Let \[ F=\{a+b\sqrt[3]{2}+c\sqrt[3]{4}:a,b,c\in\mathbb Q\}. \] Prove that \(F\) is a field, that every element of \(F\) has a unique representation of this form, and find \[ (1-\sqrt[3]{2})^{-1} \] in \(F\).
Solution
Let \[ \alpha\coloneqq\sqrt[3]{2}. \]
1. The polynomial\[ m(x)\coloneqq x^3-2 \]is irreducible in \(\QQ[x]\).
Eisenstein’s criterion applies with the prime \(2\): every nonleading coefficient is divisible by \(2\), while the constant coefficient \(-2\) is not divisible by \(4\).
2. Evaluation at \(\alpha\) induces an isomorphism\[ \QQ[x]/(x^3-2) \cong F. \]
The evaluation homomorphism \[ \operatorname{ev}_\alpha:\QQ[x]\to\CC, \qquad p(x)\longmapsto p(\alpha), \] has kernel \((x^3-2)\) by step 1, since \(x^3-2\) is the minimal polynomial of \(\alpha\) over \(\QQ\). Every polynomial has a unique remainder \[ a+bx+cx^2 \] after division by \(x^3-2\). Hence the image of evaluation is precisely \[ \{a+b\alpha+c\alpha^2:a,b,c\in\QQ\}=F. \] The first isomorphism theorem gives the stated isomorphism.
3. The set \(F\) is a field.
By step 1, the ideal \((x^3-2)\) is maximal in the PID \(\QQ[x]\). Therefore \(\QQ[x]/(x^3-2)\) is a field, and step 2 identifies it with \(F\).
4. Every element of \(F\) has a unique expression\[ a+b\sqrt[3]{2}+c\sqrt[3]{4}, \qquad a,b,c\in\QQ. \]
Existence is the definition of \(F\). For uniqueness, suppose \[ a+b\alpha+c\alpha^2=0. \] Then the polynomial \(a+bx+cx^2\) lies in the kernel of \(\operatorname{ev}_\alpha\). By step 2 this kernel is \((x^3-2)\), but a nonzero polynomial of degree at most \(2\) cannot be divisible by the degree-\(3\) polynomial \(x^3-2\). Hence \(a=b=c=0\).
5. The requested inverse is\[ \boxed{ (1-\sqrt[3]{2})^{-1} = -1-\sqrt[3]{2}-\sqrt[3]{4} }. \]
Since \(\alpha^3=2\), \[ \begin{aligned} (1-\alpha)(-1-\alpha-\alpha^2) &= -(1-\alpha)(1+\alpha+\alpha^2)\\ &= -(1-\alpha^3)\\ &= 1. \end{aligned} \] The displayed inverse belongs to \(F\) by step 4.
6. Q.E.D.
Steps 3, 4, and 5 establish respectively the field property, uniqueness, and the requested inverse.