Prove that \mathbb{Z}+(3x) is a subring of \mathbb{Z}[x] and there

dooporpplauttssg 2022-05-02 Answered
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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Answers (1)

Norah Small
Answered 2022-05-03 Author has 12 answers

Step 1
There is no surjective homomorphism. Suppose that
φ:Z[x]Z+(3x).
Since
degφ(f(x))=deg(f(φ(x))=deg(f(x))deg(φ(x)) , if degφ(x)>1
then nothing in the image could be of degree 1, and the map could not be surjective. So
φ(x)=3ax+b for some a,bZ
If φ were surjective, something would have to map to 3x, and this forces a=±1
By composing with the isomorphisms of Z[x] given by maps xxc and xx, we may WLOG assume that φ(x)=3x
Step 2
The question we ask is, can we find g(x) such that
φ(g(x))=g(3x)=3x2?
Working over Q[x], we see if g(3x)=3x2, then g(x)=x2/3[x]. And because φ extends to an isomorphism of Q[x], this is the only polynomial in Q[x] that would satisfy the condition. So the map cannot be surjective.

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