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Suppose we have the conjugation isomorphism $\psi_{\alpha, \beta}: F(\alpha) \mapsto F(\beta)$ defined by $$\psi_{\alpha, \beta}(a_0+a_{1} \alpha + \cdots + a_{n-1} \alpha^{n-1}) = a_0+a_{1} \beta + \cdots + a_{n-1} \beta^{n-1}$$ (e.g. $\alpha$ and $\beta$ are conjugates). Then $\text{irr}(\alpha, F) = \text{irr}(\beta, F)$.

The notion is that we have to show that $\text{irr}(\alpha, F)$ divides $\text{irr}(\beta, F)$ and $\text{irr}(\beta, F)$ divides $\text{irr}(\alpha, F)$. But why can't we just stop and say $\text{irr}(\alpha, F)$ divides $\text{irr}(\beta, F)$? Because, by definition, if $\text{irr}(\alpha, F)$ divides $\text{irr}(\beta, F)$, wouldn't that imply that $\text{irr}(\alpha, F) = \text{irr}(\beta, F)$? Why do textbooks say you need both of the above to show equality?

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PV jones, I think you might want to go into some more detail about your notation. What are F, $\alpha$, $F(\alpha)$, etc. Your second to last line needs an edit. Also, it's best if your question title is a question which nicely summarizes what you are asking in this post. –  Bill Kronholm Apr 13 '10 at 1:05
$F$ is a field and $E$ is an algebraic extension of $F$. Also $\alpha \in E$ and $\beta \in E$. They are conjugate over $F$. –  PV jones Apr 13 '10 at 1:12
For reference, this is from Fraleigh's Abstract Algebra book. –  PV jones Apr 13 '10 at 1:15
Who says that we have to show both divisibilities? Which are these textbooks you refer to? If we had a full citation, we might be able to figure out why the authors do it the way they do. –  Gerry Myerson Apr 13 '10 at 1:17
@PV jones, our messages crossed. Which edition of Fraleigh? Which chapter, page, Theorem? I have the 7th edition, from 2003. –  Gerry Myerson Apr 13 '10 at 1:19
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