Exercise 4.3.5 (Minimal polynomial of homomorphism)

Question

Let $M:K$ and $L:K$ be field extensions, and let $\varphi: M \rightarrow L$ be a homomorphism over $K$. Show that if $\alpha \in M$ has
minimal polynomial $m$ over $K$ then $\varphi(\alpha) \in L$ also has minimal polynomial $m$ over $K$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Here we consider $\iota : K 	o L$ and $\iota' : K 	o M$ as the canonical injections $a \mapsto a$, so that $K \subset M, K \subset L$, and $\varphi \circ \iota =\iota'$ means $\varphi(a) = a$ for all $a \in K$.
\begin{center}
\begin{tikzpicture}
    
ode (M) at (1,5) {$M$};
    
ode (L) at (5,5) {$L$};
     
ode (K) at (3,3) {$K$};
    \draw[->] (M) edge node[above]{$\varphi$}(L) ;
    \draw[->] (K) edge node[left]{$\iota$} (M) ;
    \draw[->] (K) edge node[right]{$\iota'$}(L) ;
\end{tikzpicture}
\end{center}

Let $m(x) = \sum_{i=0}^d a_i x^i \in K[x]$ be the minimal polynomial of $\alpha \in M$. Then $m(\alpha) = \sum_{i=0}^d a_i \alpha^i =  0$, and $m$ is irreducible over $K$.

Since $a_i \in K, i = 1,\ldots,d$,  we have $\varphi(a_i) = a_i$. Therefore
$$\varphi(m(\alpha)) = \varphi\left(\sum_{i=0}^d a_i \alpha^i ight) = \sum_{i=0}^d \varphi(a_i) \varphi(\alpha)^i  = \sum_{i=0}^d a_i \varphi(\alpha)^i = m(\varphi(\alpha)).$$
Since $m(\alpha) = 0$, $m(\varphi(\alpha)) = \varphi(m(\alpha))=\varphi(0)  = 0$, so $\varphi(\alpha)$ is a root of $m$. Moreover $m$ is irreducible, thus $m$ is the minimal polynomial of $\varphi(\alpha)$ (Lemma 4.2.10 (iv)).
\end{proof}

Exercise 4.3.5 (Minimal polynomial of homomorphism)

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Exercise 4.3.5 (Minimal polynomial of homomorphism)

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