_2))|$
2	16	6	8
3	512	168	58
4	65536	20160	802

Acknowledgment

The author would like to thank the referee for his/her remarks which improved the quality of this paper.

Additional information

Funding

The author acknowledges the financial support from the Slovenian Research Agency (research core funding No. P1-0222).