Let $A \subseteq B$ be a ring extension. Let $f(X) \in A[X]$ be an irreducible monic polynomial such that $f(X)=g(X)h(X)$ in $B[X]$ such that polynomials $g(X),h(X)$ are monic.Then the coefficients of $g(X),h(X)$ are integral over the ring $A$.
My question is, $A\subseteq B$ , can $B$ be thought as an algebraic extension of $A$? I am not very clear when my professor mentions $A \subseteq B$ be a ring extension, what I presumed was $A$ was just a subring of $B$. But I was not able to prove the result. Any help or hint would be appreciated.
Injectivity of the "extension" homomorphism $A\to B$ is not required for this.
$g=X^n+b_{n-1}X^{n-1}+\cdots+b_0$ factors as $\prod_{i=1}^n(X-t_i)$ in $S[X]$, where $S=B[T_1,\cdots,T_n]/I$ with $I$ the ideal generated by the elements $b_{n-i}-(-1)^i\sigma_i$. The $\sigma_i$ are the elementary symmetric polynomials in $T_1,\cdots,T_n$ for $1\leq i\leq n$, and $t_i$ denotes the image of $T_i$ in $S$. Then $g(t_i)$ $=$ $0$ in $S$, so $f(t_i)$ $=$ $0$, hence the $t_i$ are integral over $A$. So $b_{n-i}$ $=$ $(-1)^i\sigma_i(t_1,\cdots,t_n)$ is also integral over $A$ for $1\leq i\leq n$.
Note that $B\subseteq S$. For we can adjoin one root of $g$ at a time, starting with $B_1$ $=$ $B[T_1]/(g(T_1))$, then write $g$ $=$ $(X-t_1)g_1$ in $B_1[X]$, pass to $B_2$ $=$ $B_1[T_2]/(g_1(T_2))$, etc, and arrive at a ring $B_n$ over which $g$ factors as a product of linear polynomials. $B_i\subseteq B_{i+1}$, because each time we divide out by an ideal generated by a monic polynomial. So $B\subseteq B_n$, and since $B_n$ is a quotient of $S$ (in fact, it is isomorphic to $S$), we have $B\rightarrowtail S$ too.
Edit: the result also follows immediately from Kronecker's theorem (also known as Dedekind Prague), by which, for $g$ $=$ $\sum_{i=0}^nb_iX^i$ and $h$ $=$ $\sum_{j=0}^mc_jX^j$ in $B[X]$, all products $b_ic_j$ are integral over the subring of $B$ generated by the coefficients of $gh$ $=$ $f$. (For we have $b_n$ $=$ $1$ $=$ $c_m$ here.) See Theorem 3.3 here.