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Let's assume we have an irreducible polynomial of degree $3$ on $\mathbb{Q}$. What are the possibility of degree of extension given by a splitting field on $\mathbb{Q}$.

That is, if $K$ is splitting field over $\mathbb{Q}$ for $p(x)$ of degree $3$. What are possibilities for $[K:\mathbb{Q}]$?


My attempt: I know that if we have $p(x)=x^3+1$ then the degree of extension is $2$ (this one will not work because later I knew $p(x)$ is reducible). But if $p(x)=x^3+2$ now the degree of extension will be $6$.

So in general what is the right way to do this?

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  • $\begingroup$ $x^3+1=(x+1)(x^2-x+1)$. $\endgroup$ Commented Dec 25, 2015 at 23:24

1 Answer 1

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Proposition: A splitting field of a polynomial of degree $n$ over $F$ is of degree at most $n!$ over $F$.

So in your case, you have an upper bound of $3! = 6$.

Sketch of the proof: Adjoin one root, $\alpha_1$. Then this generates an extension of $F_1$ of degree at most $n$. Now working in $F_1$, the polynomial is divisible by $x - \alpha_1$ leaving a polynomial of degree $n-1$. Adjoin another root to $F_1$ to generate an extension of degree at most $n-1$, and continue.

Again specializing to $n = 3$, let $p(x)$ be irreducible. Adjoining one root will give an extension of degree exactly 3. If we don't have all the roots yet, we adjoin one more root which must be of degree $2$. Because if it were degree 1, it would already have been in that first extension. Therefore, the only possibilities are degree 3 or degree 6.

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  • $\begingroup$ Thanks.. but can you give me an example (irreducible polynomial of degree 3 ) when the extension will be 3. since to me case 6 is easy to construct. $\endgroup$
    – henry
    Commented Dec 25, 2015 at 23:39
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    $\begingroup$ @henry Actually what you ask is quite hard to find. But, for example, here you can find that $x^3+x^2-2x-1$ is such a polynomial. $\endgroup$
    – Crostul
    Commented Dec 26, 2015 at 0:14

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