<s>[INST] $\mathbb C[X]/(X^2)$ is isomorphic to $\mathbb R[Y]/((Y^2+1)^2)$ This question led me to the following: 

Prove that $\mathbb C[X]/(X^2)$ is isomorphic to $\mathbb R[Y]/((Y^2+1)^2)$.

 [/INST] We want to construct an isomorphism of $\mathbb{R}$-algebras $\mathbb{R}[T,X]/(T^2+1,X^2) \cong \mathbb{R}[Y]/((Y^2+1)^2)$. By the universal properties of polynomial and quotient algebras, as well as the Yoneda Lemma, this is equivalent to a natural bijection
$\alpha : \{(a,b) \in A^2 : a^2=-1, b^2 = 0\} \cong \{c \in A : (c^2+1)^2=0\}$,
where $A$ runs through all $\mathbb{R}$-algebras. In my opinion, this reformulation catches the real content of the isomorphism: It is a really elementary statement about solutions of polynomial equations. By the way, $\mathbb{R}$ can be replaced by any ring in which $2$ is invertible (but not by an arbitrary ring, consider $A=\mathbb{Z}/4$).
Now here is a direct proof: We define $\alpha(a,b) = a+b$. Then $\alpha$ is well-defined, since $c:=a+b$ satisfies $c^2=a^2+2ab=2ab-1 \Rightarrow (c^2+1)^2=0$. Clearly $\alpha$ is natural.
In order to find or motivate the definition the inverse map, let us solve $\alpha(a,b)=c$ for $a$: We have $0=b^2=(c-a)^2=c^2-2ac+a^2=c^2-2ac-1$, hence $a=(c^2-1)/(2c)$. Here, $c$ is invertible with $1/c = -(c^3+2c)$ since $0=c^4+2c^2+1$. It follows $-2a=(c^2-1)(c^3+2c)=c^5+c^3-2c=c (-1-2c^2)+c^3-2c=-c^3-3c$, hence $a=c(c^2+3)/2$, and therefore $b=c-a=-c(c^2+1)/2$.
So let us define $\alpha^{-1}(c):=(c(c^2+3)/2,-c(c^2+1)/2)$. By construction, this is a map inverse to $\alpha$, as soon as we have shown that it is well-defined. Writing $\alpha^{-1}(c)=(a,b)$, we have $b^2=c^2 (c^2+1)^2/4=0$, and $4a^2+4=c^2(c^2+3)^2+4=(c^2+3)^2 (c^2+1)^2=0$, i.e. $a^2=-1$.
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