Suppose F, E are fields and F is a subring of E. Suppose g is an element of E and is algebraic over F, that is p(g)=0 for some nonzero polynomial p(x) in F[x]. Then there must be a nonzero polynomial m(x) smallest degree among those nonzero polynomials in F[x] with g as a root. Prove m(x) cannot be factored as a(x)b(x) for polynomials a(x), b(x) of smaller degree in F[x].
The suppose q(x) in F[x] with q(g)=0. Show q(x) is an element of m(x)F[x]
(so ker(Eg:F[x]-->E) = m(x)F[x])
Then show that the subring F[g] of E is isomorphic to the quotient ring
1) Prove m(x) cannot be factored as a(x)b(x) for polynomials a(x), b(x) of smaller degree in F[x].
Suppose, it can. Then, m(x)=a(x)b(x), and m(g)=a(g)b(g)=0. Since a(g) and b(g) are elements of the field E, and so E has no zero divisors, we have a(g)=0 or b(g)=0. But if a(g)=0, then since the degree of a(x) is less than the degree of m(x), we have a contradiction with the fact that m(x) has the ...
Ring theory is assessed in the solution.
Differential Geometry : Chain Rule and Differentiable Mappings
Chain Rule. Let M, N and Q be differentiable manifolds, and let φ : M ?> N and N ?> Q be differentiable mappings. Prove that
or simply written
[Comment: To familiarize yourself with notations in Differential Geometry try to check the form that the above equality takes when you express the (differentials of the) maps and in terms of the (Jacobian) matrices obtained by choosing coordinate systems.]