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

Question

This question led me to the following:

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

Answer 1

Try to write down a ring homomorphism ${\mathbf R}[y] \rightarrow {\mathbf C}[x]/(x^2)$ with kernel $((y^2+1)^2)$. Build it as a substitution map (evaluation of polynomials in ${\mathbf R}[y]$ at some element of ${\mathbf C}[x]/(x^2)$). Since you want to kill off $(y^2+1)^2$ but not $y^2+1$, you'd like to send $y^2 + 1$ to $x$ (or to any nonzero scalar multiple of $x$), and you need to figure out, with that goal in mind, where $y$ itself ought to be sent. (Hint: think about the first few terms of the Taylor expansion of $\sqrt{1+t}$.)

Answer 2

Both rings are commutative and contain a copy of $\mathbb C$, so it is enough to show that they are isomorphic as $\mathbb C$-algebras.

Now, as such, they are two dimensional, and there is exactly one isomorphism class of two dimensional complex local $\mathbb C$-algebras.

(Let us show this: such an algebra is necessarily commutative and generated by one element over $\mathbb C$, so it is a quotient of $\mathbb C[X]$ by the ideal generated by some polynomial. As the algebra is to be local, that polynomial has to have exactly one root in view of the CRT. The algebra is of the form $\mathbb C[X]/((X-a)^n)$ for some $a\in\mathbb C$. Up to isomorphism we can clearly suppose that $a=0$. Since our algebra has dimesion two, we must have $n=2$.)

Answer 3

Hint: $ $ Consider $\, y\mapsto x + i\ $ (comment promoted to answer per OP's request)

Edit: to answer a comment, let $\smash[t]{\ \Bbb R[y]\stackrel{\large y\,\mapsto\, x+i}\to\Bbb C[x]/(x^2)\ }$ have kernel $\,I,\,$ the ideal of polynomials $\, f(y)\in \Bbb R[y]\,$ such that $\,f(x+i) = 0\,$ in $\,\Bbb C[x]/(x^2).\,$ Note $\, f(y) = (y^2+1)^2 \mapsto (2ix)^2 = 0,\,$ thus $\, f\in I.\,$ By $\,\Bbb R[y]\,$ Euclidean $\Rightarrow$ PID, $\,I = (g),\,$ thus $\,g\mid f = (y^2+1)^2.\,$ Since $\,y^2+1 = p\,$ is irreducible, so prime in the PID $\Rightarrow$ UFD $\,\Bbb R[y],\,$ the only proper factor of $\,f = p^2\,$ is $\, p = y^2+1.\,$ But $\,p\not\in I,\,$ since $\,p(x\!+\!i) = 2ix\ne 0\,$ in $\,\Bbb C[x]/(x^2).\,$ Therefore $\, g = f = p^2,\,$ so $\, I = (g) = (f).$

The map is onto since the image $\, \supseteq\Bbb R,\, $ and $\,h =(y^3\!+3y)/2\mapsto \color{#c00}i,\,$ so $\, y-h\mapsto (x\!+\!i)\!-\!\color{#c00}i = \color{#0a0}x,\, $ hence the image contains $\, \Bbb R[\color{#c00}i,\color{#0a0}x] = \Bbb C[x]\pmod{x^2}.\,$ Or, note both have dimension $4$ over $\,\Bbb R$ $$\Bbb R[y]/((y^2\!+1)^2) \cong \Bbb R\langle 1,y,y^2,y^3\rangle \cong \Bbb R^4 \cong \Bbb R\langle 1,i,x,ix\rangle \cong \Bbb C[x]/(x^2)$$

Answer 4

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. $a^2=-1$.