Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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 $\psi(\alpha)=\beta$.

share|improve this answer
    
OF course! I keep forgetting that property of field isomorphisms. Once again, thank you very much. –  Kyle Schlitt Aug 8 '12 at 19:53
    
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
1  
@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
    
@tomasz: that's what I was initially trying to do (I showed work for this above) –  Kyle Schlitt Aug 9 '12 at 18:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.