All the roots of an irreducible polynomial are called conjugates. We sometimes say they are conjugate to each other. If this seems like a new definition, it's not. Both i and -i, complex conjugates, are roots of x2+1. In fact any two complex conjugates are roots of the same quadratic polynomial in the reals.
Let F/K be a splitting field for p(x). In other words, F contains all the roots of p(x), and is generated by those roots.
Let u and v be two such roots. We know there is an isomorphism c() from K(u) onto K(v), as described earlier. We would like to extend this to an automorphism on all of F.
Realize that a field extension need not contain all the roots of p(x). Adjoin the positive cube root of 2 to the rationals and the extension is contained in the reals. The other two roots are complex, and must be brought in via a second extension. Obviously one root cannot be mapped onto another if no other roots are present. But we can set these concerns aside, because F/K is a splitting field by assumption; it contains all the roots of p(x).
Since the dimension of K(u) = the dimension of K(v) = the degree of p, neither can properly contain the other. If K(u) = K(v), our isomorphism from K(u) into K(v) is an invertible linear transformation from an n dimensional K vector space into itself, hence it is onto, and it defines an automorphism. If K(u) ≠ K(v), we have more work to do.
Remember that the ring of polynomials over a field exhibits unique factorization. Factor p(x) over K(u). We know p(x) is irreducible over K, but it may factor in K(u). In fact it must, since u is a root. We can, at minimum, divide by x-u. In any case, we have a factorization for p(x). apply c() to the coefficients of the resulting polynomials. This gives the factorization of p(x) over K(v).
Return to K(u), where p has been factored into smaller polynomials with coefficients in K(u). We know that p(v) = 0, hence there is some ireducible polynomial q(x), a factor of p(x), with q(v) = 0.
Let c(q) = r. In other words, r is the image of q, the result of applying c() to the coefficients of q(x). Once again r is irreducible over K(v).
Let w be any root of r(x), and map v onto w. We know F contains w, since it contains all the roots of p(x). Now K(u)(v) has a basis of ui*vj with coefficients from K. Multiplication of basis elements is reduced mod p(x) and q(x). At the same time, K(v)(w) has a basis vi*wj with coefficients in K, and the same rules for multiplication. The only difference is the symbols. We have two isomorphic fields that intersect in K(v). Their dimensions, over K(v) and over K, are equal.
If w is contained in K(u,v) then we have an automorphism; otherwise we have more work to do. Repeat the above procedure. Factor q(x) in K(u,v) and let s(x) be the factor with s(w) = 0. Map s(x) to t(x) in K(v,w), let z be a root of t(x), and let c(w) = z. Then extend the isomorphism from K(u)(v)(w) onto K(v)(w)(z). This cannot continue forever, because p(x) has a finite number of roots. Eventually we produce an automorphism on E/K, where E is K adjoin some of the roots of p(x). The automorphism of E fixes K, and maps u onto v.
If E does not contain all the roots of p(x), factor p(x) in E and let q(x) be an irreducible polynomial. Let q(u) = 0. Apply c to the coefficients of q to get r, then set c(u) = v, where v is any root of r. We know F contains v, since F contains all the roots of p(x). This is the same procedure that built E in the first place. Apply it again and again, until c() is defined on all the roots of p(x). In other words, c() is an automorphism on F.
Remember that u and v were arbitrary. Within the splitting field of p(x), one root looks just like another. Any root can be mapped onto any other, and the isomorphism extends to all of F.
Disturbing as it may seem, when extending the rationals to include the roots of x2-2, we cannot tell sqrt(2) from -sqrt(2). Somebody could swap them, and relabel the rest of the extension accordingly, and we wouldn't know the difference.
This is not true for transcendental elements. Let K be any field, let u be an indeterminant, let F = K(u), and let v = u2, another transcendental element in F. The isomorphism c() maps u onto v and fixes K. Note that K(u,v) = K(u). Thus the map carryies K(u) into itself. If you want to think in terms of u, c() doubles the degree of every term of every polynomial. Since u has no preimage, the map is not onto; it is not an automorphism. And we can't extend it, because it is already defined on all of F. Transcendental elements in the same field extension are usually distinct; there is no automorphism mapping one to the other.
Since conjugates are indistinguishable, they should have the same multiplicity.
Let u and v be distinct roots of an irreducible polynomial p(x) over K. Embed these roots in the splitting field F/K. An isomorphism maps u to v, and if we factor p(x) in K(u), and apply the isomorphism, we have factored p(x) in K(v). The monomials x-u and x-v correspond. The number of factors x-u in one polynomial equals the number of factors x-v in the other. The multiplicity of the two roots is the same. These roots are arbitrary, hence the multiplicity of all the roots of an irreducible polynomial is the same. The number of distinct roots times the common multiplicity gives the degree of p(x). We will see that this multiplicity is almost always 1.
Although the roots u and v are indistinguishable inside their splitting field F, this may not be true in the context of a larger field E. For instance, E might be F adjoin the square root of u, which may or may not include the square root of v. If not, there is no E automorphism mapping u to v. Let's see when the automorphism can be extended beyond the splitting field.
Let c be an automorphism on F/K, inside a larger algebraic field E. If E contains s, derive p(x) over the base field K, such that p(s) = 0, and assume E contains all the conjugates of s. If this holds for all s in E, we can extend c() to all of E.
Adjoin y to F, such that y is a root of p(x) over K. Factor p(x) in E and let q be the irreducible polynomial satisfying q(y) = 0. Apply c to the coefficients of q, giving r. Let z be any root of r. We know that z is in E, since y and z are both roots of p(x). Let c(y) = z, and extend the isomorphism accordingly. If F(y) = F(z) we have a larger automorphism. If not,we have more work to do. But I've already described this "work" above.
After a few iterations, all the roots of p(x) have been folded into F, and c is an automorphism on this larger field. If E is still larger than F, find another element y in E-F and proceed as above. If E/F is finite, or countable, this process produces an automorphism on all of E, mapping u to v. If E/F has an uncountable basis, we need a new approach.
Let G be the closure of E, which is the closure of F and of K. We know that such a closure can be built. Furthermore, the automorphism on F extends to an automorphism on G. Let c be such an automorphism, and restrict c to E. The image c(s), for every s in E, is another root of the polynomial defined by s, and that root also lies in E. Thus c maps E into itself. Consider c(s), c(c(s)), c(c(c(s))), etc, until you get back to s. There are only so many roots of any given polynomial, so you always come back to s. Therefore c is onto, and is an automorphism on E.
In summary, let u and v be roots of the same polynomial over K, and let E be a normal extension of K. This could be the splitting field of p(x) (minimal), or the closure of K (maximal), or any number of intermediate extensions. In any case, there is an automorphism on E that maps u to v, whence u and v are indistinguishable inside E.
Hey! What's a normal extension?
Sorry about the forward reference; it is defined on the next page.