Since $K=F(\lambda)$ and the extension is finite, the element $\lambda$ is algebraic. Its <minimal polynomial> is <irreducible polynomial>[irreducible], and evaluation gives an isomorphism $F[X]/(m_\lambda)\cong F[\lambda]=F(\lambda)=K$. If its <polynomial degree>[degree] is $d$, the powers $1,\lambda,\ldots,\lambda^{d-1}$ form a <basis>. Comparing with $[K:F]=n$ gives $\boxed{\deg m_\lambda=n}$.

Solution

Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / A / A2 / i / Solution

 Solution (source code)