Prove: If $\mathit R$ is a commutative ring with unity and $\mathit I=(x)\subseteq R[x]$, then $R[x] / (x)\cong R$

Question

I know that to show a ring is isomorphic to another ring, I have to find a bijective ring homomorphism between the two rings. Or I could use the F.H.T. but I would also need a function to make that happen (If I'm thinking about it correctly). Do I need to just make up my own function and prove that it is a bijective ring homomorphism? If so, where do I start? Thanks!

Answer 1

Define $f: R[x] \to R$ by $f(p(x))=p(0)$. Prove this is a surjective homomorphism with kernel $(x)$ and use first isomorphism theorem

Consequence: If in addtion $R[x]$ is a PID, then the ideal $(x)$ is maximal and so $R$ is a field