Show $\mathbb R^2$ is not isomorphic $\mathbb{R}[x]/(x^2)$

Question

I need a simple way to show $\mathbb R^2$ is not isomorphic $\mathbb{R}[x]/(x^2)$. Both are not integral domains, and both are not fields, so I’m not sure how to go about it.

What structure do you have on $\mathbb{R}^2$? – Paul Mar 02 '20 at 14:31 — Paul, Mar 02 '20 at 14:31
@Paul its point wise addition and multiplication – citizenfour Mar 02 '20 at 14:31 — citizenfour, Mar 02 '20 at 14:31
It follows from this post. – Dietrich Burde Mar 02 '20 at 16:12 — Dietrich Burde, Mar 02 '20 at 16:12

score 2 · Answer 1 · answered Mar 02 '20 at 14:48

I think that there are several ways to see this.

One way is to look at ideals and for you may recall that given a quotient $A/I$ there is a one to one correspondence between ideals of $A/I$ and ideals of $A$ containing $I$ and furthermore that prime ideals correspond to prime ideals (this is a very standard fact, so I will not reproduce the proof here).

It follows that the ideal $(\bar x)\subset{\Bbb R}[x]/(x^2)$ is the only prime ideal in that quotient since $(x)$ is the only prime ideal containing $(x^2)$ in ${\Bbb R}[x]$.

On the other hand ${\Bbb R}\times{\Bbb R}$ has (at least) two prime ideals, namely $P_1={\Bbb R}\times(0)$ and $P_2=(0)\times{\Bbb R}$.

Thus the two rings cannot be isomorphic.

score 1 · Answer 2 · answered Mar 02 '20 at 15:32

The ring $\mathbb{R} \times \mathbb{R}$ has 4 ideals, namely $0$, $\mathbb{R} \times 0$, $0 \times \mathbb{R}$, and $\mathbb{R} \times \mathbb{R}$. In contrast, the ring $\mathbb{R}[x]/(x^2)$ has only 3 ideals, namely $0$, $(x)$, and $\mathbb{R}[x]/(x^2)$. Hence, they could not be isomorphic.

score 0 · Answer 3 · answered Mar 02 '20 at 14:31

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Everything you want to know about $\mathbb{R}^2$ can be understood through understanding the two elements $e_1 = (1, 0)$ and $e_2 = (0, 1)$. So let's look at their properties. Both satisfy $e_i^2 = e_i$ and together they satisfy $e_1e_2 = 0$.

One approach would be to show that no elements $a, b$, with $a^2 =a, b^2 = b, ab = 0$ can exist in $\mathbb{R}[x]/(x^2)$.

answered Mar 02 '20 at 14:31

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Or alternatively; that no nonzero element of $\Bbb{R}^2$ satisfies $x^2=0$. – Servaes Mar 02 '20 at 14:33
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@Servaes Since that is completely different, you ought to post your own solution. Unless the two of you really do work out that you want to rewrite this one to say that :) – rschwieb Mar 02 '20 at 14:38
Thank you for your help – citizenfour Mar 02 '20 at 14:41

score 0 · Answer 4 · answered Mar 02 '20 at 17:10

One possibility (among many others) is to combine the following observations:

(1) $\epsilon = x + (x^2) \in \mathbb{R}[x]/(x^2)$ satisfies $\epsilon^2 = 0$.

(2) any isomorphism must map nonzero nilpotent elements to nonzero nilpotent elements

(3) $\mathbb{R}^2$ does not have any nonzero nilpotent elements.

Show $\mathbb R^2$ is not isomorphic $\mathbb{R}[x]/(x^2)$

4 Answers4