# Finite rings of prime order must have a multiplicative identity

The standard definition of a ring is an abelian group that is a monoid under multiplication (with distributivity). However there are some books that have a weaker definition in that a ring only has to be closed under multiplication (no identity).

There is a problem in my algebra book asking me to prove that if a ring (defined in the second way) has $p$ elements, where $p$ is prime. If the multiplication is not trivial (i.e. sending everything to 0) then the ring is forced to have a multiplicative identity.

Its seems like a trivial proof, but I just can't see what I'm missing.

What I have so far: Given $R$ is an ring with p elements R is an abelian group of prime order, therefore it is cyclically generated, and of characteristic $p$ and isomorphic to $Z/pZ$. Essentially it boils down to showing $Z/pZ$ is forced to have a multiplicative identity, but I just can't see where this comes from (every resource I found seems to take this as fact). Since this is a requirement regardless of multiplicative structure I can't just use the fact that $Z/pZ - 0$ is a group under the typical multiplication.

Let $x$ be a nonzero element of the ring. Then $R=\{0,x,2x,3x,\ldots,(p-1)x\}$ where $2x$ means $x+x$ etc. Then $x^2=jx$ where $1\le j\le p-1$. Moreover $(ax)(bx)=abx^2=(abk)x$. All you need to do is to prove that for some $a$, $(abk)x=bx$ for all $b$. (It's surely enough to do this for $b=1$).

• Isn't it necessary to show that $x$ is not nilpotent? – Niel de Beaudrap Nov 2 '10 at 10:14
• The multiplication is nonzero, but if $x^2=0$.... – Robin Chapman Nov 2 '10 at 10:43

If $R$ is a ring of order $p$, then it injects into the ring $\text{End}(\mathbb{Z}/p\mathbb{Z})$ of endomorphisms of the abelian group $\mathbb{Z}/p\mathbb{Z}$. How many elements does this ring have?

• ...endomorphisms of the abelian group $\mathbb Z/p\mathbb Z$... – Mariano Suárez-Álvarez Nov 2 '10 at 14:15
• Right; I'll clarify that. – Qiaochu Yuan Nov 2 '10 at 14:21

Consider the subset $Z ⊆ R$ of elements such that left-multiplication by $z ∈ Z$ does map everything on $0$, $Z$ is trivial because $R$ has non trivial multiplication, hence every non zero element $a$ of $R$ defines an automorphism of the abelian group $R$ which sends $x$ to $ax$ (because the kernel of each map is trivial) , hence we can define a map from $R\setminus\{0\}$ to the automorphisms group $\operatorname{Aut}R$ this map is injective, or $\operatorname{Aut}R$ has $p-1$ element then our map send a non zero element $a$ of $R$ on the identity morphism of $R$, we obtain $ax=x$ for every element $x$ of $R$.

Go on a hunt for all the ways in which R can fail to be isomorphic to ℤ/pℤ, and demonstrate the failure of any of them to hold.

Consider the subset Z ⊆ R of elements such that left-multiplication by z ∈ Z does map everything to zero. We may easily show that Z is an ideal in R:

• For y,z ∈ Z, we have (y+z)r = yr + zr = 0+0 = 0 for any r ∈ R, so y+r ∈ Z;
• For any z ∈ Z and a,r ∈ R, we have (az)r = a(zr) = 0 and (za)r = z(ar) = 0; so az, za ∈ Z.

Well, really, we only needed the first of these two criteria, because it shows that (Z,+) must be a subgroup of the abelian group (R,+), which is of prime order. We know that Z ≠ R; so it follows that Z = {0}.

Okay then. If none of the non-zero elements map everything to zero via multiplication, perhaps some of them map some things to zero? Select a ∈ R \ {0} arbitrary, and consider the set Sa = {b ∈ R | ab = 0}. Well, again we find that Sa is an abelian subgroup of R, and again we know that Sa is not allowed to be all of R; so we know that Sa = {0}. That is, R contains no zero divisors.

Well, are there any non-zero elements such that multiplication is not invertible? Well, no: any a,b ∈ R such that ar = br for some non-zero r ∈ R would have the property (a − b)r = 0, so that a = b; so multiplication by elements of R \ {0} yield bijections on R \ {0}.

$\rm R = \{0,a,2a,\cdots,(p-1)a\}\$ for $\rm\ 0 \ne a\in R\:$. So $\rm\ a^2 = k\:a\$ and $\rm\ k \ne 0\:$ via multiplication nontrivial. Now $\rm\ j = 1/k\ \:mod\ p\ \Rightarrow\ ia\ ja\ = \ ijka\ =\ ia\ \Rightarrow\ ja = 1\:$.