Find the field of fractions and the integral closure of a subring of $\mathbb Z[x]$.

Question

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

1. Find the field of fractions of $R$.
2. Find the integral closure of $R$ in it's field of fractions.
3. Show that $R$ cannot be generated as a ring by 1 and a polynomial $f(x) \in R$.

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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?

1. 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 $\mathbb{Q}(x)$.

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

3. 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.

Answer 1

(1) seems ok.

For (2), notice that $x$ is a root of the monic polynomial $g(t)=t^3-x^3$ which is defined over $R$, so that $R$ is not integrally closed.

For (3), even if you have an isomorphism from $\mathbb{Z}[t]\to R$, it doesn't mean that $R$ as a subring of $\mathbb{Z}[x]$ must be the whole ring. For example, $\mathbb{Z}[x^2]$ is isomorphic to $\mathbb{Z}[x]$. (hint: if $t\mapsto f(t)$ then $g(t)\mapsto g(f(t))$. Can such a homomorphism be onto?)

Answer 2

I'm guessing your argument for $2$ is that you can use factorization in $\mathbf{Z}[x]$ to factor in $R$.

The problem is that when you use factorization in $\mathbf{Z}[x]$, the factors will may not be in $R$! In particular, consider the problem of factoring of $x^8$.

Answer 3

2. 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 $\mathbb Z[x]$.

3. 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.