Natural irrationalities: Let $L$ be the splitting field of a polynomial over $K$. If $L$ contains an element $x$ that lies in some radical extension $R:K$, then there is a radical extension $R':K$ with $x \in R' \subset L$
@LeakyNun well, Cardano knew how to solve the cubics using roots of unity, but I don't think a proof that it is impossible to do it without roots of unity was known prior to Galois theory (I may be wrong)
now, if the correlation matrix is singular, what does that mean...
"What data produce singular correlation matrix of variables?
What multivariate data must look like in order its correlation or covariance matrix be singular matrix described above? It is when there is linear interdependances among the variables. If some variable is an exact linear combination of the other variables, with constant term allowed, the correlation and covariance matrces of the variables will be singular. "
@LeakyNun that's just not true (casus irreducibilis gives a counterexample). You need $L=\Bbb C(x_1, \dots, x_n)$ and $L= \Bbb C(t_1, \dots, t_n)$ where the $t_i$ are symmetric polynomials in the $x_i$. That's what I meant. Abel proved only theorems on general polynomials and no theorems about specific extensions of $\Bbb Q$
it'll be something like: "given a maximally entangled quantum state and three non-commuting observables, the three random variables corresponding to these observables will be linearly dependent."
