Is $(x)\otimes_{k[x]/(x^2)}(x)$ zero? 
I am trying to decide if $(x)\otimes_{k[x]/(x^2)}(x)$ is zero. 

So I considered $x \otimes x$ which I rewrote as $1 \otimes x^2 = 1 \otimes 0 = 0$. But then I realized that $1$ does not live in $(x)$. Now I am unsure how to proceed. Also it's beginning to feel like $x \otimes x \neq 0$. Thanks.
 A: Let $A = k\left[x\right]/\left(x^2\right)$, and let us denote the projection of $x \in k\left[x\right]$ onto $A$ by $x$ (by abuse of notation). I assume that your $\left(x\right)$ means the $A$-submodule of $A$ spanned by $x$. And your question is: Is $\left(x\right) \otimes_A \left(x\right) = 0$ ?
I claim that the answer is "No". Indeed, there is an $A$-bilinear map $\left(x\right) \times \left(x\right) \to k\left[x\right]/\left(x\right)$ (note that the right hand side of this is an $A$-module, in the obvious way) sending every $\left(xa,xb\right)$ to the projection of $ab$. You need to check that this map is well-defined (any given element of $\left(x\right)$ can be written as $xa$ for several different polynomials $a$, and you need to check that they all lead to the same $ab$). But once this is proven, it follows that there is a nonzero $A$-bilinear map from $\left(x\right) \times \left(x\right)$ to an $A$-bimodule (indeed, this map is clearly nonzero, as it sends $\left(x,x\right)$ to $1$). But this map factors through $\left(x\right) \otimes_A \left(x\right)$. Thus, $\left(x\right) \otimes_A \left(x\right)$ cannot be $0$ (or else it would factor through $0$, which flies in the face of its being nonzero).
Actually it is easy to see that $\left(x\right) \otimes_A \left(x\right) \cong k\left[x\right] / \left(x\right)$, and also the $A$-module $\left(x\right)$ itself is isomorphic to $k\left[x\right] / \left(x\right)$.
A: The following observations are made en passant in the accepted answer. I think they deserve a full proof.

Let $R=K[X]/(X^2)$, and $x$ be the residue class of $X$ modulo the ideal $(X^2)$. Then $$\operatorname{Ann}_R(x)=Rx.$$ 

Since $x^2=0$ we have $Rx\subseteq\operatorname{Ann}_R(x)$. For the converse, let $f(x)\in\operatorname{Ann}_R(x)$. Then $f(x)x=0\Rightarrow f(X)X\in(X^2)\Rightarrow f(X)\in(X)\Rightarrow f(x)\in Rx$.
Since $Rx\simeq R/\operatorname{Ann}_R(x)$ we get $Rx\simeq R/Rx$, and this gives us $$Rx\otimes_RRx\simeq R/Rx\otimes_RR/Rx\simeq R/Rx.$$
