5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 at 15:36
  • 2
    $\begingroup$ @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 at 15:44
  • $\begingroup$ Sir, one last query, is i can say $a= x(x^{2})^{-1}$ ? $\endgroup$ – Akash Patalwanshi Apr 11 at 3:10
  • 1
    $\begingroup$ @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 at 3:55
  • 1
    $\begingroup$ @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 at 4:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.