well defined mapping

Question

Which of the following mappings is well-defined?

a) $f: \mathbb{Z}_m \rightarrow \mathbb{Z}_m, \overline{x} \mapsto \overline{x^2}$

b) $g: {\mathbb{Z}}_m \rightarrow {\mathbb{Z}}_m, \overline{x} \mapsto \overline{2^x}$

I think the second one is not well defined because of the counter example in $\mathbb{Z}_5$

$5 \mapsto \overline{2^5} = \overline{2}$ but $10 \mapsto \overline{2^{10}} = \overline{4}$

But how do I proof/disproof a)

thank you

Answer 1

For b), you were right. To look at a, we say $x = a + m\cdot q$ and see if the result is independent of $m$, i.e. if $f(a+m\cdot q) \equiv f(a)$: $$f(a+m\cdot q) = (a + m\cdot q)^2 = a^2 + 2m\cdot q + m^2 q^2 \equiv a^2 = f(a) \qquad (\text{mod }m)$$ This concludes the proof, that $f$ is well-defined

Answer 2

For a) let $x,y\in\mathbb Z$ such that $\overline x=\overline y$ in $\mathbb Z_m$ so we have $$f(\overline x)=\overline {x^2}=\overline x^2=\overline y^2=\overline{ y^2}=f(\overline y)$$ hence $f$ is well-defined since it doesn't depend on the representation $x$ of $\overline x$

Answer 3

$m\mid x-y\Rightarrow m\mid x^{2}-y^{2}=\left(x-y\right)\left(x+y\right)$.

In other words:

$\overline{x}=\bar{y}\Rightarrow\overline{x^{2}}=\overline{y^{2}}$.

So well defined.