Determine the maximal ideals of $\mathbb R^2$ by noting $\mathbb R^2 \cong \mathbb R[x]/(x^2-1)$

Question

This is a letter of an exericse in Artin Algebra and has been asked and answered here as well as by Brian Bi here and Takumi Murayama here. I have a different approach.

An earlier exercise is to prove $\mathbb R^2 \cong \mathbb R[x]/(x^2-1)$. One way to do that is with the surjective homomorphism $\varphi:\mathbb R[x] \to \mathbb R^2$ defined by $\varphi(x)=(1,-1)$. This solution is in the answer here and in the answer by Takumi Murayama.

1. The maximal ideals in $\mathbb R[x]/(x^2-1)$ include the principal ideals $([x-1+(x^2-1)])$ and $([x+1+(x^2-1)])$.

2. The corresponding maximal ideals in $\mathbb R^2$ are principal ideals generated by

$$\overline\varphi([x-1+(x^2-1)])=(\overline\varphi \circ \pi) (x-1) = \varphi(x-1)=(0,-2)$$

and

$$\overline\varphi([x+1+(x^2-1)])=(\overline\varphi \circ \pi) (x+1) = \varphi(x+1)=(2,0)$$

where $\overline \varphi$ is the isomorphism from $R[x]/(x^2-1) \to \mathbb R^2$.

3. The principal ideal generated by $(0,\pm 2)$ is the same as the one generated by $(0,1)$. Also, the one by $(\pm 2,0)$ is the same as the one by $(1,0)$.

4. We know that $((0,1))=0\times \mathbb R, ((1,0))=\mathbb R \times 0$.

5. After proving that $([x-1+(x^2-1)])$ and $([x+1+(x^2-1)])$ are the only maximal ideals in the quotient ring, we can say $((0,1))=0\times \mathbb R, ((1,0))=\mathbb R \times 0$ are the only maximal ideals in $\mathbb R^2$.

Is this correct?

To recap, I answered the maximal ideals of $\mathbb R[x]/(x^2-1)$ instead of for $\mathbb R^2$. Then I converted the ideals from $\mathbb R[x]/(x^2-1)$ to $\mathbb R^2$.

Answer 1

Yes, your argument is fine (modulo that you still need to show that $(x+1)$ and $(x-1)$ are the only maximal ideals of $\Bbb R[x]/(x^2 - 1)$). If you have an isomorphism $f : R\to S,$ then the maximal ideals of $S$ are precisely the images via $f$ of maximal ideals of $R.$

I only have two comments: you can probably remove some of the steps and some of the notation, but if you really want to spell everything out, that's fine. Secondly, there's a minor typo in step 2: $\overline{\varphi}\circ\pi(x+1) = \varphi(x + 1),$ not $\varphi(x-1).$

Another comment not related to the correctness of the proof: as in your first link, it is also not too difficult to argue directly that if you have a finite product of fields $A = k_1\times\dots\times k_n,$ the maximal ideals are precisely the ideals $$\mathfrak{m}_i := \{(x_1,\dots, x_n)\in A\mid x_i = 0\},$$ and you could generalize without too much trouble to a direct product of finitely many commutative rings. For infinite products of fields, the answer is stranger, as you might expect.