# Reduction homomorphism between Polynomials

Let $R$ be a ring with an arbitrary ideal $I$. We consider the canonical coefficient reducing homomorphism between polonomial rings $\phi: R[X] \to (R/I)[X]$ defined by $\sum c_i X^i \to \sum \bar{c_i} X^i$ where $\bar{c_i}:= c_i mod I$ in canonical way. I want to know why $I R[X] =ker(\phi)$. Obviously we have $I R[X] \subset ker(\phi)$ (ideal property) but the other inclusion isn't clear to me.

• It might be worthwhile to clarify that $I R[X]$ is the ideal of $R[X]$ generated by $I$ - not $\{ xy : x \in I, y \in R[X] \}$ where the latter is not necessarily closed under addition. – Daniel Schepler Sep 30 '17 at 18:52

Anything in the kernel is $$f=a_0+a_1X+a_2X^2+\cdots+a_nX^n$$ where $n\in\Bbb N$ and $a_0,\ldots,a_n\in I$. Each $a_j X^j\in I R[X]$ and $f$ is a finite sum of these monomials.
Another point of view is this: start from the short exact sequence: $$0\longrightarrow I\longrightarrow R\longrightarrow R/I\longrightarrow 0,$$ which you can tensor by the free $R$-module $R[X]$ to obtain the short exact sequence: $$0\longrightarrow I\otimes_R R[X]\longrightarrow R[X]\longrightarrow R/I[X]\longrightarrow 0,$$ and observing the image of $I\otimes_R R[X]\longrightarrow R[X]$ is the ideal $IR[X]$.