Wikipedia says there are two. But then any finite ring is also an abelian group under addition and any finite abelian group of order p is isomorphic to $\mathbb{Z_p}$. So shouldn't it be only one. Which one is other?
Asked
Active
Viewed 224 times
0
-
A ring $R$ can be isomorphic as an additive group to $\mathbb{Z}_p$ without being isomorphic to $\mathbb{Z}_p$ as a ring. – Matthew Towers Mar 16 '15 at 19:09
-
1Similar to http://math.stackexchange.com/questions/1043373/any-ring-of-prime-order-commutative – Krish Mar 16 '15 at 19:22
-
Related: https://math.stackexchange.com/questions/1825661 – Watson Dec 11 '16 at 18:47
1 Answers
2
You can always define $a \cdot b=0$ for any $a,b$, when you have an abelian group. This gives you a (very boring) ring structure.
Let us show that any other ring structure is isomorphic to the usual ring structure of $\mathbb Z/p\mathbb Z$:
Let $R$ be a ring with abelian group $\mathbb Z/p\mathbb Z$.
Since any element in the abelian group $\mathbb Z/p\mathbb Z$ is a sum of the form $1 + \dotsb + 1$, the ring structure of $R$ is uniquely determined by $x := 1 \cdot 1$.
If we have $x \neq 0$, $x$ generates the abelian group $\mathbb Z/p\mathbb Z$, hence there is $a$ with $ax=1$, which shows $1 \cdot a = 1$.
This gives rise to an isomorphism of rings $\mathbb Z/p\mathbb Z \to R, 1 \mapsto a$.
MooS
- 31,390
-
Hmmm... so even if we add commutativity property to ring still there are two rings of order p. And no noncommutative ring of order p. How can you be sure there are none other? – Bhaskar Vashishth Mar 16 '15 at 19:13
-