Suppose F and H are fields of size $\;q=p^{r}\;$containing$\;GF(p)\;$as subfield.$\;\alpha\;$is a primitive element of F and $\;\beta\;$ is a primitive element of H.$\;m(x)\;$is the minimal polynomial of $\;\alpha\;.\;$Non-zero Elements of both fields satisfy the equation$\;x^{q-1}=1\; .\;\;\;m(x)\;$is divisor of $\;x^{q-1}-1.\;$Hence there is an element in H (say)$\;\beta^{t}\;\;$which is a root $\;\;m(x)\;.\;$ I want to show that there exists a field isomorphism $\;\phi:F\to H \;$ which carries zero to zero and $\;\alpha\;$to $\;\beta^{t}\;$ Can you help to prove it? I have tried $\;\phi(\alpha^{j} )=\beta^{tj}\;$ but I could not show that $\;\phi\;$ preserves addition.( This argument is presented by Vera Pless in the book Introduction to Theory of Error correcting codes.)
Abstract algebra, Field extension
1 Answers
You exploit the two isomorphisms $$ \begin{align} f :\ &K[x]/(m) \to F\\ &u + (m) \mapsto u(\alpha) \end{align} $$ which holds because $m$ is the minimal polynomial of $\alpha$ over the field $K$ with $p$ elements, and $$ \begin{align} g :\ &K[x]/(m) \to H\\ &v + (m) \mapsto v(\beta^{t}) \end{align} $$ which holds because $m$ is the minimal polynomial of $\beta^{t}$ over $K$.
Note that $$ \begin{cases} f(x + (m)) = \alpha, & \text{or $f^{-1}(\alpha) = x + (m)$,}\\ g(x + (m)) = \beta^{t}.\\ \end{cases} $$
If you compose $g \circ f^{-1}$, you get an isomorphism $F \to H$ such that $g \circ f^{-1} (\alpha) = g(x + (m)) = \beta^{t}$.
Explicitly, if $n$ is the degree of $m$, $$ g \circ f^{-1} : a_{0} + a_{1} \alpha + \dots + a_{n-1} \alpha^{n-1} \mapsto a_{0} + a_{1} \beta^{t} + \dots + a_{n-1} (\beta^{t})^{n-1}, $$ for $a_{i} \in K$.