Problem: Let $R$ be a ring and consider the ring homomorphisms $$ \phi: R[X] \rightarrow R: \sum_{i=0}^n a_i X^{i} \mapsto a_0 $$ and $$ \psi : R[X] \rightarrow R[Y] : \sum_{i=0}^n a_i X^{i} \mapsto \sum_{i=0}^n a_i Y^{2i}. $$
a) Explain how these morphisms induce an $R[X]$-module structure on $R$ and on $R[Y]$.
b) Show that $$R \otimes_{R[X]} R[Y] \cong R[Y]/(Y^2). $$
Attempt: a) Define the action of $R[X]$ on $R$ as $$ R[X] \times R \rightarrow R: (f(X), r) \mapsto a_0 r $$ where $f(X) = \sum_{i=0}^n a_i X^{i}. $
Furthermore, define the action of $R[X]$ on $R[Y]$ by $$ R[X] \times R[Y] \rightarrow R[Y] : ( \sum_{i=0}^n a_i X^{i}, \sum_{j=0}^m b_j Y^{j}) \mapsto \left( \sum_{i=0}^n a_i Y^{2i} \right) \left( \sum_{j=0}^m b_j Y^j \right). $$
b) For this part, I was not sure. I first tried to define a map $$ \chi: R \times R[Y] \rightarrow R[Y] / (Y^2) : (r, \sum_{i=0}^n a_i Y^{i}) \mapsto \sum_{i=0}^n ra_i Y^{i} + (Y^2) $$ and I think it is $R[X]$-bilinear, and then use the universal property of tensor product. But I didn't know how to find the inverse of $\chi$.
I also tried to look at an exact sequence of the form $$ 0 \rightarrow (Y^2) \rightarrow R[Y] \rightarrow R[Y] / (Y^2) \rightarrow 0 $$ and then try to tensor with $R$ maybe, but I could not prove the stated isomorphism in this way.
Help/advice is appreciated.
You can show the latter isomorphism in your comment by noting Max's comment. Otherwise, you can proceed as you did before by describing explicitly the isomorphism.
Your attempt in part b is correct, just need to check the $R[X]$-bilinearity of the map which is basically by definition. One comment here is that we should write the map $\chi$ as $$\chi: R \times R[Y] \rightarrow R[Y] / (Y^2) : (r, \sum_{i=0}^n a_i Y^{i}) \mapsto r(a_0 +a_1 Y^{1}) + (Y^2)$$ The inverse of $\chi$ (the tensor product version of $\chi$) is the obvious one, i.e, $$\phi: R[Y]/(Y^2) \rightarrow R \otimes_{R[X]} R[Y] : a_0 +a_1 Y^{1} + (Y^2)\mapsto 1 \otimes (a_0+a_1y) $$
From this, it is clear that the composition map $R[Y]/(Y^2) \rightarrow R \otimes_{R[X]} R[Y] \rightarrow R[Y] / (Y^2)$ is the identity. In order to check that the map $R \otimes_{R[X]} R[Y] \rightarrow R[Y] / (Y^2) \rightarrow R \otimes_{R[X]} R[Y]$ is also the identity, notice the following \begin{align*} &\ \ \ \ \ a\otimes (b_0+b_1y+b_2y^2+\ldots +b_{2n}y^{2n}+b_{2n+1}y^{2n+1})\\ &=a\otimes (b_0+b_2y^2+\ldots +b_{2n}y^{2n})+a\otimes (b_1y+b_3y^3+\ldots +b_{2n+1}y^{2n+1})\\ &= a\otimes (b_0+b_2x+\ldots +b_{2n}x^{n}).1+a\otimes (b_1+b_3x+\ldots +b_{2n+1}x^{n}).y\\ &= a.(b_0+b_2x+\ldots +b_{2n}x^{n})\otimes1+a.(b_1+b_3x+\ldots +b_{2n+1}x^{n})\otimes y\\ &= a.b_0\otimes 1 +a.b_1\otimes y\\ &= a \otimes (b_0+b_1y) \end{align*} (the $.$ sign is the $R[X]$-multiplication in the modules $R$ and $R[Y]$). Actually this is the reason why we have the isomorphism. I hope this is clear enough.