Adjoining an element to a ring $R$ that is not a field.

Question

Let $R = \mathbb{Z}/(p^{k}\mathbb{Z})$, where $p$ be any prime number, and $k > 1$ be any integer. Now let us consider an equation $x^r = p$ in $R$ and $\pi$ be the root of this equation, where $r \neq 1$. Now, it is clear that $\pi \notin R$, because if we assume $\pi \in R$, then $$\pi^r = p \\ \implies v_p(\pi^r) = v_p(p) \\ \implies v_p(\pi) = \frac{1}{r}$$

Now note that $R = \{x \in \Bbb{Z}_p : x = a_0 + a_{1}p + \cdots a_{k-1}p^{k-1} ~\&~ 0 \leq a_{i} < p \}$. So $v_{p}(R^\times) = \Bbb{N} \cup \{0\}$. Hence, our assumption is wrong and $\pi \notin R$.

I want to create an extension ring by adjoining $\pi$ to this ring $R$. I am a little confused with how the ring looks after adjoining an element. I read from Artin's algebra, and according to this book, we can denote the extension ring as $R[\pi]$ where $R[\pi] = \{f(\pi): f \in R[x]\}$.

Can we write something like $R[\pi]$ is the same as the ring $R[x]/\langle x^{r} - p\rangle$? My question is, are these two notations the same? If isomorphic, then how can I prove this isomorphism?

One can consider one homomorphism $\phi :R[x] \to R[\pi]$, by $\phi(f(x)) = f(\pi)$. Can we show that $\operatorname{Ker}(\phi)=\langle x^{r} - p\rangle$?

I want to understand how to adjoin an element to a ring that is not a PID or a field (for example $\mathbb{Z}/(p^{k}\mathbb{Z})$).

Edit: After all this discussion and from the answer, I can write that: If we assume $\pi = x + \langle x^r - p \rangle$, then my $\pi$ will not be ambiguous.

Answer 1

The notation $R[\pi]$ is unclear because you haven't defined the meaning of this symbol $\pi$; this could be interpreted as referring to the polynomial ring. The unambiguous way to adjoin an element satisfying $x^r = p$ is to write $R[x]/(x^r - p)$, although you could probably also informally write $R[\sqrt[r]{p}]$ and be understood.

because we can't claim that, the kernel is an ideal so the principal ideal and generated by the smallest power polynomial.

This is irrelevant. This ring is just defined as the quotient by a principal ideal. It doesn't matter that there are some other ideals that are not principal.

Note that whether this constitutes an "extension ring" of $R$ is less clear if $R$ is not a field, since in general there is no guarantee that the natural map from $R$ is injective. In fact if you try to adjoin some arbitrary elements satisfying some arbitrary relations you may get the zero ring. So you have to check that it's nonzero if you want to get any use out of it.