What is the inverse element in a ring?

Question

Let $\Bbb L$ be a positive integer and define $\Bbb Z_L$ and $\Bbb Z^*_L$ as $$\Bbb Z_L=\{ 0,1,...,L-1\},\qquad\Bbb Z^*_L=\{i\in \Bbb Z_L\mid \gcd(i,L)=1\}$$ where $\Bbb Z_L$ is the ring of integers modulo $L$. Then my textbook says that each element of $\Bbb Z^*_L$ has an inverse element in $\Bbb Z_L$. Why is this true?

Answer 1

First of all, for the future, it is always a good thing to add some attempt of yours to a question, to show that you already put some effort.

Secondly, try to be a bit more precise. I assume that you are looking at $Z_{L}^*$ as endowed with the multiplication, so you are looking for a multiplicative inverse.

Now, asking for $gcd(i,L)=1$ implies that (by Bezout theorem) there exist $a,b\in Z$ such that $$aL+bi=1.$$ If you reduce modulo $L$ this relation, it tells you that $\bar{b}\bar{i}=\bar{1}$, that is to say, that $\bar{b}$ is the inverse of $\bar{i}$ in $Z_{L}^*.$

Answer 2

This is a nice counting argument - I'll sketch how it goes (it's a good exercise to fill in the remaining steps).

Suppose I have a ring $R$. Then for any $a\in R$ we have a "multiplication by $a$" function $$m_a:R\rightarrow R: x\mapsto ax.$$ (Technically I should call this left multiplication, but in the current context our ring is commutative so it doesn't matter.)

Now in general there's no reason for $m_a$ to be injective (and it's a good idea at this point to cook up some injective and non-injective examples). However, we do learn something interesting from a non-injectivity:

Suppose $m_a(x)=m_a(y)$. Then $a$ is a zero divisor - that is, there is some nonzero $b$ such that $ab=0$ (in the sense of $R$).

Do you see why? HINT: think about $ax-ay$ ...

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Now finite rings - like those in the question - have a nice property: any map from a finite ring to itself is injective iff it is surjective. (This isn't a property of rings, just a property of finite sets; and this applies to all maps, homomorphisms and non - indeed, the $m_a$s aren't ring homomorphisms in general.)

So looking at $Z_L$ as in the question, if $a\in Z_L$ there are two possibilities for how $m_a$ behaves:

• Option $1$: $m_a$ is surjective. Then in particular $1$ is in the range of $m_a$; do you see why this means that $a$ is invertible?

• Option $2$: $m_a$ is not surjective, hence not injective. Then by the above point we have that $a$ is a zero divisor. So now the problem boils down to showing:

No element of $Z_L^*$ is a zero divisor.

And this will follow from just writing out the definitions.

Answer 3

This is true because of Bézout's identity: if $\gcd(i,L)=1$, there exist intergers $u,v$ such that $$ui+vL=1,$$ hence, denoting $[\,.]$ the congruence classes modulo $L$, one has $\;[u][i]=[1]$, so that $[u]=[i]^{-1}$.

Answer 4

You can conclude the invertibility using the cancelling rule for modular equivalences:

• $(\star)$: If $(k,L)=1$, then $ka \equiv kb \mod L \Leftrightarrow a \equiv b \mod L$

Now, take $i \in Z^*_L$, $i\neq 1$ and consider the powers $i^m$ for $m \in \mathbb{N}$.

Since $Z^*_L$ is finite, there are $m,n \in \mathbb{N}$ with $m>n$ such that $$i^m \equiv i^n \mod L \Rightarrow i^{m-n}\cdot i^{n} \equiv i^n \mod L \stackrel{(\star)}{\Rightarrow} \boxed{i^{m-n} \equiv 1 \mod L}$$