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.
Best Answer
You definitely have the right idea, but what you've written isn't quite right. If we let $\pi:R \longrightarrow R/I$ be the canonical quotient map, then your map $\phi$ can be expressed as $\phi(a_0 + a_1 x + \cdots + a_n x^n) = \pi(a_0) + \pi(a_1)x + \cdots \pi(a_n)x^n$. As you point out, $\phi$ is a surjective ring homomorphism. By the First Isomorphism Theorem, we're done if we can show that the kernel of $\phi$ is $I[x]$.
Suppose that $\phi$ maps $a_0 + a_1x + \cdots + a_nx^n$ to zero. Recall that a polynomial is zero if and only if each of its coefficients is zero. This means that $\pi(a_i) = 0$ for each $i$. By the definition of $\pi$, this implies that $a_i \in I$ for each $i$, and we conclude that our polynomial is in $I[x]$.