How to build $\mathbb{C}$

Question

I've defined $\mathbb{C}$ as $\mathbb{R} [X]/ (X^2+1)$, how do I show that $\mathbb{Q} [X]/ (X^2+1)$ is a subset of $\mathbb{C}$? And is $i \in \mathbb{Q} [X]/ (X^2+1)$? And can we see $\mathbb{Q} [X]/ (X^2+1)$ as $\mathbb{Q}^2$ just as we can see $\mathbb{C}$ as $\mathbb{R}^2$?

Answer 1

You have a canonical map $\mathbf{Q}\rightarrow\mathbf{R}$ that induces $\mathbf{Q}[X]\rightarrow\mathbf{R}[X]$ and this one send the ideal $(X^2+1)$ of $\mathbf{Q}[X]$ in the ideal $(X^2+1)$ of $\mathbf{R}[X]$ ensuring that $\mathbf{Q}[X]\rightarrow\mathbf{R}[X]$ that passes to the quotient to give a map $\mathbf{Q}[X] / (X^2 + 1 )\rightarrow\mathbf{R}[X] / (X^2 +1 )$, which gives you the inclusion of $\mathbf{Q}[X] / (X^2 + 1 ) = \mathbf{Q}[i] = \mathbf{Q}(i)$ in $\mathbf{R}[X] / (X^2 + 1 ) = \mathbf{C}$. As $i^2 = -1$, $\mathbf{Q}(i)$ is indeed of dimension $2$ over $\mathbf{Q}$, and is then isomorphic (in the category of $\mathbf{Q}$-vector spaces) to $\mathbf{Q}^2$.