In order to demonstrate the root extension theorem, I need to prove that if an element of F[x]/(p(x)) is represented as $\overline a =a+(p(x))$ where $p(x)=a_0+a_1x+\cdots+a_nx^n$, then$\overline a_0+\overline a_1\overline x+\cdots+\overline a_n\overline {x^n}=0$ implies that $a_0+a_1\overline x+\cdots+a_n\overline {x^n}$=0. It seems easy, but I've tried a lot and I couldn't understad why we remove the overline of the coefficients.

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    $\begingroup$ you seem to have forgotten a "$=0$" before and after the "implies" $\endgroup$ – Glougloubarbaki Oct 21 '12 at 0:22
  • $\begingroup$ @Glougloubarbaki yes, thank you $\endgroup$ – user42912 Oct 21 '12 at 0:26
  • $\begingroup$ I deleted my comments, and posted them as an answer. Sorry. $\endgroup$ – Rankeya Oct 21 '12 at 3:38

In my opinion this is just notation because you are identifying $F$ as a subfield of $F[x]/(p(x))$. So, elements $\overline{a_i}$ are identified with $a_i$.

Technically, what this proof is showing is that if $\overline{F}$ is the image of $F$ under the natural injective ring map $F \rightarrow F[x]/(p(x))$, then the polynomial $\overline{a_n}T^n + \dots + \overline{a_1}T + \overline{a_0} \in \overline{F}[T]$ has the root $\overline{x}$, where now $F[x]/(p(x))$ is an extension of $\overline{F}$ (in the strict sense of an inclusion of fields). But, $F[x]$ and $\overline{F}[T]$ are isomorphic, and $\overline{a_n}T^n + \dots + \overline{a_1}T + \overline{a_0}$ is the image of $p(x)$ under this isomorphism.

If you look at Lang's Algebra (Proposition 2.3 of chapter V), there he actually constructs an extension of $F$ (if by extension you mean a strict inclusion of fields and not just an injective map of fields) in which $p(x)$ has a root.

  • $\begingroup$ Maybe I should add that the edition of Lang's Algebra that I have is the revised third edition. $\endgroup$ – Rankeya Oct 21 '12 at 3:42

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