$\mathbb{R}[x]/\langle x^2+1\rangle$ and $\mathbb{R}[x]/\langle x^3 +x+1\rangle$ are isomorphic?

Question

$\mathbb{R}[x]/\langle x^2+1\rangle$ and $\mathbb{R}[x]/\langle x^2 +x+1\rangle$ are isomorphic or not?

I guess these are isomorphic as they are isomorphic to the field of complex number. But how can I answer about the following question:

$\mathbb{R}[x]/\langle x^2+1\rangle$ and $\mathbb{R}[x]/\langle x^3 +x+1\rangle$ are isomorphic?

Answer 1

The rings $\mathbb{R}[x]/\langle x^2+1\rangle$ and $\mathbb{R}[x]/\langle x^2 +x+1\rangle$ are isomorphic, because they're both isomorphic to $\mathbb{C}$ (easy proof).

$\mathbb{R}[x]/\langle x^2+1\rangle$ and $\mathbb{R}[x]/\langle x^3 +x+1\rangle$ are $\mathbb{R}$-algebras, the former has dimension $2$ and the latter has dimension $3$. So they aren't certainly isomorphic as $\mathbb{R}$-algebras.

As far as only the ring structure is concerned, they're not isomorphic because one is a field and the other one isn't.

Answer 2

The Fundamental Theorem of Algebra implies that every finite field extension of $\mathbb R$ has degree at most $2$ and so is isomorphic to $\mathbb R$ or $\mathbb C$.

This settles the isomorphism class for $\mathbb R[x]/(f(x))$ when $f$ is irreducible of degree at most $2$.

There are no other irreducible polynomials over $\mathbb R$. In particular, a cubic polynomial is never irreducible and so the quotient ring in this case then:

• $\mathbb R \times \mathbb C$, if the cubic has exactly one real root. This is the case for $x^3 +x+1$.

• $\mathbb R \times \mathbb R \times \mathbb R$, if the cubic has three distinct real roots.

• $\mathbb R \times A$, if the cubic has three real roots, one being a double root. Here $A = \mathbb R[x]/(x^2)$.

A similar analysis applies to higher degree polynomials.

Answer 3

No, they are not isomorphic.

Pure Algebra tells us that the dimension of each is equal to the degree of the monic polynomial mentioned in its definition.

Let $\alpha$ be the real root of $X^3+X+1$, so that this polynomial factors as $(X-\alpha)(X^2+\alpha X+\alpha^2+1)$, where the second, quadratic, factor has negative discriminant. Call one its (complex) roots $\beta$. Then we map \begin{align} \Bbb R[X]&\>\longrightarrow\>\Bbb C\oplus\Bbb R\\ f(X)&\>\mapsto\>\bigl(f(\beta),\,f(\alpha)\bigr)\>. \end{align} You see that the only way a polynomial $f$ can go to zero is if $f(\beta)=f(\alpha)=0$, in other words if and only if $(X^3+X+1)\big|f(X)$, so that the kernel of our map is $(X^3+X+1)$. This shows that we’ve established an isomorphism between $\Bbb C\oplus\Bbb R$ and $\Bbb R[X]/(X^3+X+1)\,$.