From the following question:
Let $R,S_1,...,S_n$ be commutative rings, and suppose that $f_i: R\to S_i$ is an integral ring homomorphism for each $i$. Show that ring homomorphism $f:R\to \prod_{i=1}^n S_i$ for which $f(r)=(f_1(r),...,f_n(r))$ is integral. (1)
and the answer below:
If you can use the fact that the composition of integral homomorphisms is integral, I suggest you decompose $f$ as $f=\tilde{f} \circ\Delta,$ where $$\tilde{f}:R^n \rightarrow \prod_{i=1}^nS_i \text{ is given by }\tilde{f}(r_1, \dots, r_n)=(f_1(r_1), \dots, f_n(r_n)), \text{ and }$$ $$\Delta: R \rightarrow R^n \text{ is given by }r \mapsto (r, r, \dots, r).$$
To prove the integrality of $\tilde{f}$, I consider the case when $n=2$ and take $(b_1,b_2) \in B_1 \times B_2$. We have $b_1^{n} + f_{1 n-1}b_1^{n-1} + \dots + f_{11}b_1 + f_{10} =0$ and $b_2^{m} + f_{2 m-1}b_2^{m-1} + \dots + f_{21}b_2 + f_{20} =0$ for some $f_i \in Imf_1$ and $f_j \in Imf_2$.
Then, WLOG, let us say that $n > m$, so I tried to write an integral equation like $(b_1,b_2)^n + (f_1 n-1, 0)(b_1,b_2)^{n-1} \dots$. so that the second components of the coefficients are $0$ for $m+1 , \dots, n$. Is it correct? Plus, is it easier prove $(1)$ instead of showing the integrality of $\tilde{f}$?
I assume $B_i=S_i$.
I do not see any reason for your last integral equation to be true. You could just take your second equation $b_2^m+f_{2m-1}b_2^{m-1}+\dots +f_{21}b_2+f_{20}$ and multiply it by $b_2^{n-m}$. Then you can combine the two equations.
Another approach would be to remember that the set of elements which are integral form a subring. Thus it suficces to prove the assertion for elements of the form $(0,\dots,0,b_i,0,\dots,0)$, where the claim is obvious by assumption.