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## Tag: algebraic field extensions

### 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.

### Field extensions…..things that books won’t explicitly point out.

This is going to flesh out some real details.

First, something exceedingly important. When doing elementary algebra for the first time (say in grade 4), one often asks “what is $x$?” The answer to that is “$x$” is just a useful construct that is temporarily holding place, waiting for an actual number to pop in and satisfy the equation. $x$ is not an actual number. There is no number in the real number line called “$x$“. OK.

Say $F$ is a field, and we have the polynomial ring $F[x]$. Then if $a\in F$ satisfies the equation $f(x)=0$ ($f(x)\in F[x]$), then $x$ has the same sense as that used in your grade 4 algebra. $x$ doesn’t actually belong to the field $F$. It is just waiting for $a\in F$ to pop in and satisfy the equation. $x$ does technically belong to the polynomial ring $F[x]$, but we’re concerned only with $F$ here. We can think of the whole of $F[x]$ as a useful construct to help us deal with our problems. But just you wait.

If $p(x)\in F[x]$ is irreducible over $F$, then the **field** $F[x]/\langle p(x)\rangle$ contains the element $x$ (technically, the element is $x+\langle p(x)\rangle$). Yes. There is an equation in  $\frac{F[x]}{\langle p(x)\rangle}[X]$ such that when you punch in $x$, the equation is satisfied. So $x$ has moved on from being an imaginary filler for another element to actually being an element of the field itself.

Now we shall discuss some fundamental facts concerning field extensions:

1. If $f(x)\in F[x]$ is irreducible in $F[x]$, then there exists a field extension such that $f(x)$ has a root in that extension- The irreducibility property is important here. The proof proceeds by creating a field $F[x]/\langle f(x)\rangle$, and saying this is the field in which $f(x)=0$. But $F[x]/\langle f(x)\rangle$ is a field, and not a ring. $\frac{F[x]}{\langle f(x)\rangle}[X]$ is the corresponding ring of the field. Assume $f(x)(\in F[x])=a_0+a_1x+a_2x^2+\dots+a_nx^n$. Then this is mapped isomorphically to $a_0+a_1X+a_2X^2+\dots+a_nX^n$ in $\frac{F[x]}{\langle f(x)\rangle}[X]$, and is satisfied by $x+\langle f(x)\rangle \in \frac{F[x]}{\langle f(x)\rangle}$. Hence, $\frac{F[x]}{\langle f(x)\rangle}$ is the field extension of $F$ which contains a root of the irreducible (in $F$) polynomial $f(x)$.

2. Say we have $F(\alpha)$ such that $\alpha$ is algebraic over $F$. Then every element in $F(\alpha)$ is algebraic over $F$. The proof of this is based on the pigeonhole principle, and is quite clever. Any standard textbook on algebra will have this proof.