# Conjugation Isomorphism

• Apr 12th 2010, 05:03 PM
Sampras
Conjugation Isomorphism
Theorem. 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 $\text{irr}(\alpha, F) = \text{irr}(\beta, F)$?
• Apr 12th 2010, 07:24 PM
tonio
Quote:

Originally Posted by Sampras
Theorem. 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 $\text{irr}(\alpha, F) = \text{irr}(\beta, F)$?

If you already know that the corresponding irreducible polynomials are (monic, of course) and of the same degree, then yes: it is enough to show that one divides the other. Perhaps in this problem though they do not assume this...

Tonio