I need to find the splitting field of the polynomial $f(x)=x^4-2$ over $\mathbb Z_3[x]$ and $\mathbb Z_7[x]$. I know that if it's irreducible over $\mathbb Z_3[x]$, then $\mathbb Z_3 [x]/ \langle f(x) \rangle$ is the splitting field.

The problem is that I don't know how to determine whether it's irreducible or not. I know a result (Lemma 27.5 here: http://people.virginia.edu/~mve2x/3354_Fall2010/lecture27.pdf) which says that if $F$ is a field and a quadratic polynomial doesn't have roots in $F$, then it's irreducible over $F[x]$. Here, $f(x)$ doesn't have roots in $\mathbb Z_3$, but $f(x)$ has grade $4$, not $2$. Is there a similar result which says that if a polynomial of grade 4 has no roots in $F$, then it's irreducible over $F[x]$?

Also, $f(x)$ has the roots $2$ and $5$ in $\mathbb Z_7$, so $f(x)=(x-2)(x-5)g(x)$ for some quatratic polynomial $g(x) \in \mathbb Z_7[x]$, does that help me to find the splitting field?

Thanks in advance.

  • $\begingroup$ Since you're talking about fields, I assume that by $\Bbb Z_3$ you mean $(\Bbb Z/3\Bbb Z)$, not the $3$-adic integers. $\endgroup$ – Greg Martin Jun 14 '14 at 23:53
  • $\begingroup$ @GregMartin That's right. $\endgroup$ – Twink Jun 14 '14 at 23:56

Suppose $\alpha$ is a root somewhere of $f(x)=x^4-2$ over $\Bbb Z_3$.
Then $\alpha^4=2$, so $\beta:=\alpha^2$ has $\beta^2=2$. As $\beta\notin\Bbb Z_3$, we have the quadtratic extension (the $9$ element field) $F:=\Bbb Z_3(\beta)$ so far, which must be included in the splitting field (as $\beta=\alpha^2$).

Then $\ f(x)=x^4-\beta^2=(x^2-\beta)(x^2+\beta)\ $ over $F$, so we still have to find $\sqrt{\pm\beta}$ either within $F$, or we have to join them.

Now, we are lucky as we have $(\beta+1)^2=\beta^2+2\beta+1=2+2\beta+1=-\beta$, and similarly, $(\beta-1)^2=\beta$. So, $F$ already contains all four roots ($\pm\beta\pm1$) of $f$.

Over $\Bbb Z_7$, notice that $5=-2$, so you already got $f(x)=(x-2)(x+2)g(x)$. Moreover, knowing that $2^4=4^2\equiv 2$, we can also write it as $$f(x)=x^4-4^2=(x^2-4)(x^2+4)\ =(x^2-4)(x^2-3)$$ so we get $g(x)=x^2-3$, and this doesn't have a root yet in $\Bbb Z_7$, so we have to join one, and that will yield to the quadratic splitting field.

  • $\begingroup$ Thank you so much. I just have two questions: $\beta \notin \mathbb Z_3$ for the polynomial $x^2-2$ doesn't have a root in $\mathbb Z_3$, is that the reason? And also, why do we have to join only one root of $x^2-3$? What about the other one? $\endgroup$ – Twink Jun 15 '14 at 1:36
  • $\begingroup$ And a last question, if I want to find the Galois groups $\text{Gal}(\mathbb Z_3(\beta)/\mathbb Z_3)$ and $\text{Gal}(\mathbb Z_7(\gamma)/\mathbb Z_7)$ (where $\gamma$ is a root of $x^2-3$ in $\mathbb Z_7$), do you think it's necessary to find explicitly $\beta$ and $\gamma$? And in that case, is that possible to do?. $\endgroup$ – Twink Jun 15 '14 at 1:37

It is possible for a quartic polynomial with no roots in a field to not be irreducible over that field; an example is $x^4+x^3+x+2=(x^2+1)(x^2+x+2)$ over $\Bbb Z_3$. But it would have to be the product of two quadratics. In this case, if $x^4-2=(x^2+ax+b)(x^2+cx+d)$, then (by coefficients of $x^3$) $a=-c$, hence (by coefficients of $x$) $b=d$, which is impossible since $b^2\equiv2\pmod3$ has no solution. So $x^4-2$ is indeed irreducible over $\Bbb Z_3$.

And since there is only one field (up to isomorphism) of any given prime power order, adjoining any of the roots of $x^4-2$ to $\Bbb Z_3$ yields the splitting field, which is isomorphic to $\Bbb F_{27}$. (Indeed, in general the splitting field of an irreducible polynomial over $\Bbb F_q$ of degree $d$ is always $\Bbb F_{q^d}$.)

In the second case, the splitting field of $x^4-2$ is the same as the splitting field of the irreducible quadratic $g(x)$, which is $\Bbb F_{49}$.

  • $\begingroup$ No, it is going to be $\Bbb F_9$ for the first one, and $f$ is not irreducible: $f(x)=((x-1)^2-2)((x+1)^2-2)$. $\endgroup$ – Berci Jun 15 '14 at 0:07
  • $\begingroup$ Ah, I could spot it: should be $b^2\equiv -2\pmod3$, which has solution $\pm1$. $\endgroup$ – Berci Jun 15 '14 at 0:18

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