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I know that given a field $\mathbb{K}$, the one variable polynomial ring $\mathbb{K}[x]$ is an euclidean domain. This helps to figure out how the quotient $\dfrac{\mathbb{K}[x]}{(f(x))}$ (where $f(x) \in \mathbb{K}[x]$) is made: its elements are the polynomials $h(x)$ such that $\text{deg}(h(x))<\text{deg}(f(x))$, because of the euclidean division. If we have a generic $A[x]$ where $A$ is a commutative ring with $1$, what can we say about $\dfrac{A[x]}{(f(x))}$ with $f(x) \in A[x]$?

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Let $c \in A$ be the leading coefficient of $f$. If $c$ is a unit, then every element of $A[x]/(f)$ can still be represented as a polynomial in $x$ of degree less than $\deg(f)$, by exactly the same argument. However, if $c$ isn't a unit, the situation is a bit more complicated.

For example, consider $\newcommand{\ZZ}{\mathbb{Z}}\ZZ[x]/(2x - 1)$. If $g \in \ZZ[x]$ has odd leading coefficient, then for any $h \in \ZZ[x]$, we have $\deg(g + (2x - 1)h) \geq \deg(g)$. In particular, any system of representatives of $\ZZ[x]/(2x - 1)$ requires polynomials of arbitrarily high degree. (One can also see this by observing that $\ZZ[x]/(2x - 1) \cong \ZZ[\frac{1}{2}]$, and one can have arbitrarily high powers of $2$ in the denominator of elements of $\ZZ[\frac{1}{2}]$.)

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