Is it true that no polynomial (with the exception of constant polynomial) have an inverse in $\mathbb Z/p\mathbb Z$ (where p is prime)?

Question

I was solving the following exercise:

Find the inverse of $p(x) = 1 + x$ in $R[x]$ over $\mathbb Z/5\mathbb Z$ or show that it does not exist.

and finding that it does not exist because if there is a polynomial $q(x)$ such that $p(x)q(x)=1$. Which will contradict the proposition:

Let $R$ be a commutative ring. For every pair of non-zero polynomials $p$ and $q$ in $R[x]$, if the leading coefficient of $p$ or $q$ is not a zero divisor in $R$ then $\deg(pq)=\deg(p)+\deg(q)$

my question is it safe to assume that there does not exist an inverse of polynomial of $\deg>0$ over $\mathbb Z/p\mathbb Z$ where $p$ is a prime number?

Answer 1

Yes it is true, this is because $\deg(PQ)=\deg(P)+\deg(Q)$ for nonzero polynomials, so both summands must be $0$.

Answer 2

Well, for any polynomials $f,g\in R[x]$, where $R$ is an integral domain (i.e., has no zero divisors), the degree formula $$\deg(f\cdot g) = \deg(f)+\deg(g)$$ holds, since the highest terms do not cancel.

Thus if $f$ is a unit in $R[x]$, then $f\cdot g =1$ for some $g\in R[x]$. Then $0 =\deg (1) = \deg(f\cdot g) =\deg(f) + \deg(g)$. Since the degrees are nonnegative it follows that $\deg(f)=0$ and so $f$ is invertible in $R$. Done.