Yesterday I had a discussion with a friend about computing minimal polynomials. For example, say you are given algebraic numbers $\alpha$ and $\beta$ as specified by minimal polynomials which have degrees $n$ and $m$. How do you compute the minimal polynomial of $\alpha+\beta$, for example?

The method I was taught, which seems to be fairly standard (e.g., see Dummit and Foote Chapter 13.2) is the following:

Multiply $\alpha+\beta$ by $\alpha^i\beta^j$ for each $i=0{,}~1{,}~\dotsc{,}~n-1$ and $j=0{,}~1{,}~\dotsc{,}~m-1$, and use the minimal polynomials of $\alpha$ and $\beta$ to reduce the resulting expressions to linear combinations of $\alpha^i\beta^j$ (again with $0\leq i<n$ and $0\leq j<m$). In other words, what you are doing is computing a matrix $M$ which satisfies

\[ (\alpha+\beta)\left[\alpha^i\beta^j\right]_{i,j} = M\left[\alpha^i\beta^j\right]_{i,j} \]

where $[\alpha^i\beta^j]_{i,j}$ is a column vector containing the $\alpha^i\beta^j$.

From this we see that $\alpha+\beta$ is an eigenvalue of $M$ and therefore is a root of the characteristic polynomial $p_M$ of $M$. If $p_M$ is irreducible then it will be the minimal polynomial of $\alpha+\beta$, but in general the minimal polynomial will divide $p_M$, and so it will be necessary to factor $p_M$.

However, once $p_M$ has been factored, how does one tell which factor is the required minimal polynomial? The obvious answer is to evaluate each factor at $\alpha+\beta$ and see which one gives zero. “You could do that numerically”, my friend says, and I respond with “…or symbolically”. But then he asks if I will always be able to determine if the symbolic expression is zero or not. Well, I hadn’t thought of that, and I admitted I wasn’t sure, but claimed “anything you can do numerically I can do symbolically!”

I spent several hours yesterday trying to solve that problem, but eventually had to go to bed. This morning, after looking in Cohen I found the following passage:

…it does not make sense to ask which of the irreducible factors $\alpha+\beta$ is a root of, if we do not specify embeddings in $\mathbb{C}$…

Wait, what? I got excited as I realized I had just uncovered a misconception of mine!

Note that if the conjugates of $\alpha$ are $\alpha_i$ then they are all “symbolically identical”: $\mathbb{Q}(\alpha_i)$ is isomorphic for each conjugate. From that, I had assumed that the values of $\alpha_i+\beta_j$ would also be symbolically identical for all conjugates of $\alpha$ and $\beta$. Not true! As a trivial counterexample, if $\alpha$ is a root of $x^3-2$ and $\beta$ is a root of $x^3+2$ then two possible values for $\alpha+\beta$ are $0$ and $\sqrt[3]{2}\sqrt{3}i$, and these have very different algebraic behaviour.

So the whole problem of computing the minimal polynomial of $\alpha+\beta$ was not well defined unless you specify *which* roots $\alpha$, $\beta$ you are talking about—for example, by specifying them numerically.