- #1
coquelicot
- 299
- 67
Hello,
Below are two results with their proof. Of course, there may be several ways to prove these results, but I just need some checking. Can someone check carefully if the math is OK ? (but very carefully, because if there is a failure, I will be murdered :-) ) ? thx.
Claim 1: Let ##L/K## be a field extension. Then ##P \in L[X]## is algebraic over ##K[X]##, if and only if the coefficients of ##P## are algebraic over K. Furthermore, the conjugates of ##P## over ##K[X]## obtain by replacing each coefficients ##P_i## of ##P## by a certain conjugate of ##P_i## over ##K##.
Proof: Assume that ##P = P_kX^k + \cdots + P_1 X + P_0##.
If ##P## is algebraic over ##K[X]##, there exists an algebraic relation of the form ##a_nP^n + a_{n-1}P^{n-1} + \cdots+ a_1P + a_0 =0## where ##a_i\in K[X]## and ##a_i## are not all identically equal to 0. Whence it is easy to see that the the free coefficient ##P_0## of ##P## fulfils an algebraic relation of the form ##\alpha_n P_0^n + \cdots + \alpha_0## where the ##\alpha_i## are the free coefficients of ##a_i##. So ##P_0## is algebraic over ##K## and a fortiori over ##K[X]##.
It follows immediately that the polynomial ##(P - P_0)/X = P_kX^{k-1} + \cdots+ P_1## is algebraic over ##K[X]##. The same argument as above then shows that ##P_1## is algebraic over ##K##, and so on, ##P_2, \cdots, P_k## are algebraic over ##K##.
Conversely, if the coefficients ##P_i## of ##P## are algebraic over ##K##. Let
##h(X,Y) = (Y-P)(Y-P')\cdots(Y-P^{(m)})##, where ##P^{(i)}## are the polynomials obtained from ##P## by replacing the coefficients of ##P## by their conjugates over ##K##, in all the possible ways. By the fundamental theorem of symmetric functions, ##h\in K[X, Y]## and ##h(X, P(X)) = 0##. This proves that ##P## is algebraic over ##K[X]##. Also, this makes clear the last assertion of Claim 1.
Claim 2: If ##\sigma_1(X1,\ldots, X_n), \ldots, \sigma_n(X_1, \ldots, X_n)## are the n elementary polynomials over an integral domain ##A##, and if P is a polynomial in n variables such that ##P(\sigma_1, \ldots, \sigma_n) = 0##, then ##P = 0##.
proof: Let ##P(S_1,\ldots, S_n)## be a polynomial of minimal degree in ##S_1## such that ##P(\sigma_1, \ldots, \sigma_n) = 0##.
Then substituting 0 for ##X_2,\ldots, X_n## in the relation above leads to ##P(X_1, 0, \ldots, 0) = 0##, since
##\sigma_1 = X_1+X_2+\cdots+X_n##,
##\sigma_2 = X_1X_2+X_1X3 + \cdots## etc.
Hence ##S_1## divides ##P##, or ##P = S_1 Q##, and there holds ##Q(\sigma_1, \ldots, \sigma_n) = 0##. By the minimality of the degree in ##S_1##, there must holds ##P = Q = 0##.
Below are two results with their proof. Of course, there may be several ways to prove these results, but I just need some checking. Can someone check carefully if the math is OK ? (but very carefully, because if there is a failure, I will be murdered :-) ) ? thx.
Claim 1: Let ##L/K## be a field extension. Then ##P \in L[X]## is algebraic over ##K[X]##, if and only if the coefficients of ##P## are algebraic over K. Furthermore, the conjugates of ##P## over ##K[X]## obtain by replacing each coefficients ##P_i## of ##P## by a certain conjugate of ##P_i## over ##K##.
Proof: Assume that ##P = P_kX^k + \cdots + P_1 X + P_0##.
If ##P## is algebraic over ##K[X]##, there exists an algebraic relation of the form ##a_nP^n + a_{n-1}P^{n-1} + \cdots+ a_1P + a_0 =0## where ##a_i\in K[X]## and ##a_i## are not all identically equal to 0. Whence it is easy to see that the the free coefficient ##P_0## of ##P## fulfils an algebraic relation of the form ##\alpha_n P_0^n + \cdots + \alpha_0## where the ##\alpha_i## are the free coefficients of ##a_i##. So ##P_0## is algebraic over ##K## and a fortiori over ##K[X]##.
It follows immediately that the polynomial ##(P - P_0)/X = P_kX^{k-1} + \cdots+ P_1## is algebraic over ##K[X]##. The same argument as above then shows that ##P_1## is algebraic over ##K##, and so on, ##P_2, \cdots, P_k## are algebraic over ##K##.
Conversely, if the coefficients ##P_i## of ##P## are algebraic over ##K##. Let
##h(X,Y) = (Y-P)(Y-P')\cdots(Y-P^{(m)})##, where ##P^{(i)}## are the polynomials obtained from ##P## by replacing the coefficients of ##P## by their conjugates over ##K##, in all the possible ways. By the fundamental theorem of symmetric functions, ##h\in K[X, Y]## and ##h(X, P(X)) = 0##. This proves that ##P## is algebraic over ##K[X]##. Also, this makes clear the last assertion of Claim 1.
Claim 2: If ##\sigma_1(X1,\ldots, X_n), \ldots, \sigma_n(X_1, \ldots, X_n)## are the n elementary polynomials over an integral domain ##A##, and if P is a polynomial in n variables such that ##P(\sigma_1, \ldots, \sigma_n) = 0##, then ##P = 0##.
proof: Let ##P(S_1,\ldots, S_n)## be a polynomial of minimal degree in ##S_1## such that ##P(\sigma_1, \ldots, \sigma_n) = 0##.
Then substituting 0 for ##X_2,\ldots, X_n## in the relation above leads to ##P(X_1, 0, \ldots, 0) = 0##, since
##\sigma_1 = X_1+X_2+\cdots+X_n##,
##\sigma_2 = X_1X_2+X_1X3 + \cdots## etc.
Hence ##S_1## divides ##P##, or ##P = S_1 Q##, and there holds ##Q(\sigma_1, \ldots, \sigma_n) = 0##. By the minimality of the degree in ##S_1##, there must holds ##P = Q = 0##.
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