How can I prove that $\mathbb Z[x]/(1+x^2)\mathbb Z[x]$ is a free module with basis $\{1,\bar x\}$? 1) How can I prove that $\mathbb Z[x]/(1+x^2)\mathbb Z[x]$ is a free $\mathbb{Z}$-module with basis $\{1,\bar x\}$?
I wanted to prove that $$\mathbb Z[x]/(1+x^2)\mathbb Z[x]\cong \mathbb Z^2,$$
but it looks complicate. 
2) Is $\mathbb Z[x]$ a free $\mathbb Z-$module? I would say yes and that $\{1,x,x^2,...\}$ is a basis, but how can I prove it? 

1) Over any commutative ring $R$, the quotient ring $R[X]/(f(X))$ of $R[X]$ by a monic polynomial is a finitely generated free $R$-module, with rank equal to the degree of the polynomial. 
Denoting by $x$ the class of $X$ in the quotient, you just have to prove that any $x^n$, with $n\ge \deg f$ lies in the submodule generated by $\;1, x,\dots,x^{\deg f}$ (simple induction), and that these elements are linearly independent.
2) You're perfectly right. It is part of the definition: $R[X]$ is the free algebra of the monoid $\mathbf N$. In particular, as an $R$-module it is simply $R^{(\mathbf N)}$ (functions from $\mathbf N$ to $R$ with finite support).
