Show there is no ring map $R=\mathbb{Z}[\sqrt{-3}]\to\mathbb{Z}[i]=S$ such that $1_R\mapsto 1_S$

Question

Let $R=\mathbb{Z}[\sqrt{-3}]$, and let $S=\mathbb{Z}[i]$. For sake of contradiction, assume $\varphi:R\to S$ is a ring map with $\varphi(1_R)=1_S$. Note that $\mathbb{Z}[\sqrt{-3}]$ is a free abelian group on the generators $1,\sqrt{-3}$. Thus $\varphi$ is completely determined by its action on these generators. Clearly $1_R=1\in\mathbb{Z}[\sqrt{-3}]$ and $1_S=1\in\mathbb{Z}[i]$. So $\varphi(1)=1$. Furthermore, $\varphi(\sqrt{-3})^2=\varphi(\sqrt{-3}^2)=\varphi(-3)=-3\varphi(1)=-3$, so $\varphi(\sqrt{-3})=\pm\sqrt{-3}$. This can't happen since $\pm\sqrt{-3}\not\in\mathbb{Z}[i]$. Therefore $\varphi$ cannot exist. Is this correct?

Answer 1

This looks great to me!

Depending on what this is for, you may have a bit of routine lifting to do: what you have is $\phi(\sqrt{-3})^2 \in \mathbb{Z}[i]$ with $\phi(\sqrt{-3})^2= -3$, implying there is an element whose square is $-3$ in $\mathbb{Z}[i]$. As you already know, there is no such element. But you may be required to show this by showing you cannot find $a,b \in \mathbb{Z}$ with $(a+bi)^2= -3$. This is just a matter of expanding and relating parts, which gives $a^2 - b^2= -3$ and $2ab= 0$; obviously, there are no integers $a,b$ satisfying these equations.