If a field $Q$ were to be extended to include roots of the quadratic polynomial $x^2$$-2=0$, the extended field $Q$($\sqrt2$) would include elements of the form $a$ + $b$$\sqrt 2$. However, extending the same field $Q$ to include roots of the cubic polynomial $x^3$$-x-1=0$ involves including elements of the form $a+by+c$$y^2$ where $a, b, c$ are elements of the original $Q$ and $y$ is a root of the cubic.
While it is somehow intuitive that extending the field to include roots of a cubic means including elements with a quadratic form, is there a general rule that says this must be the case? How can I prove that extending $Q$ by merely $a+by$ is insufficient in the latter case?
indeed, there is a general rule. If $\alpha$ is algebraic over $\mathbb{Q}$, then $\mathbb{Q}(\alpha)$ is a vector space of dimension $n$ where $n$ is degree of minimal polynomial of $\alpha$.
You can show that $1, \alpha, \alpha^{2}, \cdots, \alpha^{n-1}$ are linearly independent using the fact that minimal polynomials are irreducible. Further if we write $\mathbb{Q}(\alpha) \cong \mathbb{Q}[X]/(p(x))$, where $p(x)$ is the minimal polynomial of $\alpha$ , we observe that this is a spanning set as well, hence a basis.
All that holds true for arbitrary fields as well. I have skipped the proofs of my statements but they pretty much can be worked out without much pain.