Prove or disprove that the rings $R$ and $S$ are isomorphic.
$R = \mathbb{Q}[x]/\langle x^4 – 2 \rangle$ and $S = \mathbb{Q}(\sqrt2)[x]/ \langle x^2 – 2x + 1 + \sqrt2 \rangle$.
$\textbf{May use the following fact without a prove:}$
Suppose that $E$ is an extension of a field $F$, $\alpha \in E$ and $\alpha$ is algebraic
over F.
Let $m_{\alpha, F}(x)$ be the minimal polynomial of $\alpha$ over $F$.
Define $\psi_{\alpha, F} : F[x]/ \langle m_{\alpha, F}(x) \rangle \rightarrow F(\alpha)$ which by $\psi_{\alpha, F} (f(x) + \langle m_{\alpha, F}(x) \rangle) = f(\alpha)$, where $f(x) ∈ F[x]$.
Then $\psi_{\alpha, F}$ is well-defined and is an isomorphism of fields.
$\textbf{My Attempt:}$
Want to prove that rings $R$ and $S$ are isomorphic.
Notice that $\mathbb{C}$, $\mathbb{Q}$ and $\mathbb{Q}(\sqrt2)$ are all fields.
Then, $\mathbb{C}$ is an extension of a field $\mathbb{Q}$ and $\mathbb{C}$ is an extension of a field $\mathbb{Q}(\sqrt2)$.
Let $a_1 = i\sqrt[4]{2}$ which it is a root of $x^4 – 2 \in \mathbb{Q}[x]$. So, $a_1$ is algebraic over $\mathbb{Q}$.
Since, $x^4 – 2$ is monic polynomial and $a_1^4 – 2 = 0$. Also, by Eisenstein's criterion, $x^4 – 2$ is irreducible over $\mathbb{Q}$.
So, $m_{a_1, \mathbb{Q}}(x) = x^4 – 2$.
Hence, be the fact above, $\psi_{a_1, \mathbb{Q}} : R \rightarrow \mathbb{Q}(i\sqrt[4]{2})$ is well-defined and is an isomorphism of fields.
Let $a_2 = 1 + i \sqrt[4]{2}$ which it is a root of $x^2 – 2x + 1 + \sqrt2 \in \mathbb{Q}(\sqrt2)[x]$. So, $a_2$ is algebraic over $\mathbb{Q}(\sqrt2)$.
Since, $x^2 – 2x + 1 + \sqrt2$ is monic polynomial and $a_2^2 – 2a_2 + 1 + \sqrt2 = 0$.
Also, $x^2 – 2x + 1 + \sqrt2$ is irreducible over $\mathbb{Q}(\sqrt2)$.
So, $m_{a_2, \mathbb{Q}(\sqrt2)}(x) = x^2 – 2x + 1 + \sqrt{2}$.
Hence, be the fact above, $\psi_{a_2, \mathbb{Q}(\sqrt2)} : S \rightarrow \mathbb{Q}(\sqrt2)(1 + i\sqrt[4]{2})$ is well-defined and is an isomorphism of fields.
Notice that $\mathbb{Q}(\sqrt2)(1 + i\sqrt[4]{2}) = \mathbb{Q}(\sqrt2, 1 + i\sqrt[4]{2}) = \mathbb{Q}(\sqrt2, i\sqrt[4]{2})$.
Then, $\psi_{a_2, \mathbb{Q}(\sqrt2)} : S \rightarrow \mathbb{Q}(\sqrt2, i\sqrt[4]{2})$ is well-defined and is an isomorphism of fields.
Then, the inverse of $\psi_{a_2, \mathbb{Q}(\sqrt2)}$ which is $\psi_{a_2, \mathbb{Q}(\sqrt2)}^{-1} : \mathbb{Q}(\sqrt2, i\sqrt[4]{2}) \rightarrow S$ is well-defined and is an isomorphism of fields.
$\textbf{I am now stack on prove that $\sigma : \mathbb{Q}(i\sqrt[4]{2}) \rightarrow \mathbb{Q}(\sqrt2, i\sqrt[4]{2})$ is an isomorphism of fields.}$
Since, isomorphism is transitive. Then, I can prove that $R$ and $S$ are isomorphic.
Best Answer
$$\sqrt2 = -\left( i \sqrt[4]2 \right)^2$$