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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 $(r+s-1, r-r^2, s-s^2) = \ker \psi$?

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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:

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

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  • $\begingroup$ This is a nice way of thinking about it. +1 $\endgroup$ Aug 8, 2021 at 16:29
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I fear that your approach is more difficult than necessary. Observe that $\mathbb Q[y]$ is a principal ideal domain (PID), hence every ideal is generated by a single element. Consequently, if we can find a surjective ring homomorphism $\varphi : \mathbb Q[y] \to \mathbb Q[r, s]/(r + s - 1, r - r^2, s - s^2)$ such that $(y^2 - 1) \subseteq \ker \varphi,$ then the generator of $\ker \varphi$ must divide $y^2 - 1$ and must therefore be $y - 1,$ $y + 1,$ or $y^2 - 1.$ Based on your objective, you already have a candidate for $\varphi$: let $\varphi(f(y)) = f(2 \bar r - 1),$ where $\bar r$ denotes the image of $r$ in $R_1.$ Observe that $$\bar r = \frac{2 \bar r - 1 + 1} 2 = \varphi {\left(\frac{y + 1} 2 \right)} \text{ and } \bar s = 1 - \bar r = \varphi {\left(1 - \frac{y + 1} 2 \right)},$$ hence $\varphi$ is surjective. Further, we have that $\varphi(y^2 - 1) = (2 \bar r - 1)^2 - 1 = 4 \bar r^2 - 4 \bar r + 1 - 1 = 0,$ from which it follows that $(y^2 - 1) \subseteq \ker \varphi.$ Like I mentioned before, from here, we deduce that $\ker \varphi = (y - 1)$ or $\ker \varphi = (y + 1)$ or $\ker \varphi = (y^2 - 1).$ Once you verify that the former two cases cannot happen, you may conclude that $\ker \varphi = (y^2 - 1)$ so that $R_1 \cong R_2$ via $\varphi.$

Last, you must establish that $\varphi$ and $\psi$ are function inverses. Once you have done this, then you can conclude that $R_1 \cong R_2$ via $\psi,$ as desired.

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  • $\begingroup$ Alternatively, we can construct a homomorphism in the other direction (which is directly suggested by your proof of surjectivity), and check that they are inverse to each other. So we don't need to exclude other cases for the kernel and don't need to use the fact that polynomial rings are PIDs. In fact, this approach (constructing two morphisms in both directions and check that they are inverse) is repeated in many examples, whereas the Yoneda Lemma tells us that we only need to do it once (see my answer). $\endgroup$ Aug 8, 2021 at 13:14

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