Question: is a ring of prime order a field? Must a ring of prime order contain a multiplicative identity?
My attempt: I consider the ring $\mathbb{3Z}/\mathbb{9Z}$
I saw it has only three elements $0+\mathbb{9Z}$, $3+\mathbb{9Z}$ and $6+\mathbb{9Z}$. (These are only elements of $\mathbb{3Z}/\mathbb{9Z}$ because any coset of $\mathbb{9Z}$ in $\mathbb{3Z}$ must be equal to one of the above three cosets)
Further, I saw $(3+\mathbb{9Z})(6+\mathbb{9Z})=18+\mathbb{9Z}=0+\mathbb{9Z}=\text{zero element in the ring }\mathbb{3Z}/\mathbb{9Z}$
Hence $\mathbb{3Z}/\mathbb{9Z}$ has zero divisors and hence it is not an integral domain and hence not a field.
For second part : clearly none of ( $0+\mathbb{9Z}$, $3+\mathbb{9Z}$ and $6+\mathbb{9Z}$ ) these elements is multiplicative identity (unity) in $\mathbb{3Z}/\mathbb{9Z}$ and hence the ring of prime order need not have unity.
But when I searched MSE, I saw a question with title “show that a finite ring of prime order must have a multiplicative identity” (here is link Finite rings of prime order must have a multiplicative identity )
So please tell me, am I wrong in the second part of the question? and please also verify my attempt for first part.
Please help.
Best Answer
Your example does show that a ring of prime order might not have an identity and therefore might not be a field. If you don't include the requirement of a multiplicative identity, there is a trivial ring structure on every abelian group: you can just define $x\cdot y = 0$ for every pair $x,y$. This operation is distributive over addition on both sides and is associative. It's even commutative, if you want that property. Multiplication in $3\mathbb Z/9\mathbb Z$ is trivial. Rings with this property are never fields, so your example works out and can be generalized.
The question you link to specifically assumes that this is not the case - i.e. that some product is non-zero. This lets you first write out the elements as an additive group $\{0,x,2x,\ldots,(p-1)x\}$ where multiplication here means a sum of that number of copies of $x$, not anything to do with the ring structure. You note that $(ax)(bx)=(ab)x^2$ by distributivity - which implies that $x^2$ is not $0$ if multiplication is not trivial. Then you just need to find some $a$ such that $ax^2=x$ and you will find that $ax$ is the identity - and this is not so hard since $x^2$ is just some non-zero element in a cyclic additive group of order $p$, so generates the whole additive group. If we assume a ring of prime order has non-trivial multiplication, this does indeed show that it is a field.