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.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this communityI 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.
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.
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.