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 $\mathbb{Z}/p\mathbb{Z}$. Essentially it boils down to showing $\mathbb{Z}/p\mathbb{Z}$ 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 $\mathbb{Z}/p\mathbb{Z} – \{ 0 \}$ is a group under the typical multiplication.
Best Answer
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 $b=1$).