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

Prove that $\mathbb{Z}+\left(3x\right)$ is a subring of $\mathbb{Z}\left[x\right]$ and there is no surjective homomorphism from $\mathbb{Z}\left[x\right]⇒\mathbb{Z}+\left(3x\right)$
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Norah Small

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