# Is there a general way of consistently defining an evaluation map for $R[x]/I$?

For some ring of polynomials over R, and a particular element of that ring, we define the evaluation map to be the natural thing; $P(a)$ is just what you get when you replace all the x's in the polynomial P with a's and evaluate.

Is there a consistent way of doing with when dealing with polynomials in $R[x]/I$? I imagine we need to first find a normal form for elements of $R[x]/I$, then essentially apply the same construction, but I'm not sure how to go about doing that.

consistently

In general, a homomorphism $\varphi: A \to B$ descends to a well-defined map $\overline{\varphi}: A/I \to B$ iff $I \subseteq \ker \varphi$, i.e., $\varphi(I) = 0$. So the evaluation map $\text{eval}_a$ at a point $a \in R$ induces a well-defined map on $R[x]/I$ iff $f(a) = 0$ for all $f \in I$.
This is an example of a more general phenomenon in algebraic geometry. The coordinate ring of an affine variety $V \subseteq \mathbb{A}^n$ is defined to be the ring of all polynomial functions on $V$, and is isomorphic to $k[x_1, \ldots, x_n]/\mathbb{I}(V)$, where $\mathbb{I}(V) = \{f \in k[x_1, \ldots, x_n] : \forall a \in V,\ f(a) = 0\}$ is the vanishing ideal of $V$.
From this perspective, $R[x]/I$ is identified with the ring of polynomial functions on the set $\mathbb{V}(I) = \{a \in R : \forall f \in I,\ f(a) = 0\}$. So, as above, we find that we can only evaluate these polynomials at points in $\mathbb{V}(I)$, i.e., points where all elements of $I$ vanish.