Finite fields, irreducible polynomial and minimal polynomial theorem

[Karl Karlsson]

I thought i understood the theorem below:

i) If A is a matrix in $M_n(k)$ and the minimal polynomial of A is irreducible, then $K = \{p(A): p (x) \in k [x]\}$ is a finite field

Then this example came up:

The polynomial $q(x) = x^2 + 1$ is irreducible over the real numbers and the matrix $$J=
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}$$
has $q(x)$ as minimal polynomial. $$K=\{p(J):p(x)\in \mathbb{R}[x]\}=\{aI+bJ:a,b\in \mathbb{R}\}$$
is a finite field that is isomorphic to $\mathbb{C}$.

Why can't the finite field K above not be for example $\{aI+bJ^2+cJ^3:a,b,c\in \mathbb{R}\}$, since k[x] in the theorem i) is $\mathbb{R}[x]$ ?

 

[fresh_42]

I thought i understood the theorem below:

i) If A is a matrix in $M_n(k)$ and the minimal polynomial of A is irreducible, then $K = \{p(A): p (x) \in k [x]\}$ is a finite field

Could it be that there is plenty of information missing? The minimal polynomial of $I\in M_2(\mathbb{R})$ is $m_I(x)=x-1$ which is certainly irreducible. But $K=\{p(I): p(x) \in \mathbb{R}[x]\}=\{\operatorname{diag}(r)\,|\,r\in \mathbb{R}\}=\mathbb{R}$ which is definitely not finite.

 

[martinbn]

You probably mean finite extention, not finite field. Also since A is a zero of its minial polynomial, you only need the values of polynomials mod the minimal one. In your example all powers bigger than one can be exressed by smaller ones using the minimal polynomial.

 

[Infrared]

Since $J^2=-I$ and $J^3=-J$, we have $aI+bJ^2+cJ^3=(a-b)I-cJ,$ so you're describing the same set.

 

[Karl Karlsson]

Since $J^2=-I$ and $J^3=-J$, we have $aI+bJ^2+cJ^3=(a-b)I-cJ,$ so you're describing the same set.

Oh, right... I did not think that $J^n$ would be some multiple of either I or $J$ now I see that this is the case. But in the theorem:

i) If A is a matrix in $M_n(k)$ and the minimal polynomial of A is irreducible, then $K = \{p(A): p (x) \in k [x]\}$ is a finite field.

Does $p(x)$ in the set $\{p(A): p (x) \in k[x]\}$ mean every single combination of polynomials in $k[x]$ ? For example if k was $\mathbb{R}$ would that mean that the set $\{p(A): p (x) \in k[x]\}$ would be $\sum_{k=0}^n a_k\cdot A^k$ for every $n \in\mathbb{N}$ and every $a_k\in\mathbb{R}$ ?

 

[martinbn]

Oh, right... I did not think that $J^n$ would be some multiple of either I or $J$ now I see that this is the case. But in the theorem:

i) If A is a matrix in $M_n(k)$ and the minimal polynomial of A is irreducible, then $K = \{p(A): p (x) \in k [x]\}$ is a finite field.

Does $p(x)$ in the set $\{p(A): p (x) \in k[x]\}$ mean every single combination of polynomials in $k[x]$ ? For example if k was $\mathbb{R}$ would that mean that the set $\{p(A): p (x) \in k[x]\}$ would be $\sum_{k=0}^n a_k\cdot A^k$ for every $n \in\mathbb{N}$ and every $a_k\in\mathbb{R}$ ?

Finite Field extension, not a finite field. Yes, it means values of all polynomials, when evaluated at the given matrix A. But the same argument shows, that you need the combinations of monomials of degree up to the degree of the min polynimial of A.

 

[Karl Karlsson]

Finite Field extension, not a finite field. Yes, it means values of all polynomials, when evaluated at the given matrix A. But the same argument shows, that you need the combinations of monomials of degree up to the degree of the min polynimial of A.

Thanks for explaining!