# Find a splitting field extension for $t^3-5$ over $\mathbb{Z}_7$, $\mathbb{Z}_{11}$, and $\mathbb{Z}_{13}$

I need to find a splitting field extension for $t^3-5$ over $\mathbb{Z}_7$, $\mathbb{Z}_{11}$, and $\mathbb{Z}_{13}$.

Looking at $t^3-5$ over $\mathbb{Z}_7$, I see that there are no roots of $t^3-5$ in $\mathbb{Z}_7$. From here, I'm not sure where to go. I think I need to consider $\alpha$ a root of $t^3-5$ such that $\alpha \notin \mathbb{Z}_7$, and then adjoin this alpha to $\mathbb{Z}_7$ to get a splitting field?

Hint: A splitting field for $t^3-5$ over a field $F$ is generated by one cubic root of $5$ and one cubic root of $1$. Depending on $F$, one of these roots or both may already be in $F$.

Solution:

Let $E$ be the splitting field for $t^3-5$ over the given fields.

• $\mathbb{Z}_7$ contains a cubic root of $1$ (e.g. $2$), but no cubic root of $5$ and so $E=\mathbb{Z}_7/(X^3-5)$, the field of $7^3$ elements.

• $\mathbb{Z}_{11}$ contains a cubic root of $5$ (e.g. $3$), but no cubic root of $1$ and so $E=\mathbb{Z}_{11}/(X^2+X+1)$, the field of $11^2$ elements.

• $\mathbb{Z}_{13}$ contains a cubic root of $1$ (e.g. $3$) and a cubic root of $5$ (e.g. $7$) and so $E=\mathbb{Z}_{13}$, the field of $13$ elements.

• Why is the splitting field generated by one cubic root of 1 and one cubic root of 5? I understand the cubic root of 5 being a generator, but why do we know that one cubic root of 1 is also a generator? – Ldog327 Oct 14 '16 at 2:04
• @Ldog327, because the quotient of two roots of $t^3-5$ is a cubic root of unit. Also $t^3-5=(t-\alpha)(t-\omega\alpha) (t-\omega^2\alpha)$. – lhf Oct 14 '16 at 2:05
• Okay, so now a splitting field is generated by a cubic root of 5 and an $\omega$. Isn't the cube root of 5 never in any of these fields? – Ldog327 Oct 14 '16 at 2:13
• @Ldog327, try it: find all cubes in the given fields. – lhf Oct 14 '16 at 2:15
• I see. So for example, in $\mathbb{Z}_7$ we have $6^3$ mod 7 is 5, so the splitting field is $\mathbb{Z}_7(\omega)$, correct? – Ldog327 Oct 14 '16 at 2:19

HINT

►No problem with $\Bbb Z_{13}$ in which $$t^3-5=(t-7)(t-8)(t-11)$$ ►In $\Bbb Z_{11}$ one has$$t^3-5=(x-3)g(x)$$ where $g(x)$ is quadratic irreducible. In a quadratic extension one has $\alpha\ne\beta$ but conjugates with $g(\alpha)=g(\beta)=0$. Hence the splitting field is $\Bbb F_{121}$ the finite field with $11^2$ elements.

► In $\Bbb Z_7$ the polynomial $t^3-5$ is irreducible. By similar reasoning you have an extension of $\Bbb Z_7$ in which there are three elements $\alpha, \beta, \gamma$ with $t^3-5=(t-\alpha)(t-\beta)(t-\gamma)$ where $\alpha$ is in a cubic extension of $\Bbb Z_7$ hence in $\Bbb F_{343}$ and $\beta$ and $\gamma$ in a quadratic extension of $\Bbb F_{343}$ Can you see now the corresponding splitting field? (Is it $\Bbb F_{7^6}=\Bbb F_{117649}?$)