isomorphism $\psi$ between quotient rings 
Show that there is an isomorphism $\psi$ between the rings $R_1 := \mathbb{Q}[r,s]/(r+s-1, r-r^2, s-s^2)$ and $R_2 := \mathbb{Q}[y]/(y^2-1)$ so that $\psi([r]) = [(1+y)/2]$ and $\psi(s)=[(1-y)/2].$

I could define an surjective homomorphism $\psi$ from $\mathbb{Q}[r,s]$ to $\mathbb{Q}[y]/(y^2-1)$ and show that its kernel is $(r+s-1,r-r^2, s-s^2)$. The first isomorphism theorem for rings gives an isomorphism $\psi_2 : R_1\to R_2$ so that $\psi_2 \circ q = \psi,$ where $q : \mathbb{Q}[r,s]\to R_1$ is the quotient map. Then it suffices to define $\psi$ so that $\psi(r) = [(1+y)/2]$ and $\psi(s) = [(1-y)/2]$. Define $\psi$ by $\psi(f(r,s)) = [f(\frac{1+y}2, \frac{1-y}2)],$ where each $f(r,s)\in \mathbb{Q}[r,s].$ It's easy to show that $\psi$ is a homomorphism. I can also show that $(r+s-1, r-r^2, s-s^2)\subseteq \ker \psi$ and that $\psi$ is surjective using the fact that the elements of $\mathbb{Q}[y]/(y^2-1)$ are of the form $[ay+b]$ for some $a,b \in \mathbb{Q}$. This can be shown using the division algorithm for polynomials. Then one can find $f$ so that $f(\frac{1+y}2, \frac{1-y}2) =ax+b.$ But how do I show that $(r+s-1, r-r^2, s-s^2) = \ker \psi$?

The most elegant way to solve such problems with no effort is to use universal properties and in particular the Yoneda Lemma. In fact, this Lemma tells us that two objects with the same universal properties are isomorphic. A universal property of an object $R$ is just a description of the homomorphisms $R \to S$ for any other object $S$ of the ambient category (or, of the homomorphisms $S \to R$). Here, we are in the category of commutative $\mathbb{Q}$-algebras (of course, you could also work in the category of rings, but it would make life less easy).
So if $S$ is a commutative $\mathbb{Q}$-algebra, then the universal properties of quotient rings ("fundamental theorem on homomorphisms") and polynomial algebras ("evaluation homomorphisms") show us that there are natural bijections
$\mathrm{Hom}(\mathbb{Q}[r,s]/(r+s-1,\, r-r^2,\, s-s^2),S)\\
\cong \{(a,b) \in S^2 : a+b-1=0,\, a-a^2=0,\, b-b^2=0\},$
$\mathrm{Hom}(\mathbb{Q}[y]/(y^2-1),S) \cong \{c \in S : c^2-1=0\}.$
So all we have to do is to find a natural bijection between the sets
$\{(a,b) \in S^2 : a+b=1,\, a=a^2,\, b=b^2\} \cong \{c \in S : c^2=1\}.$
But notice that $a+b=1$ means that $b$ is superfluous, we may replace it by $1-a$. The equation $b=b^2$ then holds automatic, since in general if $a$ is idempotent ($a=a^2$) then $1-a$ is idempotent. (This also shows, by the way, that the ideal $(r+s-1,r-r^2,s-s^2)$ is equal to the ideal $(r+s-1,r-r^2)$.)
So all we have to show is that the set of idempotents in $S$ is naturally isomorphic to the set of involutions in $S$ (i.e. $c^2=1$).
Well, if $a$ is idempotent, then $c := 1-2a$ is an involution, since $c^2=1-4a+4a^2=1$. Conversely, if $c$ is an involution, then $a := (1-c)/2$ is idempotent, since $a^2=(1-2c+c^2)/4=(2-2c)/4=(1-c)/2=a$. These maps are inverse to each other. This finishes the proof.
This correpondence between idempotents and involutions can also be motivated or explained with a bit of algebraic geometry. An idempotent is nothing but a section on $\mathrm{Spec}(S)$ with values in $0,1$ (in the residue fields). An involution is nothing but a section on $\mathrm{Spec}(S)$ with values in $-1,+1$. Now we just need a linear transformation which takes $\{0,1\}$ to $\{-1,+1\}$, for example $1-2x$, to get the correspondence.

Here is the whole proof in one chain of natural isomorphisms:
$\mathrm{Hom}(\mathbb{Q}[r,s]/(r+s-1,r-r^2,s-s^2),S)\\
\cong \{(a,b) \in S^2 : a+b-1=0,\, a-a^2=0,\, b-b^2=0\}, \qquad | ~ \text{ substitute } b=1-a \\
\cong \{a \in S : a=a^2,\, (1-a)=(1-a)^2=0\} \\
= \{a \in S : a=a^2\} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad ~\,\, | ~ \text{ substitute } a=(1-c)/2 \\ 
\cong \{c \in S : (1-c)/2=\bigl((1-c)/2)\bigr)^2\} \\
= \{c \in S : c^2=1\} \\
\cong \mathrm{Hom}(\mathbb{Q}[y]/(y^2-1),S).$
