## Math Genius: 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 implying 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, and 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 a ring with $$p$$ elements and
$$R$$ is an abelian group of prime order, therefore it is cyclically generated, and of characteristic $$p$$ and isomorphic to $$mathbb{Z}/pmathbb{Z}$$. Essentially it boils down to showing $$mathbb{Z}/pmathbb{Z}$$ 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 a fact). Since this is a requirement regardless of multiplicative structure, I can’t just use the fact that $$mathbb{Z}/pmathbb{Z} – { 0 }$$ 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 \$1le jle 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\$).

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

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 \$Rsetminus{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}.