I'm asking because the correction of an exercise of a course I'm following (commutative algebra) has the following conclusion:
Let I be an ideal of a commutative ring A, and $I[X]=\{\sum_{k=0}^{n}i_{k}X^{k}\mid i_{k}\in I\}\subseteq A[X]$ ideal in A[X]. Show that A[X]/I[X] is isomorphic to the ring (A/I)[X].
Let $\pi_{I}:A\rightarrow A/I$ be the natural application$^{(1)}$ and $\phi:A[X]\rightarrow(A/I)[X]\ :\ \sum_{k=0}^{n}a_{k}X^{k}\mapsto\sum_{k=0}^{n}\pi_{I}(a_{k})X^{k}$.
$\phi$ is a surjective morphism and $\ker(\phi)=I[X]$.
$\Rightarrow$ $I[X]\subseteq \ker(\phi)$ therefore by the universal quotient property there exists a natural$^{(1)}$ isomorphism $f:A[X]/I[X]\rightarrow(A/I)[X]$.
The part which causes me trouble is:
and $\ker(\phi)=I[X]$
I'm wondering where that conclusion comes from; if it is a general property of quotient rings or if it actually comes from that specific space's structure, if it comes from that space's structure in particular.
(1) translated from french: "application canonique", maybe canonical? I am unsure.
This statement $\ker\phi=I[X]$ is specific to this problem, and it is immediate:
A polynomial $p(X)=\sum_{i\le k}a_iX^i$ is in the kernel of $\phi$, by definition, iff $\phi(p)=0$, iff each coefficient of $\phi(p)$ is zero, iff each $\pi_I(a_i)=0$ by definition of $\phi$.
But $\pi_I(a_i)=0\in A/I$ means exactly $a_i\in I$.
So we get that $p\in\ker\phi\iff\forall i:a_i\in I\iff p\in I[X]$.