By H. Lausch, W. Nobauer

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**Extra resources for Algebra of Polynomials**

**Sample text**

E) f). Let (c, . , ck) be acommon solution of (S) and (S,). Then pi(c,, . , ck) = qr(cl, . , ck), for every equation p , = qi of (S,). Let (b,, .. ,bk) be any solution of ( S ) . 41, there is an isomorphism 8 : A(c,, . , ck) A@,, .. , bk)fixing A elementwise and mapping ci to b,, i = 1, . , k. Hence pi@,, . , bk) = qi(b,, . , bk) and thus (b,, . , bk) is also a solution of (S,). f ) g). e. (5') and (Sl)have some solution in common. e. ( S ) and (S1) are equivalent. g) =+a). Let 0 be the congruenceon A(X, 23) generated by the equa2 0.

Of indeterminates, and A that congruence on A(Y, 022) which is generated by {( . Y/,J, q,(xil. . , x i k ) )1 i E t, I t L}. 21. But 0 is separating, by Th. 23, whence a = b. Hence A can be embedded into C = A(Y, % ) / A ,a %-extension of A . If Z,, is the congruence class of xl,under A, then (XI1, . , 2,J E C k 1, a solution of the given system, for all IEL. We have to show that rn # 11 implies (&, .. , Xmk) # (Xfll, . , X,J. Assume the contrary, then x m t h f l t , t = 1, 2, . , k . 21 yields a polynomial chain in A(Y, %) from x,, to xnt.

We have to find, in a finite number of steps, a word of 112 representing p. By Th. 6, there is an isomorphism p : D[x,, . , x k ] D[x,, . , xk-J [xk]fixing DU {xl, . , xk}elementwise. By Th. 1I , we can find a representation p = (a,n xk)U b, where u,, b, 6D[x,, . ,xk-,], a, == b,, in a finite number of steps. By induction, in a finite number of steps, we can find representations of a,, b, in the 112 belonging to D[xl, . , xk-,]. By substituting these representations into p = (a,nx,)U b, and applying the laws of 23,we get a representation of p by a word of 9.