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.


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.

  • $\begingroup$ Sir that means, multiplication in ring $\mathbb{3Z}/\mathbb{9Z}$ is trivial? Sir what about first part of the question? $\endgroup$ – Akash Patalwanshi Apr 10 '20 at 15:36
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    $\begingroup$ @AkashPatalwanshi I edited the answer a bit - the ring you give has trivial multiplication and is a counterexample to the claim that a ring of prime order is a field (but is, in some sense, the only counterexample - and is specifically ruled out in the question you link to) $\endgroup$ – Milo Brandt Apr 10 '20 at 15:44
  • $\begingroup$ Sir, one last query, is i can say $a= x(x^{2})^{-1}$ ? $\endgroup$ – Akash Patalwanshi Apr 11 '20 at 3:10
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    $\begingroup$ @AkashPatalwanshi No! $a$ is an integer, not an element of the ring. There's some notational trickery in this answer: we write $ax$ to mean $x+x+x+\ldots+x$ with $a$ copies of $x$ and $x^2$ to mean $x$ times itself within the ring - there are two kinds of multiplication going on here. $\endgroup$ – Milo Brandt Apr 11 '20 at 3:55
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    $\begingroup$ @AkashPatalwanshi You first need to know that a cyclic group of prime order is generated by each non-identity element. Then, $x$ and $x^2$ are both non-identity elements, therefore, restating what it means to generate a group, for some integer $a$ we must have $ax^2 = x$. You could, practically, just work this out by adding $x^2$ to itself until you get $x$ and counting how many times you did this addition. $\endgroup$ – Milo Brandt Apr 11 '20 at 4:10

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