Verified answer

Consider the mapping f : P -> R[Xm+1,...,Xn] given by f(p(X1,...,Xn)) = p(a1,...,am,Xm+1,...,Xn) for all p in P. It's a ring homomorphism:

f(p(X1,...,Xn)) + f(q(X1,...,Xn)) = p(a1,...,am,Xm+1,...,Xn) + q(a1,...,am,Xm+1,...,Xn) = (p + q)(a1,...,am,X1,...,Xm) = f((p + q)(X1,...,Xn)),

f(p(X1,...,Xn)) f(q(X1,...,Xn)) = p(a1,...,am,Xm+1,...,Xn)q(a1,...,am,Xm+1,...,Xn) = (pq)(a1,...,am,Xm+1,...,Xn) = f((pq)(X1,...,Xn))

for all p, q in P. It's kernel is {p(X1,...,Xn): p(a1,...,am,Xm+1,...,Xn) = 0} = {p(X1,...,Xn): X1 - a1,...,Xm - am are factors of p(X1,...,Xm)} = p. It's onto, since given p(Xm+1,...,Xn) in R[Xm+1,...,Xn], f(p(0,0,...,0,Xm+1,...,Xn)) = p(X1,...,Xm). Hence, the first isomorphism theorem for rings gives P/p ≈ R[Xm+1,...,Xn].

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