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Suppose we have the conjugation isomorphism $\psi_{\alpha, \beta}: F(\alpha) \mapsto F(\beta)$ defined by $\psi_{\alpha, $\psi_{\alpha, \beta}(a_0+a_{1} \alpha + \cdots + a_{n-1} \alpha^{n-1}) = a_0+a_{1} \beta + \cdots + a_{n-1} \beta^{n-1}$ 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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The Conjugation IsomorphismWhy do we need to show that two irreducible polynomials divide each other?
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The Conjugation Isomorphism

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?