It is well known that if $\mathbb{F}$ is a field, then a monic polynomial $p(x) \in \mathbb{F}[x]$ has at most deg$(p(x))$ distinct roots from $\mathbb{F}$.
Now, suppose we are not working over a field, but rather over $\mathbb{Z}_n$, where $n$ is not prime. The presence of zero divisors when $n \geq 6$ allows us to quickly see that this result does not extend to these rings.
For instance, in $\mathbb{Z}_6$, $$x^2-5x = (x-2)(x-3)=x(x-5)$$ has 4 distinct zeros. This same method of using the zero divisors from our ring will allow us to construct degree 2 monic polynomials with at least 3 distinct roots.
My question is about $\mathbb{Z}_4$. The only zero divisor is 2, so if we try to apply the same method of construction as above, we get $$(x-2)(x-2)=x^2-4x+4=x^2$$ So, because of the coincidence that $2+2 = (2)(2)=4,$ we don't pick up a third distinct zero for our monic polynomial.
My question then is does the result quoted at the beginning about distinct roots of monic polynomials coincidentally extend to $\mathbb{Z}_4?$ Or can we produce a counter example? Since there are only $4$ distinct elements of $\mathbb{Z}_4$, we need only check the degree 2 and degree 3 case. Are there any hints anyone can give about either methods to construct a counterexample, or proof methods on why a counterexample is impossible? It is of course possible in principle to just write out the 16 degree 2 monic polynomials and 64 degree three monic polynomials and check the zeros, but is there a more clever way to see what is going on?
Thanks!
