there exists an isomorphy between $\mathbb{Q}[a,b]/\langle a+b-1,a-a^2,b-b^2 \rangle$ and $S=\mathbb{Q}[x]/\langle x^2-1\rangle$ Let $R=\mathbb{Q}[a,b]/\langle a+b-1,a-a^2,b-b^2 \rangle$ and $S=\mathbb{Q}[x]/\langle x^2-1\rangle$. Show that there exists isomorphism $\varphi : R \to S $ such that $$\varphi( \overline{a})=\overline{(1+x)/2}, \quad \text{and} \quad\varphi(\overline{b})= \overline{(1-x)/2}$$
Can someone give me some suggestion on this problem? I have tried looking at the elements of $R$, but it has not proved to show the result.
 A: Here is the format of the proof (fill in the blanks).
The formulas $a \mapsto (1+x)/2$ and $b \mapsto (1-x)/2$ give a map $\mathbb{Q}[a, b] \to \mathbb{Q}[x]/(x^2-1)$ because you can map polynomial ring generators to anywhere. By _____ [fill in calculations to check], $a+b-1$, $a^2-a$ and $b^2-b$ all map to $0 \in S$. Therefore the map descends to the quotient and gives $$\phi : \mathbb{Q}[a, b] / (a+b-1, a^2-a, b^2-b) \to \mathbb{Q}[x]/(x^2-1)$$ as you asked for, but it's not clear whether $\phi$ is an isomorphism.
So, next consider the map $\mathbb{Q}[x] \to \mathbb{Q}[a, b] / (a+b-1, a^2-a, b^2-b)$ given by sending $x \mapsto 2a-1$. (Found by solving $a = (1+x)/2$ for $x$.) By _____ [fill in calculations] this map sends $x^2-1$ to $0 \in R$. So it descends to the quotient and gives $$\psi : \mathbb{Q}[x]/(x^2-1) \to \mathbb{Q}[a, b] / (a+b-1, a^2-a, b^2-b).$$
Finally, check that $\phi \circ \psi$ and $\psi \circ \phi$ are the identity by checking the three equalities $\phi \circ \psi(x) = x$, $\psi \circ \phi(a) = a$, $\psi \circ \phi(b) = b$ (fill in the details _____).
