# To show a ring of order p (prime) is isomorphic to the integers mod p.

• rachellcb
In summary, if R is a finite ring of order p where p is prime, then either R is isomorphic to Z/pZ or xy=0 for all x,y in R. This is because if R is not isomorphic to Z/pZ, then it must be an integral domain with no zero-divisors, which means xy cannot equal 0 for any x,y in R.
rachellcb
If R is a finite ring of of order p where p is prime, show that either R is isomorphic to Z/pZ or that xy=0 for all x,y in R

I know that both R and Z/pZ have the same number of elements (up to equivalence) and that R isomorphic to Z/pZ implies R must be cyclic (I think) but am otherwise lost on where to start. Help much appreciated!

Suppose that $(R,+,\cdot)$ is not a trivial ring. Notice that $(R,+) \cong (\mathbb{Z}/p\mathbb{Z},+)$ by the Lagrange Theorem.

Let $r \in R$ be an element generating the additive group of $R$. If $a,b \in R$ are non-zero, then it follows that $a = r + \cdots + r$ (n times) and $b = r + \cdots + r$ (m times) where neither $n,m$ divides $p$; and in particular it follows that $ab = r^2 + \cdots + r^2 \; (mn \; \mathrm{times}) = ba$. This proves two things:
1. If $r^2 = 0$, then $(R,+,\cdot)$ would be a trivial ring. This means that $r^2 \neq 0$.
2. Since $p$ is prime and $r^2 \neq 0$, this means that $ab \neq 0$. This shows that $R$ contains no zero-divisors.
Using the fact that $R$ contains no zero-divisors, you should pretty easily be able to show that $R$ is a unital ring; and therefore, $R$ is an integral domain. From there, it should be pretty simply to construct the desired isomorphism.

## 1. What does it mean for a ring to be isomorphic to the integers mod p?

When we say that a ring is isomorphic to the integers mod p, it means that there exists a bijective map between the elements of the ring and the integers mod p, where addition and multiplication operations are preserved.

## 2. How can we prove that a ring of order p is isomorphic to the integers mod p?

To prove that a ring of order p is isomorphic to the integers mod p, we can use the First Isomorphism Theorem, which states that if we have a homomorphism between two rings, and its kernel is the zero ideal, then the two rings are isomorphic.

## 3. Can a ring of order p be isomorphic to the integers mod q, where q is a different prime number?

No, a ring of order p can only be isomorphic to the integers mod p. This is because the order of a ring is determined by the number of elements it contains, and since p and q are different prime numbers, their corresponding rings will have different numbers of elements.

## 4. Why is it important to show that a ring of order p is isomorphic to the integers mod p?

Showing that a ring of order p is isomorphic to the integers mod p allows us to better understand the structure and properties of the ring. It also allows us to use the well-known properties of the integers mod p in our calculations and proofs involving the ring.

## 5. Can we generalize this result to rings of order n, where n is any positive integer?

Yes, we can generalize this result to rings of order n, where n is any positive integer, by using the Chinese Remainder Theorem. This theorem states that if we have two rings, each with relatively prime orders, then their direct product is isomorphic to the direct product of their corresponding integers mod p. Therefore, we can use this theorem to show that a ring of order n is isomorphic to a product of rings of order p, where p is a prime factor of n.

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