Determining the minimal polynomial using the characteristic polynomial of a matrix

Question

Consider a field $K$ and an extension $K(a)$. I have seen it proven that if we define $$ f_a = a\cdot \text{Id}_{K(a)}: K(a) \rightarrow K(a) $$ as the linear function defined by multiplication by $a$ and then define $p_a (t) = \text{det}(f_a -t \cdot \text{Id}_{K(a)})$ (the characteristic polynomial of the linear function), then the minimal polynomial of $a$ over $K$ is $(-1)^np_a(t)$ when $n$ is the degree of the extension. However, I am trying to apply this in a special case. Let $K=\mathbb{Q}$ and $(\sqrt[4]{2})^2 + (\sqrt[4]{2}) + 1$. I have proven that $K(a)$ is an extension of degree 4 and now I need to find the minimum polynomial. However when I apply this lemma I have the linear function: $$\begin{pmatrix} a & 0& 0 & 0\\ 0 & a & 0 & 0\\ 0& 0 & a & 0\\ 0&0 &0 & a \end{pmatrix}$$ and therefore my minimal polynomial comes out as $(t-a)^4$ but when I expand this I find that it is not in $\mathbb{Q}[t]$. I feel like I am misunderstanding the lemma. Any hints or solutions welcome :)

Answer 1

When you are viewing $\Bbb Q[a]$ as a $4$-dimensional $\Bbb Q$ vector space, then $a = (\sqrt[4]2)^2 + \sqrt[4]2 + 1$ is not a scalar since only elements of $\Bbb Q$ are. Therefore, $a$ cannot be used in matrix entries, and the matrix you write makes no sense in this setting. Also, multiplication by $a$ is not a scalar multiplication, and its matrix depends on the $\Bbb Q$-basis you are using.

Although $[1,a,a^2,a^3]$ is a basis of $\Bbb Q[a]$, expressing multiplication by $a$ on this basis (as is done in the answer by azif00) is not very helpful in finding the minimal polynomial, since finding the final column of that matrix is equivalent to finding the minimal polynomial of $a$. It is more instructive to use the basis $[1,\sqrt[4]2,(\sqrt[4]2)^2,(\sqrt[4]2)^3]$. The matrix then is $$ A=\operatorname{Mat}_{[1,\sqrt[4]2,(\sqrt[4]2)^2,(\sqrt[4]2)^3]}(x\mapsto ax) \pmatrix{1&0&2&2\cr1&1&0&2\cr1&1&1&0\cr0&1&1&1} $$ Now since you know that the minimal polynomial of $a$ has degree $4$ (since $a$ generates all of $\Bbb Q[\sqrt[4]2]$; all that checking this requires is that $a$ does not generate a subfield of degree$~2$), you know that the minimal polynomial of $A$ coincides with the characteristic polynomial$~\chi_A$, which you can compute with a bit of effort. If you don't like computing polynomial determinants, you can also compute $[1,a,a^2,a^3,a^4]$ in coordinates using this matrix (each column vector is obtained by multiplying the previous one by the matrix) giving the columns of the following matrix $$ M=\pmatrix{1&1&3&13&43\cr 0&1&2&9&36 \cr 0&1&3&8&30 \cr 0&0&2&7&24}, $$ and compute $\ker(M)$ to find relations between the columns. That kernel is generated by the vector $(7,-4,2,-4,1)\in\Bbb Q^5$, giving as minimal polynomial $X^4-4X^3+2X^2-4X+7$, which fortunately coincides with$~\chi_A$. This polynomial is irreducible in $\Bbb Q[X]$, but when lifted to $\Bbb Q[\sqrt[4]2][X]$ it has $a$ as a root; it is just a simple root however, and the polynomial does not equal $(X-a)^4$ as you suggested.

Answer 2

Indeed, if $a = (\sqrt[4]{2})^2 + (\sqrt[4]{2}) + 1$ and you write $a^4$ as $c_0+c_1a + c_2 a^2 + c_3a^3$ with $c_0,c_1,c_2,c_3 \in \mathbb Q$, then the matrix representation of $f_a$ with respect to the ordered basis $(1,a,a^2,a^3)$ (for $\mathbb Q(a)$ as a $\mathbb Q$-vector space) is given by $$\begin{pmatrix} 0 & 0 & 0 & c_0 \\ 1 & 0 & 0 & c_1 \\ 0 & 1 & 0 & c_2 \\ 0 & 0 & 1 & c_3 \end{pmatrix}$$ (since $f_a(1) = 0 + 1a + 0a^2+0a^3$, $f_a(a) = 0+0a+1a^2+0a^3$, $f_a(a^2)=0+0a+0a^2+1a^3$ and $f_a(a^3) = c_0+c_1a + c_2 a^2 + c_3a^3$). So, you still need to find $c_0,c_1,c_2$ and $c_3$.