Minimal polynomial of $\varphi:g+(f)\mapsto xg+(f)$ is $f$.

Question

Let $K$ be a field, $a$ an element in a field extension such that $a$ is algebraic over $K$. Denote by $f\in K[x]$ the minimal polynomial of $a$ over $K$. Consider the $K$-vector space $K[x]/(f)$ and the $K$-linear map $$ \varphi: V\to V,\quad g+(f)\mapsto xg+(f). $$ Show that the minimal polynomial of $\varphi$ is given by $f$.

My thoughts: I think we need to find a matrix representation of $\varphi$, then find its minimal polynomial, and show it equals $f$. To do this, we will need a basis for $V$. I know from basic field theory that we can do the following (let $n=\deg(f)$):

One way to make a basis: Note that $V\cong K(a)$, so a basis for $V$ is given by $1,a,\ldots,a^{n-1}$. I don't know what to do with this, since the map $\varphi$ is defined in terms of the cosets. Maybe I can transform this basis into some usable form? Any hints?

Answer 1

Under the isomorphism $V \cong K(a)$, $x+(f)$ corresponds to $a$ and so a much better way to look at this question is to think of $\varphi: K(a) \to K(a)$ as multiplication by $a$ map.

Note that $f(\varphi)(x)=f(a)(x)=0$ for any $x \in K(a)$ using the fact that $f$ is the minimal polynomial for $a$. Therefore $f(\varphi)=0$ and because $f$ is irreducible, this is the minimal polynomial for $\varphi$.

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The crucial insight is that while $\varphi$ is multiplication by $a$, the map $f(\varphi)$ is multiplication by $f(a)=0$, so the zero map.

Answer 2

The minimal polynomial $g$ of $\varphi$ is uniquely characterized by being a monic polynomial such that $g(\phi) = 0$ and if any other polynomial $h$ satisfies $h(\varphi) = 0$ then $g \mid h$. Now, given a polynomial $h$, what is the endomorphism $h(\varphi)$? If $h = \sum h_i x^i$ then $h(\varphi) = \sum h_i \varphi^i$. As $\varphi$ is just multiplication by $x$, this shows that $h(\varphi)$ is multiplication by $h$. As $f$ maps to zero in $K[x]/(f)$, this tells us that $f(\varphi) = 0$. Hence, $g \mid f$. Furthermore, as $f$ is the minimal polynomial of $a$, it is monic and irreducible. Hence, as $g$ is also monic, $g = 1$ or $g = f$. $g = 1$ vanishes nowhere but $g$ vanishes on $\phi$. We conclude that $g=f$.