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.

  • $\begingroup$ Do you have a concrete example in mind? Or do you need a general approach? $\endgroup$
    – Arthur
    Commented Mar 25, 2023 at 15:26
  • $\begingroup$ I would be interested in a more general approach. Although I am trying to find the minimal polynomial of $i+\sqrt{2}$ over $\mathbb{C}/ \mathbb{Q}$ $\endgroup$
    – wanymose
    Commented Mar 25, 2023 at 15:39
  • $\begingroup$ @wanymose There is no good general method. As for your $x=\sqrt{2}+i$, start by trying to find a polynomial which has $x$ as a root. Hint: start by computing $x^2$. $\endgroup$
    – Mark
    Commented Mar 25, 2023 at 15:47
  • $\begingroup$ See this duplicate for $i+\sqrt{2}$. In general, it is often enough to compute powers and then see how the minimal polynomial would look like from it. Most homework minimal polynomials will not be difficult. Probably easier than this homework. $\endgroup$ Commented Mar 25, 2023 at 16:14

1 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.


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