Can someone explain what the quotient ring $R[x]/(x^2)$ is isomorphic to? I know it's weird because it's reducible/has double root, but I'm not exactly sure what the implications of that, or how to proceed.
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It doesn't really matter, but what is R? Any ring? Or the field of real numbers? As for the question: R[x]/(x^2) is isomorphic to R[x]/(x^2). Since this answer will certainly not be enough, please specify your question. – Martin Brandenburg Aug 12 '14 at 09:46
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Hint: Intuitively, you are "killing" polynomials of degree $2$ or higher in the quotient ring. So only degree $0$ or degree $1$ polynomials will "survive", and they would have the following multiplication rule:
$(a+bx)(c+dx)=ac+(ad+bc)x$.
Can you make this formal?
voldemort
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Usually, it's the other way around: you find an isomorphism $A \to \mathbb{R}[\epsilon]/(\epsilon^2)$ because you understand $\mathbb{R}[\epsilon]/(\epsilon^2)$ and want to understand $A$.
To better understand $\mathbb{R}[\epsilon]/(\epsilon^2)$, it may be of use to note that if $f$ is a polynomial, then
$$ f(a + \epsilon) = f(a) + f'(a) \epsilon $$
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1Great answer. Unfortunately, we cannot write $f'(a) = \frac{f(a+\epsilon)-f(a)}{\epsilon}$ since $\epsilon$ is a zero divisor. – Martin Brandenburg Aug 12 '14 at 12:20
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I'm not sure if it's what you're after, but it's often called the ring of dual numbers over $\mathbb{R}$.
It's the coordinate ring of the variety* consisting of a single double point at the origin on the real number line.
*As Martin points out below, "variety" typically refers to a reduced scheme (with additional properties). By definition, a single fat point is not reduced since it's local ring has nilpotents. So I should really say "...of the non-reduced scheme consisting of..."
Derek Allums
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A little caution: Often varieties are assumed to be reduced. I would speak of a scheme here. – Martin Brandenburg Aug 12 '14 at 09:47
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