Showing that ring generated by a subset is isomorphic to a quotient ring

Question

Let $R$ be a commutative ring with $1$, $S$ a subset of $R$ and let $T$ be a subring of $R$ generated by $S$. Define $\phi: \mathbb{Z}[x_s|s \in S] \to R$ to be the algebraic extension of \begin{equation*} x_s \mapsto s. \end{equation*} Then $\phi$ is a ring homomorphism with Im$(\phi)=\mathbb{Z}[S]=T$. Now denote by $I=(f_i)$ an ideal generated by some polynomials of $\mathbb{Z}[x_s|s \in S]$. Suppose we claim that $T \cong \mathbb{Z}[x_s|s \in S]/I$. We know from the First Isomorphism Theorem that Im$(\phi)\cong\mathbb{Z}[x_s|s \in S]/ker(\phi)$ and since Im$(\phi)=T$, the only thing we still have to show to conclude the claim is that $ker(\phi)=I$. Is this correct or is there a gap somewhere in my reasoning?