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If $\alpha_1, \cdots, \alpha_n$ are distintct roots of a polynomial $p(x)$. If I want to show that $p(x)$ is irreducible over a field $F$, is it suffices to show that $deg (p) \leq [F(\alpha_1, \cdots, \alpha_n) : F]$?

if so, I have a question about an example in p.524 of dummit which stated: $[\mathbb{Q}(\sqrt[6]{2}):\mathbb{Q}]=6, [\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$. Since $(\sqrt[6]{2})^3=\sqrt{2}$, $\mathbb{Q}(\sqrt{2}) \subseteq \mathbb{Q}(\sqrt[6]{2})$ and $[\mathbb{Q}(\sqrt[6]{2}):\mathbb{Q}(\sqrt{2})]=3$. Hence $x^3-\sqrt{2}$ is irreducible over $\mathbb{Q}(\sqrt{2})$.

Why this will implies $x^3-\sqrt{2}$ is irreducible over $\mathbb{Q}(\sqrt{2})$? How about the other two roots $\sqrt[6]{2}\zeta_3$ and $\sqrt[6]{2}\zeta_3^2$? I think we have to show $[\mathbb{Q}(\sqrt[6]{2},\zeta_3): \mathbb{Q}] \geq 3$ instead, counld someone please explain, thank you so much!

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For the second one, if $x^3-\sqrt{2}$ is reducible, the splitting field must be of degree $2$ at most, but now it is of degree $3$ at least. This is my understanding. –  Anonymous Coward Jun 13 '13 at 11:49

1 Answer 1

First answering the question in your first paragraph. No, $[F(\alpha_1,\ldots,\alpha_n):F]\ge\deg p$ is not a sufficient criterion for $p(x)$ to be irreducible over $F$. As a counterexample I proffer $p(x)=(x^2-2)(x^2-3)$ over the rationals. Obviously $p(x)$ is not irreducible. Yet $[\mathbb{Q}(\pm\sqrt2,\pm\sqrt3):\mathbb{Q}]=4\ge\deg p.$

The argument in the second paragraph could benefit from the following details. Here $\mathbb{\root6\of2}=K(\root6\of 2),$ where $K=\mathbb{Q}(\sqrt2)$. For a simple algebraic extension $L=K(\alpha)/K$ we always have that $[L:K]$ equals the degree of the minimal polynomial of $\alpha$ over $K$. In the present case we know that $$ 6=[\mathbb{Q}(\root6\of2):\mathbb{Q}]=[\mathbb{Q}(\root6\of2):K][K:\mathbb{Q}], $$ because the degree of the tower of extensions is the product of the degrees of the smaller extensions. As $[K:\mathbb{Q}]$ is known to be two, we can conclude that (this time $\alpha=\root6\of2$) $$ [K(\alpha):K]=[\mathbb{Q}(\root6\of2):K]=\frac{6}{2}=3. $$ Therefore we know that the degree of the minimal polynomial of $\alpha$ is three.

On the other hand $\alpha$ is a root of the polynomial $q(x)=x^3-\sqrt2\in K[x]$. The minimal polynomial of $\alpha$ over $K$ is thus a factor of $q(x)$. But $q(x)$ is also of degree three. Therefore we can conclude that $q(x)$ is the minimal polynomial of $\alpha$ over $K$. Therefore it is irreducible in $K[x]$.

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