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Let $B$ be a ring (commutative and with identity). It is a standard fact in Algebraic Number Theory that the sum $b_{1}+b_{2}$ and the product $b_{1}b_{2}$ of integral elements $b_{1},b_{2}\in B$ over a ring $A$ are again integral over the ring $A$. The proof of this fact is due to the module-theoretic interpretation of integrality: an element $b\in B$ is integral over $A$ iff the $A$-module $A[b]$ is finitely-generated.

I don't know much about module theory, so I'm not sure if this is an obvious question:

If $f_{1}(x)$ and $f_{2}(x)$ are "integral polynomials" for $b_{1}$ and $b_{2}$ integral elements $\in B$ over $A$, (i.e. monic polynomials in $A[x]$ with $f_{1}(b_{1})=0$ and $f_{2}(b_{2})=0$) can we say something (even if the proof is not elementary) about "integral polynomials" for the sum $b_{1}+b_{2}$ or product $b_{1}b_{2}$ in terms of $f_{1}$ and $f_{2}$? Do we know how to produce such polynomials?

Shoutre
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  • It is not clear what $f_1, f_2$ are. Are they polynomials with integral coefficients? What is the relation between them and $b_1, b_2$? – Crostul Dec 21 '15 at 14:06
  • Maybe I should've made that clearer. I meant that $f_{1}$ and $f_{2}$ are monic polynomials with coefficients in $A$ such that $f_{1}(b_{1})=f_{2}(b_{2})=0$. Edited the question. – Shoutre Dec 21 '15 at 14:08
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    This is a hard question even in the case that $A$ is a field. – Crostul Dec 21 '15 at 14:11
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    General idea: Go through the details of proving that $b_1, b_2$ integral implies $A[b_1]$ and $A[b_2]$ are finitely generated. Then go through the details of showing that $A[b_1 + b_2]$ is finitely generated. Then go through the details of showing that this implies that $b_1 + b_2$ is integral. This process should give you an explicit construction of the polynomial for $b_1 + b_2$ – Alex G. Dec 21 '15 at 14:12
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    For algebraic numbers $b_1,b_2$ there is a formula for the polynomial for $b_1+b_2$ using resultants, see here. Actually that works also for the sum of integral elements, see Martin Brandenburg's answer in the duplicate. – Dietrich Burde Dec 21 '15 at 20:10
  • Dear @DietrichBurde, to make Martin Brandenburg's answer work for commutative rings, we must first pass to the universal splitting algebra so as to make the monic witnesses of integrality split, no? – Arrow Sep 21 '21 at 10:14

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