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Let $\mathbb{F}$ be a field, and let $p(x)\in\mathbb{F}[x]$ be an irreducible of degree $n$.

I know that a field extension $\mathbb{K}$\ $\mathbb{F}$ s.t $p(x)$ have a root in $\mathbb{K}$, I also know that regardless of which root we take the fields we get are isomorphic.

Does this imply that $\mathbb{K}$ have all the roots of $p(x)$ ? (or, is there a field extension of degree $n$ that have all of $p(x)$ roots ?)

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3 Answers 3

up vote 3 down vote accepted

I think you are slightly confused.

Ok, so let's just agree that there exists a field $k$ such that all the roots of $p$ lie in $k$ (see here). Ok, so what we can then do is analyze individual roots $\alpha,\beta,\gamma,\cdots,$ of $p$. To do this we can consider fields such as $F(\alpha)$ which is the smallest field containing both $F$ and $\alpha$ (i.e. the intersection of all subextensions of $k/F$ which contain $F$). What is definitively true is that $F(\alpha)\cong F[x]/(p)$. Now, this was true for ANY root $\alpha$ of $p$ and so you may conclude that, in particular, $F(\alpha)\cong F(\beta)$ for any two roots $\alpha,\beta\in k$ of $p$.

That said, while the $k$ I said contains all the roots of $p$ any extension containing A root of $p$ need not contain all the roots of $p$. For example, consider the irreducible polynomial $x^3-2\in\mathbb{Q}[x]$. Then, the complex numbers $\mathbb{C}$ are a field containing all the roots $x^3-2$--they are $\sqrt[3]{2},\sqrt[3]{2}\omega,\sqrt[3]{2}\omega^2$ where $\omega$ is a primitive third root of unity. Note then that while $\mathbb{Q}(\sqrt[3]{2})$ contains A ROOT of $p(x)$ it definitively does not contain ALL the roots of $p(x)$ since $\omega$ is complex and all the elements of $\mathbb{Q}(\sqrt[3]{2})$ are real (if this last fact isn't obvious consider that $\mathbb{R}$ is a field containing $\mathbb{Q}$ and $\sqrt[3]{2}$).

I hope this helps.

EDIT: Just like the fields $\mathbb{Q}(\sqrt[3]{2}),\mathbb{Q}(\sqrt[3]{2}\omega)$ and $\mathbb{Q}(\sqrt[3]{2}\omega^2)$ it seems you now have three isomorphic responses. Thank god that fields permit no more roots.

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"three isomorphic responses" - made me laugh :) –  Belgi Apr 17 '12 at 18:37
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Secretly it was a ploy for upvotes, the most valuable currency in the universe--glad I could help. –  Alex Youcis Apr 17 '12 at 18:37
    
Well, it worked ;) –  Belgi Apr 17 '12 at 18:44

No, $K$ will typically not have all the roots of $p(x)$. If the roots of $p(x)$ are $\alpha_1, \dots, \alpha_k$ (note $k = n$ in the case that $p(x)$ is separable), then the field $F(\alpha_1, \dots, \alpha_k)$ is called the splitting field of $p(x)$ over $F$, and is the smallest extension of $F$ that contains all roots of $p(x)$.

For a concrete example, take $F = \mathbb{Q}$ and $p(x) = x^3-2$. Then if $\sqrt[3]{2}$ denotes the real cube root of $2$, $\mathbb{Q}(\sqrt[3]{2})$ sits naturally inside of $\mathbb{R}$. Since the other 2 roots of $p(x)$ are complex numbers, they can't be contained in $\mathbb{Q}(\sqrt[3]{2})$. But the field $\mathbb{Q}(\sqrt[3]{2}, \zeta_3)$, where $\zeta_3$ is a primitive cube root of unity, is the splitting field for $p(x)$ over $\mathbb{Q}$.

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It is possible that some, but not all of the roots, are in $\mathbb{K}$. For example, consider the polynomial $x^3-2$, and let $a$ be the real cuberoot of $2$. Then you can check for yourself that $\mathbb{Q}[x]/(x^3-2)\cong \mathbb{Q}(a)=\{\alpha+\beta a+\gamma a^2:\alpha, \beta, \gamma \in \mathbb{Q}\}$ is subfield of $\mathbb{R}$, which has degree $3$ over $\mathbb{Q}$. But of course it does not contain all the roots, since the other two are complex.

Since any field extension containing $a$ is an extension of $\mathbb{Q}(a)$, there is no $\mathbb{K}$ such that $[\mathbb{K}:\mathbb{Q}]=3$ and $\mathbb{K}$ contains all the roots of $x^3-2$, since it would have dimension strictly larger than the dimension of $\mathbb{K}$. However, there is such an extension of degree six which does contain all the roots. In general, if we adjoin all the roots of a polynomial, we get an extension of degree dividing $n!$, where $n$ is the degree of the polynomial. An extension obtained by adjoining all the roots of some polynomial is called a normal extension.

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Note: I kept your notation from the answer, but the other two respondents use $K$ and $k$ for their fields, which is a little more standard. There's nothing wrong with your notation, though, aside from needing a bit more time to type out. –  Brett Frankel Apr 17 '12 at 18:36

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