I am studying the the basic theorem of field extensions with Charles C. Pinter's book: A book of Abstract Algebra. The very same theorem that is being discussed here: Example of Basic Theorem of field extensions.
I am trying to follow the proof in the book but I run into something I don't quite understand or see why is true. Hopefully someone can help me?
The part I am struggling with is where we try to prove that:
$\overline{a}_0 + \overline{a}_1\overline{x} + \cdots + \overline{a}_n\overline{x}^n = J$
where $J$ is the zero coset and: $J + a_0 = \overline{a}_0, \ldots , J + a_n = \overline{a}_n$ and $J + x = \overline{x}$.
In the book it goes:
\begin{align}
\overline{a}_0 + \overline{a}_1\overline{x} + \cdots + \overline{a}_n\overline{x}^n & = (J + a_0) + (J+a_1)(J+x) + \cdots + (J+a_n)(J+x)^n \\
& = (J + a_0) + (J+a_1x)+ \cdots + (J+a_nx^n) \\
& = J + p(x) \\
& = J \quad \quad \quad [\text{because } p(x) \in J]
\end{align}
where $p(x)$ is an irreducible polynomial in the original field.
I guess where I am struggling is how:
$(J+a_1)(J+x) = (J+a_1x)$ and $(J+a_n)(J+x)^n = (J+a_nx^n)$ ?
In advance; thank you 🙂
Best Answer
It occurs to me that you may be looking at the offending equation $$ (J+a)(J+b)=(J+ab) $$ and making too much of it. In particular, how is “$(J+a)(J+b)$” to be understood? To multiply these two cosets, you don’t take all possible products, one factor from $(J+a)$, the other from $(J+b)$; rather, you choose one element, say $a_0\in(J+a)$, and one, say $b_0\in(J+b)$, and see what coset $a_0b_0$ belongs to. That will be your product coset. But notice: I have italicized the word “choose”, so you have to verify that the resulting coset does not depend on your choice.
Accordingly, choose other $a_1\in(J+a)$ and $b_1\in(J+b)$, so that $a_1-a_0\in J$ and $b_1-b_0\in J$. Then $$ a_1b_1-a_0b_0=a_1(b_1-b_0)+(a_1-a_0)b_0\in J\,, $$ because the perenthesized quantities are in $J$. Thus $a_0b_0$ and $a_1b_1$ are in the same $J$-coset, and your choice didn’t matter. In particular, the product coset is $(J+ab)$.
(Note: because $p$ is irreducible, it happens that the set of all possible products is a coset, equal to the product coset. Not so if $p$ is reducible.)