In Fraleigh's A First Course in Abstract Algebra (Ed. 7), Example 51.4 (p. 437), he begins:
Let $E=\mathbb{Z}_p(y)$, where $y$ is an indeterminate. Let $t=y^{\,p}$, and let $F$ be the subfield $\mathbb{Z}_p(t)$ of E.
This implies, without explanation, that $\mathbb{Z}_p(y)$ is a field.
Question: Is this a field, and if so, how can we prove it? What is the inverse of $1+y$ (as a random example)?
It's unclear to me exactly what "polynomial" means in $\mathbb{Z}_p(y)$. In particular, is the polynomial $y^{\,p}$ equal to the polynomial $y$? As functions from $\mathbb{Z}_p$ to $\mathbb{Z}_p$, yes; but as formal polynomials in the indeterminate $y$, no. Fraleigh is careful to point out the distinction between these concepts in section 22 (p. 198–207), but never states which concept is intended when considering polynomials over $\mathbb{Z}_p$, where it actually matters.
The notation $R(y)$ denotes the field of rational functions in one variable $y$ with coefficients in the ring $R$. Don't get it mixed up with the $R[y]$, the ring of polynomials in $y$ with coefficients in $R$.
In fact $R(y)$ is the fraction field of $R[y]$, and elements of $R(y)$ are of the form $\frac{f(y)}{g(y)}$, where $f(y), g(y) \in R[y]$.
So the inverse of $1+y$ is $\frac{1}{1+y}$.