Algebraic field extensions: a continuation

by ayushkhaitan3437

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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