> [!NOTE] Lemma ($\mathbb{C}\cong \mathbb{R}[X]/(X^{2}+1)\mathbb{R}[X]$) > Let $\mathbb{R}[x]$ denote the [[Ring of Polynomial Forms|ring of polynomial forms]] over the [[Real numbers|real numbers]]. Let $\mathbb{R}[x]/(x^{2}+1)\mathbb{R}[x]$ denote the [[Quotient Ring|quotient ring]] of $\mathbb{R}[x]$ by $(x^{2}+1)\mathbb{R}[x].$ Let $\mathbb{C}$ denote the [[Complex Numbers|set of complex numbers]]. Then $\mathbb{R}[x]/(x^{2}+1)\mathbb{R}[x]$ is [[Isomorphism of Rings|isomorphic]] to $\mathbb{C}.$ ###### Proof Define $\phi: \mathbb{R}[x] \rightarrow \mathbb{C}$ by $\phi(f)=f(i)$. To see that $\phi$ is a [[Homomorphism of Rings|homomorphism of rings]], observe that for all $f,g\in \mathbb{R}[X]$ $\begin{align} \phi(f+g)&=(f+g)(i)= f(i)+g(i) = \phi(f)+\phi(g) \\ \phi(fg) &= (fg)(i) =f(i)g(i) = \phi(f)\phi(g) \\ \phi(1)&=1. \end{align}$ Now let $\alpha \in \mathbb{C}$. We can write $\alpha=a+b i$ where $a, b \in \mathbb{R}$. Now $\phi(a+b x)=a+b i=\alpha$. So $\phi$ is surjective. Suppose $f \in \operatorname{Ker}(\phi)$. Then $f(i)=0$. Since $i$ is a root of $f$, by [[Conjugate of Complex Root of Real Polynomial is also a Root]], $-i$ is also a root of $f$. Hence $x^2+1=$ $(x-i)(x+i)$ is a factor of $f$. Conversely every multiple of $x^2+1$ is in the kernel. So $\operatorname{Ker}(\phi)=\left(x^2+1\right) \cdot \mathbb{R}[x]$. The [[First Isomorphism Theorem for Rings|First Isomorphism Theorem]] tells us that $\mathbb{R}[x] /\left(x^2+1\right) \cong \mathbb{C}$ where the isomorphism is given by $f(x)+\left(x^2+1\right) \mapsto f(i)$. ###### Proof Let $I=(x^{2}+1)\mathbb{R}[x].$ Define $\begin{align} \phi: \mathbb{R}[x]/I&\to \mathbb{C} \\ f+I &\mapsto f(i) \end{align}$Suppose there exists $f,g\in \mathbb{R}[x]$ such that $\phi(f+I)=\phi(g+I).$ Then $f(i)=g(i)$ which gives $f(i)-g(i)=0$and $f(-i)-g(-i)=0.$ Thus $f(x)-g(x)=(x^{2}+1)h(x)$for some $h\in \mathbb{R}[x]$: that is $f(x)-g(x)\in I.$ By [[Necessary Condition for Equality of Cosets]], $f(x)+I=g(x)+I.$ Thus $\phi$ is injective. By the implications right to left, $\phi$ is [[Well-defined Function with Respect to Equivalence Relation|well-defined]]. We can check that $\phi$ is also surjective.