# What's the splitting Field of $f(x)$?

What's the splitting field of $f(x) \in \mathbb{Z}_2[x]$? I know it has something to do with splitting the field into linear factors, but I'm unsure.

Is it $$\frac{f(x)}{x^n - 1}$$ If not, what is this field?

I'm not looking for an in depth answer, just a few comments.

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A splitting field, first and foremost, is a field. So in particular it isn't $\frac{f(x)}{x^n-1}$, or anything like it. –  Chris Eagle May 2 '12 at 19:15
Let $F$ be a field, in this case $\mathbb{Z_2}$, and let $f(x)$ be a polynomial of degree $\ge 1$, with coefficients in $F$. A splitting field for $f(x)$ over $F$ is an extension field $K$ of $F$ such that $f(x)$ splits as a product of linear factors with coefficients in $K$, and such that there is no intermediate field $F\subsetneq W \subsetneq K$ such that $F(x)$ splits into linear factors over $W$. –  André Nicolas May 2 '12 at 19:16
Assuming that $f(x)$ is square-free (that is, not divisible by $g(x^2)$ for any polynomial $g(x) \in \mathbb F_2[x]$), let $n$ be the smallest positive integer such than $f(x)$ is a divisor of $x^n - 1$. Then, the splitting field of $f(x)$ is $\mathbb F_{2^m}$ where $m$ is the smallest integer such that $\mathbb F_{2^m}$ contains a primitive $n$-root of unity. Thus, $(x^n - 1) | (x^{2^m-1}-1)$ and so $m$ is the smallest integer such that $2^m \equiv 1 \bmod n$. –  Dilip Sarwate May 2 '12 at 19:42