Proof. Observe that ${\mathrm {Ann}}(x_1, x_2, \ldots , x_n) = {\mathrm {Ann}} (x_1) \times {\mathrm {Ann}} (x_2) \times \cdots \times {\mathrm {Ann}}(x_n)$ and use Lemma 2.1.

Proof. (1) follows from [Reference Esmkhani and Jafarian Amiri6, Lemma 3.1]. To see (2), observe first that ${\mathrm {zp}}(M_n(F))={\sum _{r=0}^{n}\sum _{{\mathrm {rank}}(A)=r}{{\mathrm {Ann}}(A)}}/{|F|^{2n^2}}$ . Now, $\dim _F(\ker (A))=n-r$ for any matrix A of rank r. Thus, every column of a matrix $B \in {\mathrm {Ann}}(A)$ is a linear combination of $n-r$ vectors from $\ker (A)$ and thus $|{\mathrm {Ann}}(A)|=|F|^{n(n-r)}$ . By [Reference Morrison15], the number of matrices of rank r in $M_n(F)$ is exactly $\prod _{j=0}^{r-1}{(|F|^n-|F|^j)^2}/{(|F|^r-|F|^j)}$ . Therefore,

$$ \begin{align*}{\mathrm{zp}}(M_n(F))=\frac{|F|^{n^2}+\sum_{r=1}^n{|F|^{n(n-r)}(\prod_{j=0}^{r-1}{{(|F|^n-|F|^j)^2}/{(|F|^r-|F|^j)}})}}{|F|^{2n^2}}\end{align*} $$

and thus the desired equality follows.

Proof. Denote $J=J(R)$ . By Lemma 2.1,

$$ \begin{align*} {\mathrm{zp}}(R/J) = \frac{\sum_{x+J \in R/J}{|{\mathrm{Ann}}(x+J)|}}{|R/J|^2} &\geq \frac{\sum_{x+J \in R/J}{{|{\mathrm{Ann}}(x)|}/{|J|}}}{|R/J|^2} \\ &=\frac{\sum_{x \in R}{{|{\mathrm{Ann}}(x)|}/{|J|^2}}}{|R/J|^2}=\frac{\sum_{x \in R}{|{\mathrm{Ann}}(x)|}}{|R|^2}={\mathrm{zp}}(R).\\[-41pt] \end{align*} $$

Proof. Observe that for every $x, y \in R$ , $xy \in I$ if and only if $(x+I)(y+I)=I$ , so for every $x \in R$ we have $|\{y \in R; xy \in I \}|{{\kern-1pt}={\kern-1pt}}|{\mathrm {Ann}}(x+I)| |I|$ . Therefore, $\sum _{x \in R}{|\{y \in R; xy \in I \}|}{{\kern-1pt}=} \sum _{x \in R}{|{\mathrm {Ann}}(x+I)||I|}=\sum _{x+I \in R/I}{|{\mathrm {Ann}}(x+I)||I|^2}$ , which is equal to ${{\mathrm {zp}}(R/I)}/{|R|^2}$ by Lemma 2.1.

Proof. Choose $x \in R$ . If $x \in I$ then $xy \in I$ for all $y \in R$ , and if $x \notin I$ then $xy \in I$ for all $y \in I$ . Thus, $\sum _{x \in R}{|\{y \in R; xy \in I \}|} \geq |I| |R| + |R \setminus I| |I| = |I| (2|R|-|I|)$ and ${\mathrm {zp}}_I(R) \geq |I| {(2|R|-|I|)}/{|R|^2}$ . Suppose now that $R/I$ is a field. Choose $x \in R \setminus I$ . Since $\overline {x} \in R/I$ is invertible, the fact that $xy \in I$ implies $\overline {y}=\overline {0}$ and thus $y \in I$ . So, $xy \in I$ if and only if $y \in I$ . We have proved that ${\mathrm {zp}}_I(R) = |I| {(2|R|-|I|)}/{|R|^2}$ . On the other hand, if ${\mathrm {zp}}_I(R) = |I|{(2|R|-|I|)}/{|R|^2}$ , then for every $x \in R \setminus I$ , the fact that $xy \in I$ has to imply that $y \in I$ . So, for every nonzero $\overline {x} \in R/I$ , the fact that $\overline {x}\overline {y}=\overline {0}$ implies that $\overline {y}=\overline {0}$ . We have proved that $R/I$ is a ring without nontrivial zero divisors and hence a field.

Proof. By Lemma 2.1, we have ${\mathrm {zp}}(R) = {\sum _{x \in R}{|{\mathrm {Ann}}(x)|}}/{|R|^2}$ . Obviously, $|{\mathrm {Ann}}(x)|=1$ for every x that is not a zero divisor, and $|{\mathrm {Ann}}(x)| \leq |Z(R)|$ for any nonzero zero divisor x, and thus ${\mathrm {zp}}(R) \leq {(|R \setminus Z(R)| + |R| + (|Z(R)|-1)|Z(R)|)}/{|R|^2}$ , so we have proved the desired inequality. If R is local with $J(R)^2=0$ , then $|{\mathrm {Ann}}(x)| = |Z(R)|$ for every nonzero zero divisor x, so the equality does indeed hold. Conversely, if R is not a local ring, then by [Reference McDonald14, Theorem VII.7] there exists a nontrivial idempotent $e \in Z(R)$ . However, if the equality holds, then $|{\mathrm {Ann}}(x)| = |Z(R)|$ for every nonzero zero divisor x, which implies that $e^2=0$ , a contradiction. So, R is local and thus $J(R)=Z(R)$ , so $J(R)^2=0$ .

Proof. By Lemma 2.1, we have ${\mathrm {zp}}(R) = {\sum _{x \in R}{|{\mathrm {Ann}}(x)|}}/{|R|^2}$ . Choose $x \in R$ . Let $I_1(x)=\{i: e_ixe_i \in (e_iRe_i)^*\} \subseteq N$ and $I_2(x)=\{i: 0 \neq e_ixe_i \in J(e_iRe_i)\} \subseteq N$ . Observe that $I_1(x)$ and $I_2(x)$ are disjoint sets. Note also that every x can be written as $x=\sum _{i,j=1}^n{e_ixe_j}$ . Thus, for any disjoint sets $I_1, I_2 \subseteq N$ , there exist exactly

$$ \begin{align*}\prod_{i \in I_1}{|(e_iRe_i)^*|} \prod_{j \in I_2}{(|J(e_iRe_i)|-1)} \frac{|R|}{\prod\limits_{i \in I_1 \cup I_2}{|e_iRe_i|}}\end{align*} $$

elements in R such that $I_1(x)=I_1$ and $I_2(x)=I_2$ . Choose $y \in {\mathrm {Ann}}(x)$ . For any $i \in I_1(x)$ , we have $e_ixe_iy=0$ and therefore $e_iy=0$ . Also, if $j \in I_2(x)$ then $j \notin I_1(y)$ . Thus, $y \in \sum _{k \notin I_1 \cup I_2}{e_kR} + \sum _{k \in I_2, l \neq k}{e_kRe_l} + \sum _{k \in I_2}{J(e_kRe_k)}$ . Therefore, $|{\mathrm {Ann}}(x)|$ is bounded by

$$ \begin{align*}\prod\limits_{i \notin I_1 \cup I_2}{|e_iR|} \prod\limits_{j \in I_2}\frac{|e_jR|}{|(e_jRe_j)^*|}=\frac{|R|}{\prod\limits_{i \in I_1 \cup I_2}{|e_iR|}} \prod\limits_{j \in I_2}\frac{|e_jR|}{|(e_jRe_j)^*|}=\frac{|R|}{\prod\limits_{i \in I_1}{|e_iR|}} \prod\limits_{j \in I_2}\frac{1}{|(e_jRe_j)^*|},\end{align*} $$

and (4.1) holds.

Now, if R is a basic ring with $J(R)^2=0$ then $e_kRe_l \in J(R)$ for all $k \neq l$ , so the equality holds in (4.1). Conversely, assume that the equality holds in (4.1). If R is not a basic ring, then by the Wedderburn–Artin theorem there exists a full matrix ring $M_r(F)$ as a subring of $R/J(R)$ for a field F and an integer r. Thus, there exist two orthogonal minimal idempotents $\overline {e}, \overline {f} \in M_r(F) \subseteq R/J(R)$ such that $\overline {e}=P\overline {f}P^{-1}$ for some invertible (permutation) matrix $P \in M_r(F)$ . Therefore, $\overline {e}P\overline {f}P^{-1}\overline {e}=\overline {e}$ . By [Reference McDonald14, Theorem VII.12], there exist orthogonal idempotents $e, f \in R$ such that $e+J(R)=\overline {e}$ and $f+J(R)=\overline {f}$ . Also, by [Reference McDonald14, Theorem VII.13], there exist $a \in R^*$ and $i, j \in N$ such that $e=ae_ia^{-1}$ and $f=ae_ja^{-1}$ . Choose an invertible $p \in R$ such that $p+J(R)=P$ . The equality in (4.1) now implies that $(e_ia^{-1}pae_j)(e_ja^{-1}p^{-1}ae_i)=0$ ; thus, $(epf)(fp^{-1}e)=0$ and consequently $\overline {e}P\overline {f}P^{-1}\overline {e}=\overline {0}$ , a contradiction. We have proved that $R/J$ is indeed a basic ring. Now, $J(R) = \sum _{i \in N}{J(e_iRe_i)} + \sum _{i \neq j \in N}{e_iRe_j}$ , so the fact that $J(R)^2=0$ follows immediately from the equality in (4.1).

Proof. By Lemma 2.1, we have ${\mathrm {zp}}(R) = {\sum _{x \in R}{|{\mathrm {Ann}}(x)|}}/{|R|^2}$ . Since any element in R that is not a zero divisor is invertible and ${\mathrm {Ann}}(x)$ is a nontrivial subgroup of the group $(R,+)$ ,

$$ \begin{align*} {\mathrm{zp}}(R) = \frac{|R \setminus Z(R)| + \sum_{x \in Z(R)}{|{\mathrm{Ann}}(x)|}}{|R|^2} & = \frac{2|R| - |Z(R)| + \sum_{0 \neq x \in Z(R)}{|{\mathrm{Ann}}(x)|}}{|R|^2} \\ &\geq \frac{2|R| - |Z(R)| + p(|Z(R)|-1)}{|R|^2}. \end{align*} $$

Since $|Z(R)| \geq \sqrt {|R|}$ by [Reference Koh11], this implies that ${\mathrm {zp}}(R) \geq {(2|R| + (p-1)\sqrt {|R|}-p)}/{|R|^2}$ . Obviously, if $R={\mathbb {Z}}_{p^2}$ or $R={\mathbb {Z}}_p[x]/(x^2)$ , we have ${\mathrm {zp}}(R)={(2p^2+(p-1)p-p)}/{p^2}$ , so the equality holds. On the other hand, if ${\mathrm {zp}}(R) = {(2|R|+(p-1)\sqrt {|R|}-p)}/{|R|^2}$ then $|Z(R)| = \sqrt {|R|}$ and $|{\mathrm {Ann}}(x)|=p$ for any nonzero $x \in Z(R)$ . Now, $|Z(R)| = \sqrt {|R|}$ together with [Reference Kobayashi and Koh10, Corollary 2.6] implies that R is a local ring with $J(R)^2=0$ . Since $|{\mathrm {Ann}}(x)|=p$ , we conclude that $|J(R)| = p$ . Finally, we use [Reference Akbari and Mohammadian1, Theorem 3] to see that $R={\mathbb {Z}}_{p^2}$ or $R={\mathbb {Z}}_p[x]/(x^2)$ .

Proof. This follows directly from Lemmas 2.2 and 4.3 and [Reference McDonald14, Theorem I.1].