How to find the minimal polynomial over a field extensions $L/K$

Question

I have the following question: Let $L/K$ be a field extension. For $a \in L$ how do I find the minimal polynomial?

Among all monic polynomials in $K[X]$, which have $a$ as a root, the polynom with the lowest degree is the minimal polynomial.

Answer 1

There is one general but impractical method. Let $K$ be the algebraic closure of $F$. Let $\alpha,\beta\in K$. Let $p_\alpha$ be the minimal polynomial of $\alpha$ over $F$. Note that $p_\alpha$ is the characteristic polynomial of its companion matrix, so $\alpha$ is an eigenvalue of it.

On the other hand, given any eigenvalue $\lambda\in K$ of a square $F$-matrix $A$, since $\lambda$ is a root of the characteristic polynomial of $A$, the minimal polynomial of $\lambda$ over $F$ divides the characteristic polynomial of $A$.

Suppose $\lambda,\mu$ are eigenvalues of $A,B$ respectively. We want to find respective matrices which have eigenvalue $k\lambda$ (where $k\in F$), $\lambda+\mu$, $\lambda \mu$, $\lambda^{-1}$. For $k\lambda$, we use $kA$. For $\lambda^{-1}$, we use $A^{-1}$. For $\lambda\mu$, we use $A\otimes B$. For $\lambda+\mu$, we use $A\otimes I+I\otimes B$. The $\otimes$ is the Kronecker product. I'll leave it to you to verify these actually give the correct eigenvalues.

Now, we can devise a method to find the minimal polynomial of $\alpha+\beta$, $\alpha\beta$, and so on.

1. Find the companion matrix $A$ of the minimal polynomial (or any multiple of it) of $\alpha$, and that of $\beta$.
2. Find a matrix $M$ according to the rules in the previous paragraph, so that your desired expression is the eigenvalue of $M$.
3. Find the characteristic polynomial of $M$ and factorize it to find the irreducible factor which vanishes your desired expression, and that's the minimal polynomial desired.

This is clearly impractical for elements with even slightly higher degree, but it's quite fun to see it work out. Let's try this out for $i+\sqrt{2}$. The minimal polynomial of $i$ is $x^2+1$, and the companion matrix is $$ A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}. $$ The minimal polynomial of $\sqrt{2}$ is $x^2-2$ and the companion matrix is $$ B = \begin{bmatrix} 0 & 2 \\ 1 & 0 \end{bmatrix}. $$ We compute $$ M:=A\otimes I + I \otimes B = \begin{bmatrix} 0 & 2 & -1 & 0 \\ 1 & 0 & 0 & -1 \\ 1 & 0 & 0 & 2 \\ 0 & 1 & 1 & 0 \end{bmatrix}. $$ The characteristic polynomial of $M$ is $x^4-2x^2+9$, which is irreducible, so the minimal polynomial of $i+\sqrt{2}$ is $x^4-2x^2+9$.

In practice, you probably want to try to find some algebraic relation between $\alpha+\beta$ and their minimal polynomials, and try to work from there.