Find the field of fractions and the integral closure of a subring of $\mathbb Z[x]$. Let $R$ be a subring of $\mathbb{Z}[x]$ consisting of polynomials such that the coefficients of $x$ and $x^2$ are zero. 


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*Find the field of fractions of $R$.

*Find the integral closure of $R$ in it's field of fractions.

*Show that $R$ cannot be generated as a ring by 1 and a polynomial $f(x) \in R$.



I feel like the answers I have are correct, but at the same time I feel like I'm missing something here. Do these solutions look right?


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*Let $\mathbb{K}$ be the field of fractions, then certainly $\mathbb{Q} \subset \mathbb{K}$ because of $\mathbb{Z}$. Also $\frac{1}{x^3} \in \mathbb{K}$ so $\frac{1}{x^3} \cdot x^5 = x^2 \in \mathbb{K}$, likewise $\frac{1}{x^4} \cdot x^5 = x \in \mathbb{K}$. Thus $\mathbb{K}$ must be $\mathbb{Q}(x)$. 

*$R$ is a UFD because $\mathbb{Z}[x]$ is a UFD, thus $R$ is integrally closed. i.e., it's integral closure is itself.

*Suppose it were, then consider the homomorphism, $\phi: \mathbb{Z}[t] \to R $, such that $t \mapsto f(x) $. Then $\phi$ is injective, implying $R = \mathbb{Z}[x]$, contradiction.




*Since $x$ is integral over $R$ the extension $R\subset\mathbb Z[x]$ is integral. Moreover, $\mathbb Z[x]$ is integrally closed, so the integral closure of $R$ (in its field of fractions) is $\mathbb Z[x]$.

*Actually $R=\mathbb Z[x^3,x^4,x^5]$. The question ask to prove that $R$ can't be generated as a $\mathbb Z$-algebra by a single element. If $R=\mathbb Z[f]$ with $f\in R$, then $R$ is isomorphic to a polynomial ring over $\mathbb Z$, so $R$ is integrally closed, a contradiction.
