Abstract Algebra II · Algebraic field extensions and minimal polynomials
A cubic extension has no quadratic intermediate field
Problem
Let $\alpha=\sqrt[3]{2}$ and $K=\mathbb Q(\alpha)$. Prove that the only intermediate fields $\mathbb Q\subseteq L\subseteq K$ are $\mathbb Q$ and $K$. In particular, $K$ contains no quadratic extension of $\mathbb Q$.
Hint
The polynomial $x^3-2$ is Eisenstein at $2$, so $[K:\mathbb Q]=3$.
Check your work
Work the problem yourself first. Then open it in Training to check your answer and read the full worked solution.
The answer check and full solution for this problem come with ProofAnvil Practice membership ($19 USD monthly). See membership. Or start with the free Abstract Algebra II sample problem: Try the free sample problem.
More Abstract Algebra II practice problems
- Gaussian integers form a subringRings, ideals, quotient rings, and homomorphisms
- Kernel of evaluation at a pointRings, ideals, quotient rings, and homomorphisms
- Divisibility is transitiveDomains, divisibility, and factorization
- Lcm from prime exponentsDomains, divisibility, and factorization
- Content and primitive partPolynomial rings and irreducibility
- Reciprocal polynomials preserve irreducibilityPolynomial rings and irreducibility
- Kernels and images are submodulesModules and finitely generated modules over a PID
- When finitely generated torsion means finiteModules and finitely generated modules over a PID
- A quartic whose splitting field is only quadraticFinite Galois theory