You need two conditions for a field to be a splitting field for a specific irreducible representation (in characteristic zero to begin with): It must contain the character values of the representation. For this there is of course a minimal field, the field generated by those values. However, a splitting field must also split a division algebra and for that there is no unique minimal field.

As a specific example we can take the quaternion group of order $8$ over the rational numbers. The component of the group algebra corresponding to the only faithful representation is a quaternion algebra over $\mathbb Q$ ramified at $\infty$ and $2$ and hence is split by any quadratic imaginary field for which $2$ is either ramified or non-split.

To be more concrete about the quaternion representation, the ordinary real quaternion algebra makes sense over the rationals, denoted $\mathbb H_{\mathbb Q}$. It has $\mathbb Q$-basis 1,i,j,k and the usual multiplication table. Then $\pm\{1,i,j,k\}$ is a multiplicative group that is a copy of the quaternion group $Q$ and therefore we have an algebra map $\mathbb Q[Q]\to\mathbb H_{\mathbb Q}$. It is evidently surjective so that $\mathbb H_{\mathbb Q}$ is one of the factors in the group algebra (more precisely $\mathbb Q[Q]=\mathbb Q^4\times\mathbb H_{\mathbb Q}$, where the first four factors correspond to the four one-dimensional representations). Now, for a field $K$ of characteristic zero $\mathbb H_K$ has a two-dimensional irreducible representation exactly when it is split, i.e., when it is isomorphic to the algebra of $2\times2$-matrices. That is thus exactly the condition for a two-dimensional irreducible representation to exist over $K$. Now, it is well-known that the algebra is split precisely when the reduced norm form ($N(\alpha)=\alpha\overline{\alpha}$) restricted to the purely imaginary quaternions has a non-trivial zero. As $N(ai+bj+ck)=a^2+b^2+c^2$ we get the condition that David mentions in the comments.

The situation in positive characteristic is different. As the group algebra can be defined over the prime field and the Brauer group of finite fields is trivial, one only needs for (the mod $p$) character values to be in the field so in that case there is a minimal field.

**Addendum**: The question of a minimal field for all irreducible representations has essentially the same answer. First the field has to contain all character values, then there are still division algebras to split (in characteristic $0$). The example of the quaternion group still illustrates the problem, all characters are rational-valued and one must still kill the quaternion algebra for which there is no unique minimal field.

splitting fields, and looking whether there issmallestsubfield (whether exist) in the algebraic closure $\bar{F}$. Not fixing any representation, sinceeveryirreducible representation over a splitting field remains irreducible even after extension of that splitting field, so not considering only one representation, but allirreducible representations. Is this the thing you want to know? – joseph Oct 11 '11 at 13:38notlinearly ordered, containment is given by divisibility of the exponent $n$. – Torsten Ekedahl Oct 11 '11 at 16:45