$\Bbb{R}[x]/\langle(x+1)(x^2+1)\rangle$ is isomorphic to $\Bbb{R}\times\Bbb{R}\times\Bbb{R}$

Question

How to show these to rings are isomorphic?

My approach: $$\Bbb{R}[x]/\langle(x+1)(x^2+1)\rangle = \{ax^2+ bx + c + \langle(x+1)(x^2+1)\rangle\mid a,b,c \in\Bbb{R}\}$$

I take $f : \Bbb{R}[x]/\langle(x+1)(x^2+1)\rangle \to \Bbb{R}\times\Bbb{R}\times\Bbb{R}$,

$$f(ax^2+ bx + c + \langle(x+1)(x^2+1)\rangle) = (a,b,c)$$

Now I can't get the multiplication operation right. Everything else is satisfied, i.e. addition, well-definedness, bijection.

Answer 1

Your isomorphism won’t work even for $q(x)$ where $\mathbb R[x]/\langle q(x)\rangle \cong \mathbb R^3.$

For example, if $q(x)=(x+1)(x^2-4),$ you get an isomorphism with $\mathbb R^3$ with the map:

$$f(x)+\langle q(x)\rangle\mapsto (f(-1),f(-2),f(2))$$

or $$ax^2+bx+c+\langle q(x)\rangle\mapsto (a-b+c,4a-2b+c,4a+2b+c)$$

Note that $-1,-2,2$ are exactly the roots of $q(x).$

For $p(x)=(x+1)(x^2+1),$ we don’t have three real roots. Since $p(x)$ has roots $-1,i,-i,$ e might naively try a similar map to $\mathbb R\times \mathbb C^2$:

$$f(x)+\langle p(x)\rangle \mapsto (f(-1),f(i),f(-i))$$ But the thing is, since $f$ is a real polynomial, $f(-i)$ is just the complex conjugate of $f(i).$ In particular, since conjugation is an automorphism on $\mathbb C,$ one of the copies of $\mathbb C$ is redundant.

The final isomorphism with $\mathbb R\times\mathbb C$ is;

$$f(x)+\langle p(x)\rangle \mapsto (f(-1),f(i))$$

or:

$$ax^2+bx+c+\langle p(x)\rangle\mapsto (a-b+c,c-a+bi)$$