# Proof about Field conjugation isomorphisms

I'm having an awful time making sense of a proof and I was hoping someone could help.

Theorem: Let $\alpha$ and $\beta$ be algebraic over a field $F$ with $deg(\alpha, F) = n$, as elements of a field extension $E$ of $F$.

Define the map $\psi_{\alpha,\beta}:F(\alpha)\to F(\beta)$ by $\psi_{\alpha,\beta}(c_{0} + c_{1}\alpha + ... + c_{n-1}\alpha^{n-1}) = c_{0} + c_{1}\beta + ... + c_{n-1}\beta^{n-1}$.

Then $\psi_{\alpha,\beta}$ is an isomorphism if and only if $\alpha$ and $\beta$ are conjugates over $F$.

Proof:

$(\Rightarrow)$ Assume the map is an isomorphism. Let $irr(\alpha, F) = a_{0} + a_{1}x + ... + x^{n}$. Then $a_{0} + a_{1}\alpha + ... +\alpha^{n} = 0$, so $0 = \psi_{\alpha,\beta}(0) = \psi_{\alpha,\beta}(a_{0} + a_{1}\alpha + ... +\alpha^{n}) = a_{0} + a_{1}\beta + ... +\beta^{n}$, which implies that $\beta$ is a root of $irr(\alpha, F)$.

The author then uses the same argument to show that $\alpha$ is a root of $irr(\beta, F)$, which I haven't read carefully yet or worked through. But I am already in objection.

How do we obtain $\psi_{\alpha,\beta}(a_{0} + a_{1}\alpha + ... +\alpha^{n}) = a_{0} + a_{1}\beta + ... +\beta^{n}$? The definition only gives us a formula for up to degree $n-1$ polynomials.

My partial justification:

Since $a_{0} + a_{1}\alpha + ... +\alpha^{n} = 0$, we know $-a_{0} - a_{1}\alpha - ... -a_{n-1}\alpha^{n-1} = \alpha^{n}$, so $\psi_{\alpha,\beta}(\alpha^{n}) = \psi_{\alpha,\beta}(-a_{0} - a_{1}\alpha - ... -a_{n-1}\alpha^{n-1}) = -a_{0} - a_{1}\beta - ... -a_{n-1}\beta^{n-1}$

But I can conclude that this last expression equals $\beta^{n}$ only if I assume the result which I am trying to prove: that $\beta$ is a root of $irr(\alpha, F)$.

Thus, I am stuck.

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If the mapping $\psi=\psi_{\alpha,\beta}$ is an isomorphism, then it respects multiplication, so $$\psi(\alpha^n)=\psi(\alpha)^n=\beta^n.$$ In other words, while proving "$\Rightarrow$" you are allowed to assume that $\psi$ is an isomorphism. By definition it satisfies $\psi(\alpha)=\beta$.
We don't need that. $\alpha$ has degree $n$, so $\alpha^n$ can be expressed as $\sum_{i<n} c_i\alpha^i$. – tomasz Aug 8 '12 at 19:55
@tomasz: At that point we don't know that the same combination of coefficients $c_i$ will work for powers of $\beta$ though. We are still on our way to proving that. – Jyrki Lahtonen Aug 8 '12 at 21:24