Let V be a two dimensional vector space over k. Consider the set of isomorphism classes of all rank 1 $k[x]/x^2$-submodules of $V\otimes k[x]/x^2$. Does this set has a variety structure? Is it tangent bundle of $\mathbb{P}^1$?



tangent bundle of $\mathbb{P}^1$
It may be classical to consider varieties as sets, but I think it is having counterproductive effects on communication in this case. The set you are considering is now known as the set of $k$-points of the variety.


For any commutative ring $k$, the tangent bundle of $\mathbb{P}^1\_k$ can be seen as a functor that sends any $k$-algebra $R$ to the set of rank 1 locally free $R[\epsilon]/(\epsilon^2)$ submodules of $k^2 \otimes\_k R[\epsilon]/(\epsilon^2)$ that are direct summands. This follows from the Hom-functor description of tangent bundles
$$T\_{X/k}(R) = \underline{\operatorname{Hom}}(\operatorname{Spec} k[\epsilon]/(\epsilon^2), X)(\operatorname{Spec} R) = X(\operatorname{Spec} R[\epsilon]/(\epsilon^2)),$$
together with a standard functorial description of $\mathbb{P}^1$. In contrast to the wording in your question, the description of $\mathbb{P}^1$ in terms of submodules does not mention isomorphism classes, and does have a "direct summand" requirement (see for example [this Math.SE page](https://math.stackexchange.com/questions/121105/which-functor-does-the-projective-space-represent)). There is an alternative functorial description in terms of invertible quotient modules (for example in [the stacks project](http://stacks.math.columbia.edu/tag/01ND)), and that is where you use an equivalence relation.


