Let $\mathbb{L}$ an extension field of $\mathbb{K}$ and $\alpha, \beta\in\mathbb{L}$. If they have the same minimal polynomial than $\mathbb{K}(\alpha)\simeq\mathbb{K}(\beta)$, because if: $$\begin{array}{rccccrccc} \phi:&\frac{\mathbb{K}[x]}{<p(x)>}&\longrightarrow&\mathbb{K}[\beta]&\mbox{and}&\psi:&\mathbb{K}[\alpha]&\longrightarrow&\frac{\mathbb{K}[x]}{<p(x)>}\\ &f(x)+<p(x)>&\longrightarrow&f(\beta)&&&f(\alpha)&\longrightarrow&f(x)+<p(x)> \end{array},$$ are isomorphisms, where $p(x)$ is the minimal polynomial of $\alpha$ and $\beta$, than: $\varphi:\mathbb{K}(\alpha)\longrightarrow\mathbb{K}(\beta)$, defined by $\varphi=\psi\circ\phi$, is an isomorphism which $\left.\varphi\right|_\mathbb{K}$ is the identity on $\mathbb{K}$, so $\varphi$ is an isomorphism of extensions and $\mathbb{K}[\alpha]\simeq\mathbb{K}[\beta]$ (note that $\frac{\mathbb{K}[x]}{<p(x)>}$ is a field because $<p(x)>$ is maximal ideal, then $\mathbb{K}[\alpha]$ is a field and $\mathbb{K}[\alpha]=\mathbb{K}(\alpha)$).

I believe this is right, but I was thinking if we just consider: $$\begin{array}{rccc} \sigma:&\mathbb{K}(\alpha)&\longrightarrow&\mathbb{K}(\beta)\\ &f(\alpha)&\longrightarrow&f(\beta) \end{array},$$ than $\left.\sigma\right|_\mathbb{K}$ is also the identity on $\mathbb{K}$ and $\sigma$ is an isomorphism, so I don't need that $\alpha$ and $\beta$ have the same minimal polynomial for $\mathbb{K}(\alpha)$ and $\mathbb{K}(\beta)$ are isomorphic?

I can find some conditions for $\mathbb{L}$ and $\mathbb{K}$ such that I always have a minimal polynomial that $\alpha$ and $\beta$ are roots?


  • $\sigma$ is an isomorphism because:

    Surjective: If $f(\beta)\in\mathbb{K}(\beta)\Longrightarrow f(x)\in\mathbb{K}[x]\Longrightarrow f(\alpha)\in\mathbb{K}(\alpha)\Longrightarrow\sigma(f(\alpha))=f(\beta)$.

    Injection: Let $f(\beta), g(\beta)\in\mathbb{K}(\beta)$ such that $f(\beta)=g(\beta)$. We know that by $\mathbb{K}(\beta)$ be a simple extension of $\mathbb{K}$ than for any element in $\mathbb{K}(\beta)$ has a unique expression of the form $m(\beta)$, where $m(x)\in\mathbb{K}[x]$ and the degree of $m(x)$ is less than the degree of minimal polynomial of $\beta$, so $f(x)=m(x)=h(x)\Longrightarrow f(\alpha)=g(\alpha)$.

    Homomorphism: $\sigma(f(\alpha)+g(\alpha))=\sigma((f+g)(\alpha))=(f+g)(\beta)=f(\beta)+g(\beta)=\sigma(f(\alpha))+\sigma(g(\alpha))$, $\sigma(f(\alpha)g(\alpha))=\sigma((fg)(\alpha))=(fg)(\beta)=f(\beta)g(\beta)=\sigma(f(\alpha))\sigma(g(\alpha))$.

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    $\begingroup$ What isomorphism exactly is that $\;\sigma\;$ in the last part of your question? $\endgroup$ – Timbuc Jan 4 '15 at 19:32
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    $\begingroup$ In general your map $\sigma$ won't be well definied. In order that $\sigma$ is well definied, you need that $f(\alpha) = 0$ implies $f(\beta) = 0$... $\endgroup$ – Hans Giebenrath Jan 4 '15 at 19:35
  • $\begingroup$ If I have $f(\alpha)=0\Longrightarrow f(\beta)=0$ then the Surjective, Injective and Homomorphism steps are right? $\endgroup$ – donikvep Jan 4 '15 at 20:06

Your map is in general not a morphism. Consider $\mathbb K = \mathbf Q$ and $\alpha = \sqrt 2$ as well as $\beta = \sqrt 3$. Then by definition $\sigma(\sqrt2) = \sqrt 3$. But then $$ 0 = \sqrt 3^2 - 3 = \sigma(\sqrt 2)^2 - \sigma(3) = \sigma((\sqrt 2)^2 - 3) = \sigma(-1) = - \sigma(1) = -1.$$

So $\sigma$ can not be a morphism. Note that $\sigma$ being a morphism implies that every polynomial $f \in \mathbf Q[X]$ with $f(\alpha) = 0$ will also satisfy $f(\beta) = 0$. This will give the condition on the minimial polynomial you need.

  • $\begingroup$ If I have this condition, then what I did to show that $\sigma$ is an isomorphism are right? $\endgroup$ – donikvep Jan 4 '15 at 20:23
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    $\begingroup$ I think. But you have to be careful at the step $\sigma((fg)(\alpha)) = (fg)(\beta)$, because $fg$ can have degree larger then $m$. Moreover, when you have this condition, you immeadatly get that $\alpha$ and $\beta$ have the same minimal polynomial: Let $m_\alpha$ and $m_\beta$ be the minimal polynomials. From $m_\alpha(\alpha) = 0$ we get $m_\alpha(\beta) = 0$. Thus $m_\beta$ divides $m_\alpha$. Since both are irreducible we get equality. $\endgroup$ – Hans Giebenrath Jan 4 '15 at 20:30

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