Consider a polynomial expression \(H ( \alpha ) = A + B \alpha + C \alpha ^ { 2 } + \cdot \cdot \cdot + D \alpha ^ { N }\) with rational coefficients A, \(B , C , \ldots , D\) , where α is an algebraic number, in other words a root of some polynomial with rational coefficients. Prove that if \(H ( \alpha ) \neq 0\) , then the reciprocal \(1 / H ( \alpha )\) can be expressed as a polynomial in α with rational coefficients.
Solution
Let \(m(x)\in\QQ[x]\) be the minimal polynomial of \(\alpha\) over \(\QQ\).
1. The polynomials \(m\) and \(H\) are relatively prime in \(\QQ[x]\).
The minimal polynomial \(m\) is irreducible over \(\QQ\). Hence any common divisor of \(m\) and \(H\) is either a nonzero constant or an associate of \(m\). If \(m\) divided \(H\), then evaluating at \(\alpha\) would give \[ H(\alpha)=0, \] contrary to the hypothesis. Therefore \[ \gcd(m,H)=1. \]
2. There exist polynomials \(U,V\in\QQ[x]\) such that\[ U(x)m(x)+V(x)H(x)=1. \]
By step 1, \(m\) and \(H\) are relatively prime. Bézout’s identity in the Euclidean domain \(\QQ[x]\) therefore gives the stated polynomials \(U\) and \(V\).
3. The reciprocal of \(H(\alpha)\) is a polynomial in \(\alpha\) withrational coefficients, namely\[ \boxed{\frac{1}{H(\alpha)}=V(\alpha)}. \]
Evaluate the identity in step 2 at \(x=\alpha\). Since \[ m(\alpha)=0, \] one obtains \[ V(\alpha)H(\alpha)=1. \] The hypothesis \(H(\alpha)\neq0\) permits division by \(H(\alpha)\), giving the displayed formula. Since \(V\in\QQ[x]\), this has exactly the required form.
4. Q.E.D.
Step 3 is the required conclusion.