Prove: $F[t]/H[t]$ is isomorphic to $(F/H)[t]$

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  • Let $F$ a commutative and unitary ring, and also $F$ is an integral domain.

  • Let $H$ be a proper ideal of $F$.

I want to prove: $F[t]/H[t]$ is isomorphic to $(F/H)[t]$

So my idea is to use the First Isomorphism theorem with

$$f:F[t]\to (F/H)[t]$$ $$f\left(\sum_{i=0}^n p_it^i\right)\to \sum_{i=0}^n[p_i]t^i$$

So first I must prove that $f$ is an homomorphism so

(1) $f(0_{F[t]/H[t]})=0_{(F/H)[t]}$ [Obvious]

(2) $f(P+Q)=f(P)\bigoplus f(Q)$ [I prove it]

(3) $f(PQ)=f(P)*f(Q)$ : [ I prove it]

To prove $f$ is onto (surjective) is correct just to say if $$P=\sum_{i=0}^n[p_i]t^i\in(F/H)[t]$$

Then for
$$Q=\sum_{i=0}^np_it^i\in F[t]$$ $$f(Q)=P$$

I feel I don't prove nothing ...