Let \(K\) be a field, \(M\) be a field extension of \(K\). For a nonempty subset \(S\subset M^ n\) every polynomial \(f\in M[X_ 1,\dots,X_ n]\) defines a function from \(S\) into \(M\). Denote the ring of all such functions by \(M[S]\). \(f_ 1,\dots, f_ m\in M[S]\) are called algebraically independent if for every nonzero polynomial \(g\in M[Y_ 1,\dots, Y_ m]\) there exists \(x\in S\) such that \(g(f_ 1(x), \dots, f_ m(x)) \neq 0\). The maximal number \(d(S)\) of functions in \(M[S]\) which are algebraically independent is called the algebraic dimension of \(S\) [\textit{L. van den Dries}, Ann. Pure Appl. Logic 45, 189-209 (1989; Zbl 0704.03017)]. Let \({\mathfrak L}\) be a first order language which expands the language of rings with constant symbol for each element of \(K\). We call a subset of \(M^{n+1}\) definable if it is defined by some formula of \({\mathfrak L}\). Let \(S\) be a definable subset of \(M^{n+1}\). For each \(a\in M\) let \(S_ a= \{c\in M\mid (a,c)\in S\}\). A set \(S\) is said to be algebraically bounded if there exist \(f_ 1,\dots, f_ m\in M[X,Y]\) such that for each \(a\in M^ n\) for which the set \(S_ a\) is finite and nonempty there exists \(j\) such that \(f_ j (a,Y)\neq 0\) and \(S_ a\subseteq \{c\in M\mid f_ j(a,c) =0\}\). In the present paper it is proved that definable sets over each Frobenius field are algebraically bounded. For a finitely generated \(K\) a primitive recursive procedure to compute the algebraic dimension of \(S(M)\) is established.