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.



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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.