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Let $z$ be a non zero algebraic number.
We define $ Q[z]$ as the set of the values of all rational polynomial at $z$
$Q[z]$ = {$ P(z) / P\in Q[X]$}
prove that $Q[z]$ is a field (under ordinary addition an multiplication of course)
Proving that $Q[z]$ is an additive subgroup and that it is stable under multiplication is pretty straightforward, the difficulty lies in proving that the inverse of every non zero element of $Q[z]$ is also a value of a polynomials, I've tried numerous methods but none seemed to work, so is this property even true? I've already proved that $Q[z]$ $=$ {$P(z) / deg(P)< deg(Pz)$} where Pz is the minimal polynomial with rational coefficients such as that Pz(z)=0

AymaneMaaitat
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  • I'm sure this must be a duplicate. But briefly, the usual proof goes: if $\alpha \ne 0$ then multiplication by $\alpha$ is injective (by $P_z$ being irreducible) on a vector space of finite dimension; therefore, it's also surjective. – Daniel Schepler Feb 19 '20 at 18:43
  • Could you clarify your comment please, I'm finding trouble making sense out of it. – AymaneMaaitat Feb 19 '20 at 19:09

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