I need to find the splitting field of $\; x^2+1 \in \mathbb Z_7 [x] \;$ over $\mathbb Z_7 $.

The roots of the polynomial are $-i \;$ and $i$. Therefore I would conclude that the splitting field is $\mathbb Z_7(i)$ but that would be more like a guess.

I don´t really understand splitting field over $ \mathbb Z_7$. Can seomeone give me a hint, on how to find the splitting field ?


The smallest field extension of $\Bbb F_7$ is $\Bbb F_{49}$. As $\Bbb F_{49}^\times $ is a cyclic group of order $48$, it has an element of order $4$, let's call that $i$. Then $i^4= 1$ but $i^2\ne 1$, meaning that $i^2=-1$, as desired.

Note however that even though I introduced the name $i$ for an element of $\Bbb F_{49}$, it is not "the" $i$ you know from $\Bbb C$; actually $0\in \Bbb F_7$ is not "the" $0$ you know from $\Bbb C$, either.

  • $\begingroup$ Thanks.Do you know where I could find a proof that $ \; \mathbb F_{49}^*$ is in fact a cyclic group ? $\endgroup$ – XPenguen Jan 16 '16 at 14:05

Since $x^2+1$ has no roots in $\Bbb Z_7$ and the polynomial is quadratic the splitting field of $X^2+1$ is the quadratic extension $$ \Bbb Z_7[X]/(X^2+1) $$ which you can concretely model as the degree $\leq1$ polynomials $aX+b$ with coefficients in $\Bbb Z_7$ with the multiplication rule given (unsurprisingly) by $X^2=-1$.

It follows from the general theory of finite fields that this is the quadratic extension of $\Bbb Z_7$, i.e. the unique (up to isomorphism) field with $7^2=49$ elements, usually denoted $\Bbb F_{49}$.


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