2
$\begingroup$

There's a theorem in Abstract Algebra which states that:

An element of a quotient ring $\mathbb{Z}/\langle n \rangle$ or $\mathbb{Z_n}$ that is a coset $\overline{a}$ is invertible iff $a$ and $n$ are relatively prime.

I'm having problem understanding this theorem.

My confusion is: can't there be situations where $a$ and/or $n$ are not primes but $\overline{a}$ is invertible.

I know I'm wrong but I like to know where I'm wrong.

Suppose there's an ideal of $\mathbb{Z}$ which is $\langle 6 \rangle$

Now here $n$ which is $6$ is not prime.

An element(one of the coset) of quotient ring $\mathbb{Z_6}$ is:

$$ \overline{4} = \langle 6 \rangle + 4 = \{ \cdots, -8, -2, 4, 10, 16, \cdots \} $$

Here take a number from this set:

Say $4$ but $4$ is invertible in the sense that $4 - 4 = 0$ so it's inverse is $-4$ and $4$ is not prime.

Why's this invertible?

Can anyone kindly tell me the error in my thought process?

$\endgroup$
  • $\begingroup$ you gotta think multiplicatively, so gonna make sense $\endgroup$ – janmarqz Jun 4 '13 at 19:57
  • 1
    $\begingroup$ Inverse in this context is the multiplicative inverse. $\endgroup$ – Ayman Hourieh Jun 4 '13 at 19:57
  • $\begingroup$ Also keep in mind that $a$ and $n$ being relatively prime is very different from $a$ and $n$ being prime. $\endgroup$ – Alex Becker Jun 4 '13 at 20:02
2
$\begingroup$

can't there be situations where $a$ and/or $n$ are not primes but $\overline a$ is invertible.

Relatively prime does not mean that the numbers themselves are prime. Integers $a$ and $n$ are said to be relatively prime if their greatest common divisor is $1$.

Also, the type of invertibility that the theorem you quote is talking about is multiplicative invertibility. It's true of every number that $a - a = 0$, but it's not always true that there's a $b$ such that $ab = 1 \pmod n$. That happens if and only if $a$ and $n$ are relatively prime.

$\endgroup$

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.