I have the following problem.
Consider the ring $\mathbb{Z}$ and define: $$x = \sqrt{-7}\qquad z = \frac{2+3x}{4}$$ Show that $\mathbb{Z}[x] \not\subset \mathbb{Z}[z]$ and $\mathbb{Z}[z] \not\subset \mathbb{Z}[x]$.
First of all, I describe the ring extensions: \begin{align} \mathbb{Z}[x] &= \left\{a_0 + a_1 x + a_2 x^2 +a_3 x^3 + \ldots \ | \ a_i \in \mathbb{Z}\right\}\\ &= \left\{a + b x \ | \ a,b \in \mathbb{Z}\right\} \end{align} Where we use $x^2 + 7 = 0$ ($x$ is an algebraic integer). On the other hand \begin{align} \mathbb{Z}[z] &= \left\{a_0 + a_1 z + a_2 z^2 +a_3 z^3 + \ldots \ | \ a_i \in \mathbb{Z}\right\} \end{align} admits no further simplification (we have $16z^2 - 16z + 67 = 0 \Rightarrow z$ is not an algebraic integer).
I think it is easy to show $z \not\in \mathbb{Z}[x]$. To see this, assume you can find $a,b\in \mathbb{Z}$ such that $z = a+bx$. Now consider this identity in $\mathbb{Q}(x) = \mathbb{Q}[x]$, where $\{1,x\}$ is a basis. Since $$z=\frac{2}{4} + \frac{3}{4}\!x= a +bx $$ it must be $a=\frac{2}{4}$, $b=\frac{3}{4}$, so $a, b$ are not in $\mathbb{Z}$. Thus $z\not\in\mathbb{Z}[x]$.
- Is this correct?
- Any idea for the other part?
Your proof for $z\not\in\mathbb{Z}[x]$ is correct. For $x\not\in\mathbb{Z}[z]$, a nice trick is to consider what's going on mod 3. Note that $\mathbb{Z}[z]\cong\mathbb{Z}[t]/(16t^2-16t+67)$ and $2$ is a root of $16t^2-16t+67$ mod $3$, so there is a homomorphism $\varphi:\mathbb{Z}[z]\to \mathbb{F}_3$ such that $\varphi(z)=2$. (Intuitively, this is just a rigorous version of saying that mod $3$, $z=\frac{2+3\sqrt{-7}}{4}=\frac{2}{4}=2$.)
Now suppose $x$ were in $\mathbb{Z}[z]$. Then $\varphi(x)$ would satisfy $\varphi(x)^2=-7=2$ in $\mathbb{F}_3$. Since $2$ is not a square in $\mathbb{F}_3$, this is a contradiction.