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Let $k$ be a field of characteristic zero, and let $a,b$ be algebraic over $k$, with minimal polynomials $f_a,f_b$, respectively, of the same prime degree $p \geq 3$. Denote: $f_a=T^p+a_{p-1}T^{p-1}+\cdots+a_1T+a_0$ and $f_b=T^p+b_{p-1}T^{p-1}+\cdots+b_1T+b_0$. Further assume that $a-b$ has also minimal polynomial of degree $p$ over $k$, denote it by $f_{a-b}=T^p+c_{p-1}T^{p-1}+\cdots+c_1T+c_0$. Further assume that $k(a)=k(b)=k(a-b)$.

Is there an explicit connection between the three minimal polynomials?

Thank you very much!

user237522
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  • See https://math.stackexchange.com/questions/1277753/question-in-algebraic-closed-field. The minimal polynomial will be a divisor of any polynomial that has $a + b$ as a root. – Sera Gunn Jun 18 '17 at 01:49
  • Possible duplicate of Question in Algebraic closed field – Sera Gunn Jun 18 '17 at 01:50
  • Thank you very much for your comments. Please do not delete them. I will edit my question to a more specific one. – user237522 Jun 18 '17 at 01:59
  • Also see https://math.stackexchange.com/questions/331017/enlightening-proof-that-the-algebraic-numbers-form-a-field – Sera Gunn Jun 18 '17 at 02:03
  • Thanks! The third link seems very helpful (I will read it carefully). – user237522 Jun 18 '17 at 02:16
  • If I am not wrong: The minimal polynomial of $a-b$, $f_{a-b}$, divides $r$, the resultant of $f_a$ and $f_{-b}$ (since $r(a-b)=0$), and is of degree $p^2$. However, I do not see how to find $f_{a-b}$. (The section "number theory" in https://en.wikipedia.org/wiki/Resultant is relevant). – user237522 Jun 18 '17 at 03:16

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