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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)

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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 $f(\alpha) =0$.

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