## Elements of a finite field as a sum of two squares or two cubes

Posted: February 6, 2019 in Elementary Algebra; Problems & Solutions, Groups and Fields
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It is well-known that, in a finite field, every element is a sum of two squares (Problem 1). It is however not true that every element of a finite field is a sum of two cubes. For example, in $\mathbb{Z}_7,$ we cannot write $3$ or $4$ as a sum of two cubes because $\{a^3: \ a \in \mathbb{Z}_7\}=\{0,1,6\}$ and so the only elements of $\mathbb{Z}_7$ that are a sum of two cubes are $0,1,2,5,6.$
But if, in a finite field, $\alpha^3=2$ for some non-zero element $\alpha$ of the field, then we can show that every element of the field is a sum of two cubes (Problem 2).

Problem 1. Show that every element of a finite field is a sum of two squares.

Solution. Let $F$ be a finite field. So we want to show that if $x \in F,$ then $x=a^2+b^2$ for some $a,b \in F.$ We can actually be more specific if we consider two cases. Let $|F|=q.$

Case 1$q=2n$ for some integer $n.$ Then, since $x^q=x$ for all $x \in F,$ we get $x=(x^n)^2.$ So in this case, every element of the field is a square.

Case 2$q=2n+1$ for some integer $n.$ Since $F$ is finite, the multiplicative group $F^{\times}$ is cyclic.
So $F^{\times}=\langle c \rangle.$ Let $x \in F$ and consider the sets

$A:=\{a^2: \ \ a \in F\}, \ \ \ B:=\{x-a^2: \ \ a \in F\}.$

Clearly $\{0\} \cup \{c^{2m}: \ \ 1 \le m \le n\} \subseteq A$ and $|A|=|B|.$ Thus $|A| \ge n+1$ and hence

$|A|+|B| =2|A| \ge 2n+2 > |F|.$

Therefore $A \cap B \neq \emptyset,$ i.e. there exist $a,b \in F$ such that $a^2=x-b^2$ and the result follows. $\Box$

Remark 1. Regarding the second case in the solution of Problem 1, notice that, in fact, we have

$A= \{0\} \cup \{c^{2m}: \ \ 1 \le m \le n\}$

and so $|A|=n+1.$ The reason is that if $c^k=c^{2m}$ for some integers $k,m,$ then $c^{k-2m}=1$ and hence $k-2m$ must be divisible by $|F^{\times}|=2n$ implying that $k$ is even.

Problem 2. Let $F$ be a finite field and suppose that there exists $0 \ne \alpha \in F$ such that $\alpha^3=2.$ Show that every element of $F$ is a sum of two cubes.

Solution. So we want to show that if $x \in F,$ then $x=a^3+b^3$ for some $a,b \in F.$ Let $|F|=q$ and let’s consider three cases.

Case 1: $q=3n$ for some integer $n.$ Then $x=x^q=(x^{n})^3$ for all $x \in F.$

Case 2: $q=3n+2$ for some integer $n.$ Then $x=x^{2-q}=(x^{-n})^3$ for all $0 \ne a \in F$ and clearly $0=0^3.$

So, in both cases 1 and 2, for every $x \in F,$ there exists $a \in F$ such that $x=a^3.$

Case 3: $q=3n+1$ for some integer $n.$ Since $F$ is finite, the multiplicative group $F^{\times}$ is cyclic. So $F^{\times}=\langle c \rangle.$ Let $x \in F$ and consider the sets

$A:=\{a^3: \ \ a \in F\}, \ \ B:=\{x-a^3: \ \ a \in F\}, \ \ \ C:=\{a^3-x: \ \ a \in F\}.$

Clearly $\{0\} \cup \{c^{3m}: \ \ 1 \le m \le n\} \subseteq A$ and $|A|=|B|=|C|.$ So $|A| \ge n+1$ and

$|A|+ |B|+ |C| =3|A| \ge 3n+3 > |F|.$

So at least two of the sets $A,B,C$ have non-empty intersection. If $A \cap B \neq \emptyset$ or $A \cap C \neq \emptyset,$ then $x=a^3+b^3$ for some $a,b \in F$ and we are done.
Now suppose that $B \cap C \ne \emptyset.$ So there exist $a,b \in F$ such that $x-a^3=b^3-x$ and so $2x=a^3+b^3.$ Since, as given in the problem, $\alpha^3=2$ for some $\alpha \ne 0,$ we have $2 \ne 0$ and $2^{-1}=\alpha^{-3}.$ Hence

$x=\alpha^{-3}(a^3+b^3)=(\alpha^{-1}a)^3+(\alpha^{-1}b)^3. \ \Box$

Remark 2. Regarding the third case in the solution of Problem 2, notice that, in fact, we have

$A= \{0\} \cup \{c^{3m}: \ \ 1 \le m \le n\}$

and so $|A|=n+1.$ The reason is that if $c^k=c^{3m}$ for some integers $k,m,$ then $c^{k-3m}=1$ and hence $k-3m$ must be divisible by $|F^{\times}|=3n$ implying that $k$ is divisible by $3.$

## Maximum number of elements of the union of two proper subgroups of a finite group

Posted: December 2, 2018 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Problem 1. Let $G$ be a group and suppose that $H, K$ are two subgroups of $G.$ Show that if $G=H \cup K,$ then either $H=G$ or $K=G.$

Solution. If $H \subseteq K$ or $K \subseteq H,$ then $H \cup K=G$ gives $K=G$ or $H=G$ and we are done. Otherwise, there exist $h \in H \setminus K$ and $k \in K \setminus H.$ But then $hk \in G \setminus H \cup K,$ contradiction! $\Box$

So, as a result, if $G$ is a finite group and $H,K$ are two subgroups of $G$ with $H \ne G$ and $K \ne G,$ then $|H \cup K| \ne |G|.$ That raises this question: how large could $|H \cup K|$ get? The following problem answers this question.

Problem 2. Let $G$ be a finite group and suppose that $H, K$ are two subgroups of $G$ such that $H \ne G$ and $K \ne G.$ Show that $\displaystyle |H \cup K| \le \frac{3}{4}|G|.$

Solution. Recall that $\displaystyle |HK|=\frac{|H||K|}{|H \cap K|}$ and thus $\displaystyle \frac{|H||K|}{|H \cap K|} \le |G|.$ Hence $\displaystyle |H \cap K| \ge \frac{|H| |K|}{|G|}$ and so

\displaystyle \begin{aligned}|H \cup K|=|H|+|K|-|H \cap K| \le |H|+|K|-\frac{|H| |K|}{|G|} =(a+b-ab)|G|, \ \ \ \ \ \ \ \ \ (*)\end{aligned}

where $\displaystyle a:=\frac{|H|}{|G|}$ and $\displaystyle b:=\frac{|K|}{|G|}.$
Now, since $H \ne G$ and $K \ne G,$ we have $[G:H] \ge 2$ and $[G:K] \ge 2,$ i.e. $\displaystyle a \le \frac{1}{2}$ and $\displaystyle b \le \frac{1}{2}.$ So if we let $a':=1-2a$ and $b':=1-2b,$ then $a' \ge 0, \ b' \ge 0$ and thus

$\displaystyle a+b-ab=\frac{3}{4}-\frac{a'+b'+a'b'}{4} \le \frac{3}{4}.$

The result now follows from $(*). \ \Box$

Example 1. The upper bound $\displaystyle \frac{3}{4}|G|$ in Problem 2 cannot be improved, i.e. there exists a group $G$ and subgroups $H, K$ of $G$ such that $\displaystyle |H \cup K|=\frac{3}{4}|G|.$ An example is the Klein-four group $G=\mathbb{Z}_2 \times \mathbb{Z}_2$ and the subgroups $H:=\{(0,0), (1,0)\}$ and $K:=\{(0,0),(0,1)\}.$ Then $|G|=4$ and $\displaystyle |H \cup K|=3=\frac{3}{4}|G|.$

Example 2. We showed in Problem 1 that a group can never be equal to the union of two of its proper subgroups. But there are groups that are equal to the union of three of their proper subgroups. The smallest example, again, is the Klein-four group

$\mathbb{Z}_2 \times \mathbb{Z}_2= \{(0,0), (1,0)\} \cup \{(0,0), (0,1)\} \cup \{(0,0),(1,1)\}.$

## Two elements of the same order of a group are conjugate, in somewhere!

Posted: November 24, 2018 in Elementary Algebra; Problems & Solutions, Groups and Fields
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In this post, we give a nice little application of Cayley’s theorem.

Let $G$ be a group and let $g,h \in G.$ If $g,h$ are conjugate in $G,$ i.e. $g=xhx^{-1}$ for some $x \in G,$ then clearly $g^n=1$ if and only if $h^n=1.$ So $g,h$ have the same order. The converse however is false, i.e. if $g,h \in G$ have the same order, that does not imply $g,h$ are conjugate. For example, in an abelian group, two elements are conjugate if and only if they are equal but you can obviously have distinct elements of the same order in the group, e.g. in $\mathbb{Z}/3\mathbb{Z},$ both non-zero elements have the same order $3.$

We are now going to show that although two elements of the same order of a group might not be conjugate in the group, but they are certainly conjugate in some larger group.

Problem. Let $G$ be a group and suppose that $g,h \in G$ have the same order. Show that there exists a group $S \supseteq G$ such that $g,h$ are conjugate in $S.$

Solution. By Cayley’s theorem, we can embed $G$ into the symmetric group $S:=\text{Sym}(G)$ using the injective group homomorphism $f : G \to S$ defined by $f(x)=\sigma_x \in S,$ where $\sigma_x: G \to G$ is the permutation defined by $\sigma_x(y)=xy$ for all $y \in G.$ So we only need to show that $\sigma_g, \sigma_h$ are conjugate in $S.$ Well, let $|g|=|h|=n.$ Then the cycle decomposition of $\sigma_g, \sigma_h$ are in the form

$\sigma_g=(y_1, gy_1, \cdots , g^{n-1}y_1)(y_2, gy_2, \cdots , g^{n-1}y_2) \cdots$

and

$\sigma_h=(y_1, hy_1, \cdots , h^{n-1}y_1)(y_2, hy_2, \cdots , h^{n-1}y_2) \cdots$

So $\sigma_g, \sigma_h$ have the same cycle type and hence they are conjugate in $S. \ \Box$

## Rings with no proper left ideals

Posted: November 16, 2018 in Elementary Algebra; Problems & Solutions, Rings and Modules

Suppose that $R \ne (0)$ is a ring with no proper left ideals. If $R$ has $1,$ then  $R$ is a division ring. To see this, let $0 \ne x \in R.$ Then $Rx=R$ and so $yx=1$ for some $y \in R.$ Since $y \ne 0,$ we have $Ry=R$ and hence $zy=1$ for some $z \in R.$ Then $x=zyx=z$ and so $yx=xy=1$ proving that $R$ is a division ring.

But what if $R$ doesn’t have $1 ?$ The following problem answers this question.

Problem. Let $R \ne (0)$ be a ring, which may or may not have $1.$ Show that if $R$ has no proper left ideals, then either $R$ is a division ring or $R^2=(0)$ and $|R|=p$ for some prime number $p.$

Solution. Let

$I:=\{r \in R: \ \ Rr=(0)\}.$

Then $I$ is a left ideal of $R$ because it’s clearly a subgroup of $(R,+)$ and, for $s \in R$ and $r \in I,$ we have $Rsr \subseteq Rr =(0)$ and so $Rsr=(0),$ i.e. $sr \in I.$ So either $I=(0)$ or $I=R.$

Case 1: $I=R.$ That means $sr=0$ for all $r,s \in R$ or, equivalently, $R^2=(0).$ Thus every subgroup of $(R,+)$ is a left (in fact, two-sided) ideal of $R.$ Hence $(R,+)$ has no proper subgroup (because $R$ has no proper left ideals) and thus $|R|=p$ for some prime $p.$

Case 2: $I=(0).$ Choose $0 \ne r \in R.$ So $r \notin I$ and hence $Rr=R$ because $Rr$ is clearly a left ideal of $R.$ Thus there exists $e \in R$ such that $er=r.$ Now

$\text{ann}(r):=\{s \in R: \ sr=0\},$

the left-annihilator of $r$ in $R$, is obviously a left ideal of $R$ and we can’t have $\text{ann}(r)=R$ because then $Rr=(0).$ So $\text{ann}(r)=(0).$ Since

$(re-r)r=rer-r^2=r^2-r^2=0,$

we have $re-r \in \text{ann}(r)=(0).$ Thus $re=er=r.$ Let

$J=\{x \in R: \ \ xe=x\}.$

Clearly $J$ is a left ideal of $R$ and $0 \ne r \in J.$ Thus $J=R.$ So $xe=x$ for all $x \in R.$ Now let $0 \ne r'$ be any element of $R.$ Then, by what we just proved, $r'e=r'.$ On the other hand, by the same argument we used for $r,$ we find $e' \in R$ such that $r'e'=e'r'=r'.$ Thus $r'(e-e')=0,$ i.e. $r' \in \text{ann}(e-e').$
So $\text{ann}(e-e') \ne (0)$ and hence $\text{ann}(e-e')=R,$ i.e. $R(e-e')=(0)$ and thus $e-e' \in I=(0).$
So $e=e'$ and hence $r'e=er'=r'$ for all $r' \in R.$ Thus $e=1_R$ proving that $R$ is a division ring. $\Box$

Remark. The same result given in the above problem holds if $R$ has no proper right ideals.

Example. Let $p$ be a prime number. The ring

$\displaystyle R:= \left \{\begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}: \ \ \ a \in \mathbb{Z}/p\mathbb{Z}\right\} \subset M_2(\mathbb{Z}/p\mathbb{Z})$

is not a division ring and it has no proper left (or right) ideals.

## Groups satisfying (xy)^n=x^ny^n

Posted: November 11, 2018 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , , ,

Definition 1. Let $n$ be an integer. A group $G$ is called $n$abelian if $(xy)^n=x^ny^n$ for all $x,y \in G.$
In other words, $G$ is $n$-abelian if the map $f : G \to G$ defined by $f(x)=x^n$ is a group homomorphism.

Definition 2. If $G$ is both $n$-abelian and $m$-abelian, for some integers $m,n,$ then $G$ is also $mn$-abelian because then for $x,y \in G$ we will have

$(xy)^{mn}=((xy)^m)^n=(x^my^m)^n=x^{mn}y^{mn}.$

So the set

$\text{E}(G)=\{n \in \mathbb{Z}: \ \ (xy)^n=x^ny^n, \ \ \forall x,y \in G\},$

i.e., the set of those integers $n$ for which $G$ is $n$-abelian, is a multiplicative subset of $\mathbb{Z}.$ Clearly $0,1 \in \text{E}(G).$ The set $\text{E}(G)$ is called the exponent semigroup of $G.$

Remark 1. Since $(xy)^n=x(yx)^{n-1}y,$ we have $(xy)^n=x^ny^n$ if and only if $(yx)^{n-1}=x^{n-1}y^{n-1}.$ So a group $G$ is $n$-abelian if and only if $(yx)^{n-1}=x^{n-1}y^{n-1}$ or, equivalently, $(xy)^{1-n}=x^{1-n}y^{1-n}$ for all $x,y \in G.$ So a group $G$ is $n$-abelian if and only if it is $(1-n)$-abelian. In other words, $n \in \text{E}(G)$ if and only if $1-n \in \text{E}(G).$

Example 1. Every abelian group is obviously $n$-abelian for all $n.$ It is also clear that $2$-abelian groups are abelian because $xyxy=(xy)^2=x^2y^2=xxyy$ gives $yx=xy.$ As the next two examples show, there exists a non-abelian $n$-abelian group for any $n > 2.$

Example 2. Let $n \ge 3$ be an odd integer and consider the Heisenberg group $G:=H(\mathbb{Z}/n\mathbb{Z}),$ which is a non-abelian group (why?). We show that $n, n+1 \in \text{E}(G),$ i.e. $G$ is both $n$-abelian and $(n+1)$-abelian. To see that, let

$g= \displaystyle \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} \in G.$

An easy induction shows that

$\displaystyle g^m= \begin{pmatrix}1 & ma & mb + \frac{m(m-1)}{2}ac \\ 0 & 1 & mc \\ 0 & 0 & 1 \end{pmatrix}$

for all integers $m \ge 1.$ Thus $g^n=I,$ the identity matrix, and so $g^{n+1}=g,$ for all $g \in G.$
Hence $I=(xy)^n=x^ny^n$ and $xy=(xy)^{n+1}=x^{n+1}y^{n+1}$ for all $x,y \in G$ proving that $G$ is both $n$-abelian and $(n+1)$-abelian.

Remark 2.  Notice that, in Example 2, if $n$ is even, then $\displaystyle \frac{n(n-1)}{2} \ne 0$ (as elements of $\mathbb{Z}/n\mathbb{Z})$ and so $g^n$ is not always the identity element in this case. However, there’s a way to fix this, as the next example shows. But first, notice that $H(\mathbb{Z}/n\mathbb{Z})$ can also be viewed as the group of  triples

$G:=\{(a,b,c): \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z}\}$

with multiplication defined by

$(a,b,c)(a',b',c')=(a+a',b+b'+ac',c+c').$

In the next example, we modify the above multiplication such that we get $g^n=1$ for all $g \in G.$

Example 3. Let $n \ge 3$ be any integer and consider the set $G:=\{(a,b,c): \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z}\}.$ Define multiplication in $G$ by

$(a,b,c)*(a',b',c')=(a+a',b+b'+2ac',c+c').$

It’s easy to see that $(G,*)$ is a non-abelian group. We show that $n, n+1 \in \text{E}(G).$ Let $g=(a,b,c) \in G.$ A quick induction shows that

$g^m=(ma, mb+m(m-1)ac, mc)$

for all integers $m \ge 1.$ So $g^n=(0,0,0)=1_G$ for all $g \in G$ and thus $1_G=(xy)^n=x^ny^n$ for all $x,y \in G$ proving that $G$ is $n$-abelian.
Also, since $g^{n+1}=g$ for all $g \in G,$ we have $xy=(xy)^{n+1}=x^{n+1}y^{n+1}$ and so $G$ is $(n+1)$-abelian.

Problem 1. Let $G$ be a group. Show that

i) if $n, n+1 \in \text{E}(G)$ for some integer $n,$ then $x^n \in Z(G)$ for all $x \in G$

ii) if $n, n+1,n+2 \in \text{E}(G)$ for some integer $n,$ then $G$ is abelian.

Solution. i) Let $x,y \in G.$ We have

$xyx^ny^n=xy(xy)^n=(xy)^{n+1}=x^{n+1}y^{n+1}$

and so $yx^n=x^ny.$

ii) Let $x,y \in G.$ By i), we have $yx^n=x^ny$ and $yx^{n+1}=x^{n+1}y.$ Thus

$x^{n+1}y=yx^{n+1}=yx^nx=x^nyx$

and so $xy=yx. \ \Box$

Problem 2. Let $G$ be an $n$-abelian group and let $x,y \in G.$ Show that

i) $x^{n-1}y^n=y^nx^{n-1}$

ii) $(xyx^{-1}y^{-1})^{n(n-1)}=1$

Solution. i) By Remark 1, $(ab)^{n-1}=b^{n-1}a^{n-1}$ for all $a,b \in G.$ Thus

\begin{aligned} x^{n-1}y^{n-1}=(yx)^{n-1}=((yxy^{-1})y)^{n-1}=y^{n-1}(yxy^{-1})^{n-1}=y^{n-1}yx^{n-1}y^{-1}=y^nx^{n-1}y^{-1} \end{aligned}

and the result follows.

ii) Again, using the Remark 1, we have

\begin{aligned} (xyx^{-1}y^{-1})^{n(n-1)}=((x(yx^{-1}y^{-1}))^{n-1})^n=((yx^{-1}y^{-1})^{n-1}x^{n-1})^n=(yx^{-(n-1)}y^{-1}x^{n-1})^n \\ =y^n(x^{-(n-1)}y^{-1}x^{n-1})^n=y^nx^{-(n-1)}y^{-n}x^{n-1}=1, \ \ \ \text{by i)}. \ \Box \end{aligned}

Remark 3. Let $G$ be an $n$-abelian group for some integer $n \ge 2.$ An immediate result of Problem 2, ii), is that if either $G$ is torsion-free or $G$ is finite and $\text{gcd}(n(n-1), |G|)=1,$ then $G$ is abelian.

## In finite commutative rings of odd order, number of idempotents divides number of units

Posted: October 20, 2018 in Elementary Algebra; Problems & Solutions, Rings and Modules
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All rings in this post are assumed to have the multiplicative identity $1.$

In this post, we showed that the ring $R:= \mathbb{Z}/n\mathbb{Z}$ has $2^m$ idempotents, where $m$ is the number of prime divisors of $n.$ Clearly $m$ is also the number of prime (= maximal) ideals of $R$ (recall that, in general, in commutative Artinian rings, prime ideals are maximal). Let $\varphi$ be the Euler’s totient function. The number of units of $R$ is $\varphi(n).$ If $n=\prod_{i=1}^m p_i^{n_i}$ is the prime factorization of $n,$ then $\varphi(n)=\prod_{i=}^m(p_i^{n_i}-p_i^{n_i-1}).$ If $n$ is odd, then $p_i^{n_i}-p_i^{n_i-1}$ is even for all $i$ and hence $2^m$ divides $\varphi(n).$ So if $n$ is odd, then the number of idempotents of $R$ divides the number of units of $R.$ As we are going to show now, this is a property of any finite commutative ring of odd order.

Recall that a commutative ring $R$ is called semilocal if the number of maximal ideals of $R$ is finite and it is called local if it has only one maximal ideal.

Example 1. If $p$ is a prime and $k \ge 1$ is an integer, then $\mathbb{Z}/p^k\mathbb{Z}$ is a local ring with the unique maximal ideal $p\mathbb{Z}/p^k\mathbb{Z}.$ If $n \ge 2$ is an integer, then $\mathbb{Z}/n\mathbb{Z}$ is a semilocal ring (what are its maximal ideals?).

Example 2. Generalizing the above example, Artinian rings are semilocal. This is easy to see; let $R$ be an Artinian ring and let $S$ be the set of all finite intersections of maximal ideals of $R.$ Since $R$ is Artinian, $S$ has a minimal element $I:=\bigcap_{i=1}^m M_i.$ Let $M$ be any maximal ideal of $R.$ Since $I \bigcap M \in S$ and $I$ is a minimal element of $S,$ we must have $M_1M_2 \cdots M_m \subseteq I \subseteq M.$ So $M_i \subseteq M$ for some $i$ and hence $M_i=M.$ Therefore $M_1, \cdots , M_m$ are all the maximal ideals of $R.$

Problem 1. Show that if $R$ is a local ring, then $0,1$ are the only idempotents of $R.$

Solution. Let $M$ be the maximal ideal of $R$ and let $e$ be an idempotent of $R.$ Then $e(1-e)=0 \in M.$ Thus either $e \in M$ or $1-e \in M.$ If $e \in M,$ then $1-e \notin M$ because otherwise $1=e+(1-e) \in M,$ which is false. So $1-e$ is a unit because there’s no other maximal ideal to contain $1-e.$ So $(1-e)r=1$ for some $r \in R$ and thus $e=e(1-e)r=0.$ Similarly, if $1-e \in M,$ then $e \notin M$ and thus $e$ is a unit. So $er=1$ for some $r \in R$ implying that $1-e=(1-e)er=0$ and hence $e=1. \ \Box$

Problem 2 Show that the number of idempotents of a commutative Arinian ring $R$ is $2^m,$ where $m$ is the number of maximal ideals of $R.$

Solution. Since $R$ is Artinian, it has only finitely many maximal ideals, say $M_1, \cdots , M_m$ (see Example 2). The Jacobson radical of $R$ is nilpotent, hence there exists an integer $k \ge 1$ such that

$(0)=\left(\bigcap_{i=1}^m M_i \right)^k=\prod_{i=1}^m M_i^k=\bigcap_{i=1}^m M_i^k.$

Thus, by the Chinese remainder theorem for commutative rings, $R \cong \prod_{i=1}^m R/M_i^k.$ Since each $R/M_i^k$ is a local ring, with the unique maximal ideal $M_i/M_i^k,$ it has only two idempotents, by Problem 1, and so $R$ has exactly $2^m$ idempotents. $\Box$

Problem 3. Let $R$ be a finite commutative ring. Show that if $|R|,$ the number of elements of $R,$ is odd, then the number of idempotents of $R$ divides the number of units of $R.$

Solution. Since $R$ is finite, it is Artinian. Let $\{M_1, \cdots , M_m\}$ be the set of maximal ideals of $R.$ By problem 2, the umber of idempotents of $R$ is $2^m$ and $R \cong \prod_{i=1}^m R/M_i^k$ for some integer $k \ge 1.$
Since $|R|$ is odd, each $|M_i|$ is odd too because $(M_i,+)$ is a subgroup of $(R,+)$ and so $|M_i|$ divides $|R|.$ Also, units in a local ring are exactly those elements of the ring which are not in the maximal ideal. So the number of units of each $R/M_i^k$ is $|R/M_i^k|-|M_i/M_i^k|,$ which is an even number because both $|R|$ and $|M_i|$ are odd. So the number of units of $R,$ which is the product of the number of units of $R/M_i^k, \ 1 \le i \le m,$ is divisible by $2^m,$ which is the number of idempotents of $R. \ \Box$

Remark 1. The result given in Problem 3 is not always true if the number of elements of the ring is even. For example, $\mathbb{Z}/2\mathbb{Z}$ has one unit and two idempotents. However, the result is true in $\mathbb{Z}/2^n\mathbb{Z}, \ n \ge 2,$ which has $\varphi(2^n)=2^n-2^{n-1}$ units and two idempotents.
Can we find all even integers $n$ for which the result given in Problem 3 is true in $\mathbb{Z}/n\mathbb{Z}$? Probably not because this question is equivalent to finding all integers $n$ such that $2^m \mid \varphi(n),$ where $m$ is the number of prime divisors of $n,$ and that is not an easy thing to do.

Remark 2. The result given in Problem 3 is not necessarily true in noncommutative rings with an odd number of elements. For example, consider $R:=M_2(\mathbb{F}_3),$ the ring of $2\times 2$ matrices with entries from the field of order three. Then $R$ has $(3^2-3)(3^2-1)=48$ units (see Problem 3 in this post!) but, according to my calculations, $R$ has $14$ idempotents and $14$ does not divide $48.$

## Fermat’s last theorem for finite rings

Posted: October 3, 2018 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Problem. Let $R$ be a finite ring with $1$ and let $R^*=R \setminus \{0\}.$ Show that for every integer $n \ge 1$ there exist $x,y,z \in R^*$ such that $x^n+y^n=z^n$ if and only if $R$ is not a division ring.

Solution. Suppose first that $R$ is a division ring. Then, since $R$ is a finite ring, $R$ is a finite field, by the Wedderburn’s little theorem. Let $|R|=q.$ Then $x^{q-1}=1$ for all $x \in R^*$ and so $x^{q-1}+y^{q-1}=z^{q-1}$ has no solution in $R^*.$
Conversely, suppose that $R$ is not a division ring (equivalently, a field because $R$ is finite). So $|R| > 2$ and hence the equation $x+y=z$ has solutions in $R^*$ (just choose $y=1, \ z \ne 0,1$ and $x=z-1$).
Let $J(R)$ be the Jacobson radical of $R.$ Since $R$ is finite, it is Artinian and so $J(R)$ is nilpotent.
So if $J(R) \neq (0),$ then there exists $a \in R^*$ such that $a^2=0$ and so $a^n=0$ for all $n \ge 2.$ Therefore the equation $x^n+y^n=z^n, \ n \ge 2,$ has a solution $x=a, y=z=1$ in $R^*.$
If $J(R)=(0),$ then by the Artin-Wedderburn’s theorem,

$\displaystyle R=\prod_{i=1}^k M_{n_i}(F_i),$

for some finite fields $F_i.$ Since $R$ is not a field, we have either $n_i >1$ for some $i$ or $n_i=1$ for all $i$ and $k \ge 2.$ If $n_i > 1$ for some $i,$ then $M_{n_i}(F_i),$ and hence $R,$ will have a non-zero nilpotent element and we are done. If $n_i=1$ for all $i$ and $k \ge 2,$ then

$x=(1,0,0, \cdots ,0), \ y = (0,1,0, \cdots , 0), \ z = (1,1,0, \cdots , 0)$

will satisfy $x^n+y^n=z^n. \ \Box$