Finding the smallest field containing multivariate polynomial evaluations of the roots of an irreducible polynomial

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$newcommandQmathbb Q$Suppose I have an irreducible polynomial $finQ[x]$ and suppose I know the roots say $r_1,dots,r_nin bar Q$. I want to know if there is an easy way to "compute" the smallest number field $K$ such that we have
$$a_0(r_1,dots, r_n) + a_1(r_1,dots, r_n)x + dots + a_k(r_1,dots, r_n)x^k in K[x]$$
where $a_0,dots, a_k in Q[x_1,dots, x_n]$. I am actually only interested in knowing $[K:Q]$.



I suppose this is "strongly" dependent on how $a_0,dots, a_k$ are constructed. In my particular case, I want to prove that $[K:Q]$ is actually always even. I guess I can go into details of the $a_i's$ but I am not sure if this will help in solving $[K:Q]$. For me, the multivariate polynomials $a_0,dots,a_k$ are not symmetric but they are certainly not random (i.e. there are non-trivial permutations of the $n$-indeterminates that preserve the $a_i$'s). I do not know if any of these information helps.



Even a specific example that could give me a start on how to solve this kind of problem is very much appreciated.




algebraic-number-theory extension-field algebraic-numbers




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  • Do you want this to work for any $a_i in mathbbQ[x_1, cdots x_n]$ or just for specific ones?
    – Stefan4024
    2 days ago










  • I have specific ones in mind. For any $a_i$ will lead to $K$ being the splitting field of $f$.
    – quantum
    2 days ago














up vote
0
down vote

favorite












$newcommandQmathbb Q$Suppose I have an irreducible polynomial $finQ[x]$ and suppose I know the roots say $r_1,dots,r_nin bar Q$. I want to know if there is an easy way to "compute" the smallest number field $K$ such that we have
$$a_0(r_1,dots, r_n) + a_1(r_1,dots, r_n)x + dots + a_k(r_1,dots, r_n)x^k in K[x]$$
where $a_0,dots, a_k in Q[x_1,dots, x_n]$. I am actually only interested in knowing $[K:Q]$.



I suppose this is "strongly" dependent on how $a_0,dots, a_k$ are constructed. In my particular case, I want to prove that $[K:Q]$ is actually always even. I guess I can go into details of the $a_i's$ but I am not sure if this will help in solving $[K:Q]$. For me, the multivariate polynomials $a_0,dots,a_k$ are not symmetric but they are certainly not random (i.e. there are non-trivial permutations of the $n$-indeterminates that preserve the $a_i$'s). I do not know if any of these information helps.



Even a specific example that could give me a start on how to solve this kind of problem is very much appreciated.




algebraic-number-theory extension-field algebraic-numbers




share|cite|improve this question





















  • Do you want this to work for any $a_i in mathbbQ[x_1, cdots x_n]$ or just for specific ones?
    – Stefan4024
    2 days ago










  • I have specific ones in mind. For any $a_i$ will lead to $K$ being the splitting field of $f$.
    – quantum
    2 days ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite











$newcommandQmathbb Q$Suppose I have an irreducible polynomial $finQ[x]$ and suppose I know the roots say $r_1,dots,r_nin bar Q$. I want to know if there is an easy way to "compute" the smallest number field $K$ such that we have
$$a_0(r_1,dots, r_n) + a_1(r_1,dots, r_n)x + dots + a_k(r_1,dots, r_n)x^k in K[x]$$
where $a_0,dots, a_k in Q[x_1,dots, x_n]$. I am actually only interested in knowing $[K:Q]$.



I suppose this is "strongly" dependent on how $a_0,dots, a_k$ are constructed. In my particular case, I want to prove that $[K:Q]$ is actually always even. I guess I can go into details of the $a_i's$ but I am not sure if this will help in solving $[K:Q]$. For me, the multivariate polynomials $a_0,dots,a_k$ are not symmetric but they are certainly not random (i.e. there are non-trivial permutations of the $n$-indeterminates that preserve the $a_i$'s). I do not know if any of these information helps.



Even a specific example that could give me a start on how to solve this kind of problem is very much appreciated.




algebraic-number-theory extension-field algebraic-numbers




share|cite|improve this question













$newcommandQmathbb Q$Suppose I have an irreducible polynomial $finQ[x]$ and suppose I know the roots say $r_1,dots,r_nin bar Q$. I want to know if there is an easy way to "compute" the smallest number field $K$ such that we have
$$a_0(r_1,dots, r_n) + a_1(r_1,dots, r_n)x + dots + a_k(r_1,dots, r_n)x^k in K[x]$$
where $a_0,dots, a_k in Q[x_1,dots, x_n]$. I am actually only interested in knowing $[K:Q]$.



I suppose this is "strongly" dependent on how $a_0,dots, a_k$ are constructed. In my particular case, I want to prove that $[K:Q]$ is actually always even. I guess I can go into details of the $a_i's$ but I am not sure if this will help in solving $[K:Q]$. For me, the multivariate polynomials $a_0,dots,a_k$ are not symmetric but they are certainly not random (i.e. there are non-trivial permutations of the $n$-indeterminates that preserve the $a_i$'s). I do not know if any of these information helps.



Even a specific example that could give me a start on how to solve this kind of problem is very much appreciated.





algebraic-number-theory extension-field algebraic-numbers





share|cite|improve this question












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  • Do you want this to work for any $a_i in mathbbQ[x_1, cdots x_n]$ or just for specific ones?
    – Stefan4024
    2 days ago










  • I have specific ones in mind. For any $a_i$ will lead to $K$ being the splitting field of $f$.
    – quantum
    2 days ago
















  • Do you want this to work for any $a_i in mathbbQ[x_1, cdots x_n]$ or just for specific ones?
    – Stefan4024
    2 days ago










  • I have specific ones in mind. For any $a_i$ will lead to $K$ being the splitting field of $f$.
    – quantum
    2 days ago















Do you want this to work for any $a_i in mathbbQ[x_1, cdots x_n]$ or just for specific ones?
– Stefan4024
2 days ago




Do you want this to work for any $a_i in mathbbQ[x_1, cdots x_n]$ or just for specific ones?
– Stefan4024
2 days ago












I have specific ones in mind. For any $a_i$ will lead to $K$ being the splitting field of $f$.
– quantum
2 days ago




I have specific ones in mind. For any $a_i$ will lead to $K$ being the splitting field of $f$.
– quantum
2 days ago















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