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This question led me to the following:

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

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  • $\begingroup$ does the right hand ring contain any element whose square is $-1$? $\endgroup$ – user29743 Jun 8 '13 at 17:16
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    $\begingroup$ @countinghaus Yes: $\frac{1}{2}y(y^2+3)$. $\endgroup$ – user26857 Jun 8 '13 at 17:18
  • $\begingroup$ thanks! this is a cool question $\endgroup$ – user29743 Jun 8 '13 at 17:18
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    $\begingroup$ Hint: $\ Y \mapsto X+ i\ $ $\endgroup$ – Key Ideas Jun 8 '13 at 18:01
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    $\begingroup$ More generally, one may ask: For which polynomials $p,q$ over a field or even commutative ring $K$ do we have $K[x,y]/(p(x),q(y)) \cong K[y]/(q(p(y))$? This here is the special case $K=\mathbb{R}, p(x)=x^2+1, q(x)=x^2$. YACP proved it in the other thread when $q(x)=x^k$, $p$ is irreducible and $p' \neq 0$. Note that both $K$-algebras have dimension $\mathrm{deg}(p) \mathrm{deg}(q)$ over $K$, so that it suffices to find a surjective or injective homomorphism. $\endgroup$ – Martin Brandenburg Jun 11 '13 at 8:43
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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 $\,\Bbb R$ $$\Bbb R[y]/((y^2\!+1)^2) \cong \Bbb R\langle 1,y,y^2,y^3\rangle \cong \Bbb R^4 \cong \Bbb R\langle 1,i,x,ix\rangle \cong \Bbb C[x]/(x^2)$$

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  • $\begingroup$ Is there a "nice" way to see that the kernel is precisely $(y^2 + 1)^2$. Clearly $(y^2 + 1)^2$ is in the kernel. $\endgroup$ – messi Jun 10 '13 at 7:17
  • $\begingroup$ @messi: I will answer this in my answer. $\endgroup$ – Martin Brandenburg Jun 10 '13 at 14:57
  • $\begingroup$ @messi See the edit above for one simple direct way. $\endgroup$ – Key Ideas Jun 10 '13 at 15:49
  • $\begingroup$ @key Ideas, very good, that is really "nice" and now the solution is complete. $\endgroup$ – messi Jun 10 '13 at 16:04
  • $\begingroup$ @messi Thanks. See also the answers by KCd and Mariano for some motivation. My original motivation was a combination of both. I didn't say more explicitly in the original comment since I was sure the OP could discover this based on knowledge exhibited in prior answers. $\endgroup$ – Key Ideas Jun 10 '13 at 16:28
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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 $\sqrt{1+t}$.)

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  • $\begingroup$ This only shows how to construct the map in one direction. $\endgroup$ – Martin Brandenburg Jun 10 '13 at 16:10
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    $\begingroup$ @MartinBrandenburg: I was giving the main idea to get things started. Once you have the substitution map in hand, you can check the kernel is $((y^2+1)^2)$, and then by a dimension argument it has to be surjective too. $\endgroup$ – KCd Jun 10 '13 at 18:10
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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 $n=2$.)

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  • $\begingroup$ +1 Well said. Classifying rings of dual numbers is not so difficult, and provides nice examples pedagogically. $\endgroup$ – Key Ideas Jun 8 '13 at 19:10
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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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    $\begingroup$ +1 Very nice exposition. I think it would prove very helpful to many readers if you expanded a bit on the Yoneda viewpoint, so that readers not proficient in category theory might still be able to grasp the gist of the matter. $\endgroup$ – Key Ideas Jun 10 '13 at 17:05
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    $\begingroup$ Notice that this is nothing but language on top of KeyIdeas's old comment «Hint: consider $Y\mapsto Y+i$» and YACP's comment «Yes, [a root of $-1$ is] $\frac{1}{2}y(y^2+3)$». The real content of nothing is just language. :-) $\endgroup$ – Mariano Suárez-Álvarez Jun 10 '13 at 17:30
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    $\begingroup$ @Mariano: It's definitely more than just language. Also note that my proof is constructive, no dimension arguments etc. are needed. $\endgroup$ – Martin Brandenburg Jun 10 '13 at 21:03

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