Let F be any field and suppose that £ is a separable algebraic extension of F. For elements aEE, we let dga denote the degree of the minimal polynomial of a over F. Let a, BEE, dga = m, dgB = n and suppose {m, n) = l. It is easy to see that [F(a, B)'.F]=mn, and by a standard theorem of field theory (for instance see Theorem 40 on p. 49 of [l]), there exists an element yEE such that F(a, 8) = F{y) and thus dgy = mn. In fact, the usual proof of this theorem produces (for infinite F) an element of

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... the form y= a+\B, with \£F. In this paper we show that in many cases the choice of X£F is completely arbitrary, as long as A 5^0. In Theorem 63 on p. 71 of [l], it is shown that if n>m and n is a prime different from the characteristic of F, then dgia+/3) =mn. The present result includes this. Theorem. Let E^F be fields as above and let a, BEE with dga = m, dgB = n and {m, n) = 1. Then dg(a+X/3) =mnfor aWK^O, \EFunless the characteristic, ch(F) =p, a prime, and (a) p\mn or pm for every prime q\ n, then p\ n. Proof. First we reduce the problem to one of group representations. We may assume without loss that £ is a finite degree Galois extension of F and let G be the Galois group. Then G transitively permutes the sets of roots A = {a,-| 1 ^iSm} and B= {Bj\ l^j^n} of the minimal polynomials of a and 8. Let FC£ be the linear span of A^JB over F. Then V is a G-module over F and in the action of G on V there exists orbits A and B with \A\ =m, IB] =»and {n, n) = 1. We show by induction on | G\ that if aEA and BEB, then a+8 lies in an orbit of size mn, unless ch(F) =p and (a), (b) and (c) hold. This will clearly prove the theorem when applied to X/3 in place of 8. Let H = Ga and K = G$, the stabilizers in G of a and 8. Then \G:H\ =m, \G:K\ =n and since {n, n) = l, a standard argument yields | G'.HC\K\ =mn and H and K act transitively on B and A respectively. It follows that G is transitive on AXB and thus all elements of V of the form at+B, are conjugate under the action of G. Suppose that a+B does not have exactly mn conjugates. Then not all at+Bj are distinct and we may assume that a+B = aa+Bb, where