If $R$ is a commutative ring with unity and $I$ is an ideal of $R$ then show that $(R/I)[x]\cong R[x]/I[x]$.
My effort: Define $\phi :R[x]\to (R/I)[x]$
$\phi(a_0+a_1x+a_2x^2+\cdots +a_nx^n)=(a_0+I)+(a_1+I)x+(a_2+I)x^2+\cdots +(a_n+I)x^n$
Obviously $\phi $ is a ring homomorphism and surjective.
$\ker \phi =\{a_0+a_1x+a_2x^2+\cdots +a_nx^n:\phi(a_0+a_1x+a_2x^2+\cdots +a_nx^n)=I\}$
So $(a_0+I)+(a_1+I)x+(a_2+I)x^2+\cdots +(a_n+I)x^n=I\implies a_i\in I\forall i$
So $\ker \phi=I[x]$
Is the proof correct? Please help.