### Algebraic field extensions: a continuation

We shall talk about the algebraic extension $F(\alpha,\beta)$. We shall assume that both $\alpha$ and $\beta$ are algebraic over the field $F$.

Assume $\deg(F(\alpha),F)=m$. Hence, the basis for the vector space $F(\alpha)$ over $F$ is $1,\alpha,\alpha^2,\dots,\alpha^{m-1}$. Now say $\deg(F(\alpha,\beta),F)=n$. Then the basis of $F(\alpha,\beta)$ over $F(\alpha)$ is $1,\beta,\beta^2,\dots,\beta^{n-1}$.

It is known that the basis of $F(\alpha,\beta)$ over $F$ is $\begin{pmatrix} 1 & \beta&\beta^2 & \ldots & \beta^{n-1} \\ \alpha & \alpha\beta & \alpha\beta^2 & \ldots & \alpha\beta^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \alpha^{m-1} & \alpha^{m-1}\beta & \alpha^{m-1}\beta^2 & \ldots & \alpha^{m-1}\beta^{n-1}\end{pmatrix}$ (arranged as a matrix).

None of these can be dependant on each other (by definition). Also, they are $mn$ in number.

Now let us first construct $F(\beta)$ from $F$. Say $\deg(F(\beta): F)=s$. Then the basis of $F(\beta)$ over $F$ is $1,\beta,\beta^2,\dots,\beta^{s-1}$. Now let us assume $\deg(F(\alpha,\beta))$ over $F$ is $r$. The basis matrix of $F(\alpha,\beta)$ over $F$ will hence be $F(\alpha,\beta)$ over $F$ is $\begin{pmatrix} 1 & \beta&\beta^2 & \ldots & \beta^{s-1} \\ \alpha & \alpha\beta & \alpha\beta^2 & \ldots & \alpha\beta^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \alpha^{r-1} & \alpha^{r-1}\beta & \alpha^{r-1}\beta^2 & \ldots & \alpha^{r-1}\beta^{s-1}\end{pmatrix}$.

It is possible that not a single term in the two matrices are the same. However, $mn=rs$.

Remember.
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