what the spilt field of x^3-8? I think I cannot split the real number,is that correct? what about x^3-2 ,is the extension dimension will be either 6 or 3? In the beginning I think it will be 3 extension dimension ,since we have three roots. Then I look for the answer it is 6 extension dimension why?
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I can't understand what you're saying. The splitting field of $x^3-8$ is not a subset of the real numbers. – Git Gud May 01 '14 at 01:19
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The polynomial $x^3-8$ has the rational root $2$. Divide the polynomial by $x-2$. We get $x^2+2x+4$, an irreducible quadratic. So the splitting field of $x^3-8$ has degree $2$ over the rationals.
For the polynomial $x^3-2$, adjoin $\sqrt[3]{2}$ to the rationals. Since $x^3-2$ is irreducible of degree $3$, the field $E=\mathbb{Q}(\sqrt[3]{2})$ has degree $3$ over the rationals. The splitting field of $x^3-2$ is not $E$, since $x^3-2$ has some complex roots. Imagine dividing $x^3-2$ by $x-\sqrt[3]{2}$. We get an irreducible polynomial of degree $2$, so its splitting field $F$ has degree $2$ over $E$. It follows that the degree of $F$ over $\mathbb{Q}$ is $(3)(2)$.
André Nicolas
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The answer above says that the dimension (over $\mathbb{Q}$) of the splitting field of $x^3-8$ is $2$. For this fied is just the splitting field of $x^2+2x+4$. – André Nicolas May 01 '14 at 02:37