Proof: Lemma 1: If φ:R→S is a surjective ring homomorphism and R is generated by a subset X (i.e. every element of R is writable as a sum of products with subtraction [or a series of operations] of elements from X), then S is generated by φ(X). Proof: Let s be an arbitrary element of S. For some r∈R, we have s=φ(r) and since r is a series of operations of elements from X, due to the homomorphism's nature we have s is a series of operations of elements from φ(X). ◻
We can see that Z[x] is generated by {x,1}, since the latter generates Z (all potential coefficients) and the former generates xn for all n∈Z+. This implies Q[x] is generated by two elements, and in particular Q is generated by two elements (since we have a surjective homomorphism φ:Q[x]→Q by p(x)↦p(0)). Letting ab and cd denote these generators, we have ad⋅1bd=ab and bc⋅1bd=cd, so that in fact Q is generated by one element; further refer to it as 1z. Letting p∤z be prime, we must have 1p is generated by a series of operations of 1z despite the fact that every such operation can be condensed into some fraction yzn, which cannot simplify to 1p. ◻
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