Prove that Z+(3x) is a subring of Z[x] and there is no surjective homomorphism from Z[x]→Z+(3x)

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zevillageobau · Accepted Answer

Step 1There is no surjective homomorphism. Suppose thatφ:Z[x]→Z+(3x).Sincedegφ(f(x))=deg(f(φ(x))=deg(f(x))deg(φ(x))&amp;nbsp;, if&amp;nbsp;degφ(x)&amp;gt;1then nothing in the image could be of degree 1, and the map could not be surjective. Soφ(x)=3ax+b&amp;nbsp;for some&amp;nbsp;a,b∈ZIf&amp;nbsp;φ&amp;nbsp;were surjective, something would have to map to 3x, and this forces&amp;nbsp;a=±1By composing with the isomorphisms of&amp;nbsp;Z[x]&amp;nbsp;given by maps&amp;nbsp;x↦x−c&amp;nbsp;and&amp;nbsp;x↦−x, we may WLOG assume that&amp;nbsp;φ(x)=3xStep 2The question we ask is, can we find g(x) such thatφ(g(x))=g(3x)=3x2?Working over&amp;nbsp;Q[x], we see if&amp;nbsp;g(3x)=3x2, then&amp;nbsp;g(x)=x23∉ℤ[x]. And because&amp;nbsp;φ&amp;nbsp;extends to an isomorphism of&amp;nbsp;Q[x], this is the only polynomial in&amp;nbsp;Q[x]&amp;nbsp;that would satisfy the condition. So the map cannot be surjective.

Prove that \(\displaystyle{\mathbb{{{Z}}}}+{\left({3}{x}\right)}\) is a subring of

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