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I have the following exercise:

Find splitting field for the polynomial $x^2 + 1$ over $\mathbb{Z_3}$.

My solution:

At first, we should try to solve the equation $x^2 + 1 = 0$, thus $x^2 = 2$ and we need $\sqrt2$. Add this root to our new field and we have $\{0, 1, 2, \sqrt2, 2\sqrt2, 1+\sqrt2, 2 + \sqrt2, 1+2\sqrt2, 2 + 2\sqrt2 \}$ and that's our splitting field where roots of $x^2 + 1 = 0$ are $\sqrt2$ and $2\sqrt2$.

Is it correct or not? And I think there is no exact algorithm how to build a splitting field. How to do it properly?

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    $\begingroup$ Since $f$ is irreducible over $\mathbb Z_3$, the splitting field is a degree $2$ extension of it, hence a finite field of $9$ elements. $\endgroup$
    – awllower
    Commented Feb 20, 2013 at 9:28
  • $\begingroup$ @awllower What if it would be reducible? $\endgroup$
    – Tikhon Belousko
    Commented Feb 20, 2013 at 9:34
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    $\begingroup$ Then decompose that polynomial into irreducible ones, and find the splitting fields of them respectively, thus finding the splitting field of the original one. It would be extension of extensions. Of course the situation is similar. But here the reason that I mention its irreducibility is that if $f$ is reducible, then the splitting field would be the ground field itself, since $f$ is of degree $2$. Thanks for the attention. $\endgroup$
    – awllower
    Commented Feb 20, 2013 at 9:38
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    $\begingroup$ Yes, it is correct what you did. $\endgroup$
    – DonAntonio
    Commented Feb 20, 2013 at 15:09
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    $\begingroup$ Only, I wouldn't write $\sqrt{2}$, which is normally reserved for a real number. Note that since your equation also reads $x^{2} = -1$, using your notation you would write that its root is $\sqrt{-1}$. Just give the root a neutral-sounding name like $\alpha$, and build the splitting field as you did. Here the polynomial has degree $2$, so once you have added a root, you have also the other one. If the polynomial has higher degree, you might have to repeat the argument for other remaining irreducible factors of degree $> 1$. $\endgroup$
    – Andreas Caranti
    Commented Feb 22, 2013 at 11:31

1 Answer 1

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With help of commenters I found out that technically we can operate the way as I suggested in my question. Namely, take a root (e.g. $\sqrt2$ as in my example) and start to extend our field adding all linear combinations $\{a + \sqrt2b : a,b\in \mathbb{Z}_p \}$ to our new field.

In more common way if you want to construct a splitting field for $\mathbb Z_p$ and polynomial $f$ of degree $k$ we need to take a root $\beta \notin \mathbb Z_p$ of this polynomial. And than our splitting field will be the following set $\{a_0 + a_1 \beta^1 + \ldots + a_l\beta^l: a_i \in \mathbb Z_p \text{ and } l : b^l \in \mathbb{Z}_p \}$. If derived field $ GF(p^k)$ is not a splitting field we should commit one more iteration of the extension that is take a root $\gamma \notin GF(p^k)$ and construct the set of all linear combinations of this root and its degrees. So you need to repeat this step until you won't get a splitting field.

For further reading I'd recommend this useful and simple article.

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  • $\begingroup$ @YACP I've fixed several mistakes. So, please, take a look and provide your opinion... $\endgroup$
    – Tikhon Belousko
    Commented Jul 8, 2013 at 8:50
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    $\begingroup$ The part where $\beta^l\in\mathbb Z_p$ is still unclear to me. In fact, "taking a root of $f$" means nothing unless you construct a field extension of $\mathbb Z_p$ where $f$ has indeed a root. How can one build such a field extension? After making this clear, I'm sure that your answer will be ok. $\endgroup$
    – user26857
    Commented Jul 8, 2013 at 14:12
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    $\begingroup$ As one of the upvoter of your answer I also recommend you to look at this question and its answers. Finite fields are special in the sense that adjoining a single root of an irreducible polynomial immediately gives the splitting field. So there is never need for a second iteration (unless your polynomial has several irreducible factors of coprime degrees). This is because all the extensions of finite fields are Galois, i.e. normal and separable. $\endgroup$
    – Jyrki Lahtonen
    Commented Jul 13, 2013 at 15:49

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