I'm fully aware we can prove this by showing $I[x]$ is the kernel of the homomorphism from $R[x]$ to $(R/I) [x]$.
but is the following proof valid?
Suppose we have $p,q \in R[x],\not\in I[x]$ then we can express $p(x) = I_p(x) + R_p(x)$ where $I_p(x)$ is the sum of terms in $p(x)$ with coefficients in $I$, while $R_p(x)$ is the sum of terms with coefficients that are not in $I$, $R_p(x)$ is not zero by assumption $p(x) \not \in I[x]$. Similiarly express $q(x) = I_q(x) + R_q(x)$.
then $p(x)*q(x) = I_p(x)I_q(x)+I_p(x)R_q(x)+R_p(x)I_q(x)+R_p(x)R_q(x)$ the first three terms will have coefficients in I as every term contains product that has a factor in $I$, the coefficient of the highest power in the last term will not be in $I$, hence we can again isolate $R_p(x)R_q(x) = A(x) + B(x)$ with $A(x)$ with coefficients in I(x), and $B(x)\not \in I[x]$ with B(x) non-zero.
Hence $pq$ cannot be in $I(x)$ otherwise $pq-(I_p(x)I_q(x)+I_p(x)R_q(x)+R_p(x)I_q(x))-A(x) = B(x)$ would be in I[x]$ contradiction.
This is valid?
Thank you in advance
