6.7 Algebraic
[0C7]A number \(𝛼∈ℝ\) is said algebraic if there exists a polynomial \(p(x)=a_ 0+a_ 1x+\cdots + a_ n x^ n\) with rational coefficients such that \(p(𝛼)=0\). Otherwise \(𝛼\) is said transcendental.
[0C8]Given a commutative ring \(A\), the set of polynomials \(p(x)=a_ 0+a_ 1x+\cdots + a_ n x^ n\) with coefficients \(a_ i∈ A\) is usually denoted by \(A[x]\); this set, endowed with the usual operations of sum and product of polynomials, is a commutative ring.
- E210
[0C9]Given \(p(x)=a_ 0+a_ 1x+\cdots + a_ n x^ n\), \(p∈ℚ[z]\) such that \(p(𝛼)=0\), build a polynomial \(q∈ℤ[z]\) such that \(q(𝛼)=0\).
So the definition of algebraic can be given equivalently with polynomials with integer coefficients.
- E210
[0CB]Given \(𝛼≠ 0\) and \(p(x)=a_ 0+a_ 1x+\cdots + a_ n x^ n\), \(p∈ℚ[z]\) such that \(p(𝛼)=0\), build a polynomial \(q∈ℚ[z]\) such that \(q(1/𝛼)=0\).
So if \(𝛼≠ 0\) is algebraic then \(1/𝛼\) is algebraic.
Hidden solution: [UNACCESSIBLE UUID ’0BR’]
- E210
[0CC]Given \(p(x)=a_ 0+a_ 1x+\cdots + a_ n x^ n\), \(p∈ℚ[z]\) such that \(p(𝛼)=0\), given \(b∈ℚ\) build a \(q∈ℚ[z]\) such that \(q(b𝛼)=0\).
So if \(𝛼\) is algebraic then \(b𝛼\) is algebraic.
- E210
[0CD]Given \(p(x)=a_ 0+a_ 1x+\cdots + a_ n x^ n\), \(p∈ℚ[z]\) such that \(p(𝛼)=0\), given \(b∈ℚ\) build a \(q∈ℚ[z]\) such that \(q(b+𝛼)=0\).
So if \(𝛼\) is algebraic then \(b+𝛼\) is algebraic.
- E210
[0CF]Difficulty:*.More generally, given \(p(x)=a_ 0+a_ 1x+\cdots + a_ n x^ n\), \(p∈ℚ[z]\) \(q(x)=b_ 0+b_ 1x+\cdots + b_ m x^ m\), \(q∈ℚ[z]\), and given \(𝛼,𝛽\) such that \(p(𝛼)=0=q(𝛽)\), construct a polynomial \(r∈ ℚ[z]\) such that \(r(𝛼+𝛽)=0\).
(Hint: use the theory of the resultant [ 44 ] ).
So if \(𝛼,𝛽\) are algebraic then \(𝛼+𝛽\) is algebraic.
Hidden solution: [UNACCESSIBLE UUID ’0CG’]
- E210
[0CH]Show that if \(𝛼\) is algebraic then \(𝛼^ 2\) is algebraic. Hidden solution: [UNACCESSIBLE UUID ’0CJ’]
- E210
[0CK]If \(𝛼,𝛽\) are algebraic, prove that \(𝛼𝛽\) is algebraic. Hidden solution: [UNACCESSIBLE UUID ’0CM’]