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In general for $f(x)$ irreducible in $K[x]$ (for $K$ a field) we look for a field extension of $K$ having a root of $f(x)$ as $K[x]/\langle f(x) \rangle $.

In the case of $x^3-x+1$ over $\mathbb{F}_3$, I have found $\mathbb{F}_3[x]/(x^3-x+1)$ to be the splitting field of $x^3-x+1$.

I do not think this would be the case for all such polynomials and all such fields.

Could someone help me to know what is special in this case??

Mathmo123
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    Dear @Praphulla, What is special is that your field is finite. Adjoining a single root of an irreducible polynomial over a finite field always results in an extension over which the polynomial splits. – Keenan Kidwell Jul 28 '13 at 15:04
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    In general this is false: $\Bbb Q(\sqrt[3]{2})=\Bbb Q[X]/(X^3-2)$ is not the splitting field of $X^3-2$ – Andrea Mori Jul 28 '13 at 15:09
  • @keenan kidwell :I am sorry. I am unable to understand why does this happen. Can you please elaborate this. :) –  Jul 28 '13 at 15:11
  • @AndreaMori : Yes yes. I know that. If not then study of Galois extn would be not so interesting :) –  Jul 28 '13 at 15:15
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    @PraphullaKoushik Every finite field $F_{3^n}$ is normal over $F_3$ because it is the splitting field of $X^{3^n} - X$, so any polynomial $f \in F_3[X]$ with a root in $F_{3^n}$ splits there. – Cocopuffs Jul 28 '13 at 15:16
  • @Cocopuffs : Ah, I like this :) –  Jul 28 '13 at 15:18
  • This will always be the splitting field, when $K$ is finite. See this recent question. – Jyrki Lahtonen Jul 28 '13 at 20:29
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    ALL: I accidentally voted to close this as a duplicate. Ignore that, please! I guess it's hardly a surprise to anyone that I subscribe to finite-fields tag, and got an e-mail about this question. I have apparently jumped to a certain conclusion. Sorry, all. – Jyrki Lahtonen Jul 28 '13 at 20:32
  • @JyrkiLahtonen I believe that it's now possible to retract a vote for closing. – egreg Jul 28 '13 at 21:53
  • Thanks, @egreg. I didn't know that. Useful. – Jyrki Lahtonen Jul 29 '13 at 05:41
  • @Cocopuffs - Dear Cocopuffs: You wrote: "any polynomial $f\in\mathbb F_3[X]$ with a root in $\mathbb F_{3^n}$ splits there". I think you meant "any irreducible polynomial $f\in\mathbb F_3[X]$ with a root in $\mathbb F_{3^n}$ splits there". – Pierre-Yves Gaillard Aug 11 '14 at 16:44

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Your observation is true, if $K$ is a finite field.. It is also true, if $K=\mathbb{R}$, because the only algebraic extension of that field is algebraically closed. If $\deg f(x)=2$, then it also holds because the sum of the two roots is in $K$, so if you join one, you automatically also join the other.

Over other fields it may or may not hold. For example over $\mathbb{Q}$ the claim does not hold, when $f(x)=x^3-2$, because exactly one of the roots of that polynomial is real, so joining that real root won't give you the rest. On the other hand, if $f(x)=x^3+x^2-2x-1$, then $\mathbb{Q}[x]/(f(x))$ is the splitting field of $f(x)$, because the zeros of $f(x)$ are $2\cos 2\pi/7$, $2\cos4\pi/7$ and $2\cos8\pi/7$. If you join one of those roots, you will also join the others because $$ \cos2\alpha=2\cos^2\alpha-1. $$ I think that, in a sense that I cannot make precise, it is rare for a field $K$ to have this property for all irreducible polynomials $f(x)$.

Community
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Jyrki Lahtonen
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you can show this fact for finite fields because all extensions are Galois everywhere! Namely, the splitting field of an irreducible $p$ of degree $n$ will be contained in $F_{p^n}$ since $x^{p^n}-x$ contains all irreds of degree $n$. So the splitting field of $p$ will be a subfield of $F_{p^n}$ which corresponds to a subgroup of $Z/nZ$ which is normal, hence the extension is Galois. Since in a Galois extension having one root means you have all roots, your field is splitting.

Elden Elmanto
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