What's $R[x]/(x^2)$ isomorphic to?

Question

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.

Answer 1

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..."

Answer 2

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

$$ f(a + \epsilon) = f(a) + f'(a) \epsilon $$

Answer 3

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?