I just wanted to know how in his answer for the ring $\mathbb{R}[x]/(x^2)$, user Georges Elencwajg calculated the size for the ring. I was unable to comment there as I just made my account on stack exchange.


1 Answer 1


Recall that $\mathbb{R}[x]$ is the ring of polynomials in $x$, with coefficients in $\mathbb{R}$. Then if we quotient by the ideal $(x^2)$, we are essentially saying that for any $n\geq 0$, $x^{n+2} \equiv 0 \text{ mod } (x^2)$. This means that the image of any polynomial under the canonical projection $\pi: \mathbb{R}[x] \to \mathbb{R}[x]/(x^2)$ is of the form $ax+b$. Now simply regard this ring as a vector space, with basis $e_1 = 1$ and $e_2 = x$, then it's clear that it is two dimensional over $\mathbb{R}$.

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    $\begingroup$ Hey thanks a lot for the detailed answer. Unfortunately can not give you an upvote but you have my sincere thanks. $\endgroup$
    – acevans
    Feb 16, 2017 at 1:44
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    $\begingroup$ @acevans Note that although you cannot upvote, you should still be able to 'accept' an answer. $\endgroup$
    – johnnycrab
    Feb 16, 2017 at 2:05
  • $\begingroup$ @johnnycrab ahh I see the link now. Thank you $\endgroup$
    – acevans
    Feb 16, 2017 at 16:26

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