Isomorphism of quotient fields

Question

Consider $\mathbb R $ and $\mathbb Q$ with usual meanings.Which of the following rings are isomorphic?

a. $\mathbb Q[x]/\langle x^2+1\rangle $ and $\mathbb Q[x]/\langle x^2+x+1\rangle$

b. $\mathbb R[x]/\langle x^2+1\rangle $ and $\mathbb R[x]/\langle x^2+x+1\rangle$

My attempt:I know that $\mathbb Q[x]/\langle x^2+1\rangle$, $\mathbb Q[x]/\langle x^2+x+1\rangle$,$\mathbb R[x]/\langle x^2+1\rangle$ and $\mathbb R[x]/\langle x^2+x+1\rangle$ are all fields as the polynomials are all irreducible. Also any element of $\mathbb Q[x]/\langle x^2+1\rangle$ is of the form $ax+b+\langle x^2+1\rangle$. I tried with some familiar mappings like $f(ax+b+\langle x^2+1\rangle)=ax+b+\langle x^2+x+1\rangle$ but did not get the required result. Please give some required hints.

Answer 1

Hint:

$\mathbb{Q}[x]/\langle x^2+1\rangle \cong \mathbb{Q}(i)$, $\mathbb{Q}[x]/\langle x^2+x+1\rangle\cong \mathbb{Q}(\omega)$ where $\omega$ is the cube root of unity,

$\mathbb{R}[x]/\langle x^2+1\rangle \cong \mathbb{C}$, $\mathbb{R}[x]/\langle x^2+x+1\rangle \cong \mathbb{C}$.

Answer 2

$\mathbb{Q}[x]/(x^2+1)$ and $\mathbb{R}[x]/(x^2+1)$ both contain square roots of -1. Therefore so must any isomorphic field. So a starting point for answering these questions would be to ask,

Does $\mathbb{Q}[x]/(x^2+x+1)$ contain a square root of -1?
Does $\mathbb{R}[x]/(x^2+x+1)$ contain a square root of -1?

Answer 3

An idea for a hard way:

The elements of $\;K_1:=\Bbb Q[x]/\langle x^2+1\rangle\;$ can be expressed as $\;a\omega +b\;$ , with $\;\omega\in\Bbb A:=\overline{\Bbb Q}\;,\;\;s.t.\;\; \omega^2=-1\;\;,\;\;a,b,\in\Bbb Q\;$ .

Likewise for $\;K_2:=\Bbb Q[x]/\langle x^2+x+1\rangle\;$, whose elements are of the form $\;a\phi+b\;,\;\;\phi\in\Bbb A\;,\;\;s.t.\;\;\phi^2=-\phi-1\;,\;\;a,b\in\Bbb Q\;$

Now, suppose there's an isomorphism $\;\tau:K_1\to K_2\;$. In particular, we get that $\;\tau(a\omega+b)=\phi\;$ , for some $\;a,b\in\Bbb Q\;$ . But then, as this isomorphism fixes $\;\Bbb Q\;$ , we get

$$\phi=a\tau\omega+b\implies 0=\phi^2+\phi+1=a^2\tau\omega^2+2ab\tau\omega+b^2+a\tau\omega+b+1$$

and since $\;\omega^2=-1\;$ , we get:

$$0=-a^2+a(2b+1)\tau\omega+b^2+b+1\implies\tau\omega=\frac{a^2-b^2-b-1}{a(2b+1)}\in\Bbb Q$$

which of course is impossible (why?). Observe that $\;2b=-1\;$ also leads to a contradiction.

An idea for a simpler, faster way:

Any $\;\Bbb Q-$ homomorphism $\;\tau:K_1\to F\;,\;\;F\;$ any field containing $\;\Bbb Q\;$ , fulfills $\;\tau(\omega)=\pm\omega\;$ (why?), which would lead to

$$\phi=\tau(a\omega+b)=\pm a\omega+b\implies -\phi-1=\phi^2=a^2\omega^2\pm2ab\omega+b^2=-a^2\pm2ab\omega+b^2\implies$$

$$\mp a\omega-b-1=-\phi-1=-a^2\pm2ab\omega+b^2\implies(\pm2ab\pm a)\omega=a^2-b^2-b-1$$

and we're back at the end of the first way.

The gist here is that under $\;\Bbb Q-$ homomorphisms, roots of an irreducible rational polynomial are mapped to roots of the same polynomial.

The second case is much simpler: both quotients of $\;\Bbb R[x]\;$ are extensions of degree two of $\;\Bbb R\;$ , and there is only one single such extension up to isomorphism: $\;\Bbb C\;$ , so those two fields must be isomorphic.

Answer 4

In $\mathbb C$, we have $x^2+x+1=0$ precisely if $x = \dfrac{-1\pm\sqrt{-3}}2$. So the mapping from $\mathbb C$ to $\mathbb R[x]/\langle x^2+x+1\rangle$ given by $\dfrac{-1+\sqrt{-3}}2\mapsto x$ is an isomorphism.

Similarly the mapping $i\mapsto x$ from $\mathbb C$ to $\mathbb R[x]/\langle x^2+1\rangle$ is an isomorphism. Therefore $\mathbb R[x]/\langle x^2+x+1\rangle\cong\mathbb R[x]/\langle x^2+1\rangle$.

Answer 5

(a) The ring $\mathbb{Q}[X]/(X^2 + 1)$ contains an element $x$ with $x^2 = -1$, and this property is preserved by ring isomorphism (since such a map must map $-1$ to $-1$). Show that $A = \mathbb{Q}[X]/(X^2 + X + 1)$ does not contain such an element by, for example, just writing out a basis of $A\subset \mathbb{C}$ over $\mathbb{Q}$.

(b) Both rings are $2$-dimensional field extensions of $\mathbb{R}$ in $\overline{\mathbb{R}} = \mathbb{C}$, which has $[\mathbb{C}:\mathbb{R}] = 2$.