Proving that $\sqrt[3] {2} ,\sqrt[3] {4},1$ are linearly independent over rationals

Solution 1:

Let $\alpha=\sqrt[3] {2}$ and suppose $a+b\alpha+c\alpha^2=0$ with $a,b,c\in \mathbb Z$, which we may assume coprime.

Then $a\alpha+b\alpha^2+2c=0$ and $a\alpha^2+2b+2c\alpha=0$.

This means that the matrix below is singular $$ \pmatrix{ a & b & c \\ 2c & a & b \\ 2b & 2c & a} $$ Its determinant must be zero: $$ a^3-6 a b c+2 b^3+4 c^3=0 $$ This implies that $a$ is even: $a=2A$. So $$ 4A^3-6 A b c+ b^3+2 c^3=0 $$

This implies that $b$ is even: $b=2B$. So $$ 2A^3-6 A B c+ 4B^3+ c^3=0 $$ This implies that $c$ is even. This contradicts they being coprime.

Solution 2:

Suppose $1,\sqrt[3]2,\sqrt[3]4$ are linearly dependant. This means that there is a nonzero polynomial $P \in \Bbb Q[X]$ of degree at most $2$ such that $P(\sqrt[3]2)=0$.

However, we know that $Q = X^3-2$ also satisfies $Q(\sqrt[3]2) = 0$. By taking the greatest common divisor of $P$ and $Q$, we obtain a strict divisor $R$ of $Q$ (because the degree of $R$ is less than the degree of $Q$).

By Eisenstein's criterion, $Q$ is irreducible, which contradicts the existence of $R$.

For your second question :

If you know about the quadratic reciprocity law and Dirichlet's theorem about primes in arithmetic progression, you can show that the family of $\sqrt p$ are linearly independant :

If a relation existed between the square roots of a family of primes $p_0, \ldots, p_n$, then you can express $\sqrt{p_0}$ in terms of all the others. Such a formula means that $X^2 - p$ has a root in $\Bbb Q[T_1,\ldots,T_n]/(T_1^2-p_i)\ldots(T_n^2-p_n)$. If we have a prime $q$ such that (1) $q$ doesn't divide any denominator in the coefficients of the root, and (2) $p_i$ has a square root mod $q$ for $1 \le i \le n$ but not for $i=0$, you get a contradiction by looking at the root modulo $q$.
But such a prime exists because by the quadratic reciprocity law, (2) is equivalent to a modular condition, and then Dirichlet's theorem shows that there are infinitely many primes satisfying it, so there is one that satisfy (1).

In fact this shows even more : the family of $\sqrt n$ with $n \in \Bbb Z$ and $n$ squarefree, is linearly independant.