A question about isomorphism

Question

Let $p$ be a prime and let $\Bbb Z(p)$ be the ring of integers mod $p$. Define a function $f:\Bbb Z(p)\to \Bbb Z(p)$ by $$f(a)=\begin{cases}0, &a=0\\ a^{-1},&a \ne0\end{cases}$$ Find all primes $p$ for which $f$ is a ring isomorphism.

I have checked that $f$ is a homomorphism, and it is surjective obvious. But I don't know how to find $p$ to make it injective.

The problematic part is the homomorphism condition. Generally, you don't have $(a+b)^{-1} = a^{-1} + b^{-1}$. — Daniel Fischer, Jan 21 '14 at 00:08
$(a+b)^{-1}$ is the inverse of (a+b) under addition or multiplication? — user121819, Jan 21 '14 at 00:19
Multiplication, the inverse for addition is usually denoted $-a$. — Daniel Fischer, Jan 21 '14 at 00:20
$-(a+b)$ is the additive inverse, $(a+b)^{-1}$ is the multiplicative inverse. Just like in the rational numbers. — Thomas Andrews, Jan 21 '14 at 00:29

score 4 · Accepted Answer · answered Jan 21 '14 at 00:35

4

This only occurs for $p = 2, 3$. If $p > 2$, then for $a \neq 0$, $2(a^{-1}) = 2f(a) = f(2a) = (a+a)^{-1} = (2a)^{-1}$, so $4a = a$, i.e. $3a = 0$ for all $a \neq 0$, so the characteristic must divide $3$.

One can also check that for $p = 2, 3$, $f = \text{id}$.

answered Jan 21 '14 at 00:35

zcn

15,640

why $2f(a)=f(2a)$? – user121819 Jan 21 '14 at 01:05
$f$ is a ring homomorphism, in particular it preserves addition: $f(2a) = f(a + a) = f(a) + f(a) = 2f(a)$ – zcn Jan 21 '14 at 01:07
You're welcome! If you find this answer helpful, please mark it as answered, to avoid staying in the unanswered queue – zcn Jan 21 '14 at 01:11

A question about isomorphism

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