# About irreducible polynomial over field & characteristic or minimal polynomial of matrix

Let $$F$$ be a field and $$K$$ be a finite extension of $$F$$, and let $$\alpha\in K$$.

Consider a linear map $$T:K\to K$$ is defined by $$T(\beta)=\alpha\beta$$ for all $$\beta\in K$$, where $$K$$ viewed as a finite dimensional vector space over $$F$$.

Then, using the fact that $$\alpha$$ is an eigenvalue of the representation matrix $$A$$ of $$T$$, we obtain the characteristic polynomial $$f(x)=\det(A-xI)$$ of $$A$$, where $$I$$ is the identity matrix with suitable size.

At this point, i have a question.

Q1) Is it true that $$f(x)=\text{irr}(\alpha,F)$$ or $$p(x)=\text{irr}(\alpha,F)$$, where $$p(x)$$ is the minimal polynomial of $$A$$?

Q2) What is the relationship between $$\text{irr}(\alpha,F)$$ and the minimal polynomial of $$A$$?

I would be very grateful if you could give me some textbooks or brief explanations.

(ps. my algebra level is quite elementary)

The minimal polynomial of $$T$$, hence also of its matrix $$A$$, coincides with the minimal polynomial of $$\alpha$$.
The key is, if $$f(x)=a_0+a_1x+\dots+a_nx^n\ \in F[x]$$, then we have $$f(T)(b)=(a_0\mathrm{id}+a_1T+\dots+a_nT^n)(b)=a_0b+a_1T(b)+a_nT^n(b)=a_0b+a_1\alpha b+\dots +a_n\alpha^nb=f(\alpha)b$$ So, if $$\alpha$$ is a root of $$f$$, then $$f(T)=0$$, and if $$f(T)=0$$, then in particular $$f(T)(1)=0$$ which yields $$f(\alpha) =0$$.