Computing the quotient $\mathbb{Q}_p[x]/(x^2 + 1)$

Question

I am trying to compute the quotient $\mathbb{Q}_p[x]/(x^2 + 1)$ where $\mathbb{Q}_p$ represents the $p$-adic numbers. I already proved that if $p \equiv 2,3\mod4$, then $\mathbb{Q}_p[x]/(x^2 + 1) \cong \mathbb{Q}_p(i)$. Moreover, I also proved that if $x \equiv 1 \mod 4$, then $x^2 + 1$ has a root in $Q_p$ using Hensel's lemma. Online, I found that in this case $\mathbb{Q}_p[x]/(x^2 + 1) \cong \mathbb{Q}_p \times \mathbb{Q}_p$. Why? A similar problem was posted here:Decomposition of the tensor product $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}[i]$ into a product of fields. However, they do not address my question. Thanks in advance!

Answer 1

This follows from the Chinese Remainder Theorem: Let $\alpha$ be a root of $x^2 + 1$ in $\mathbb{Q}_p$. Then in $\mathbb{Q}_p[x]$, the polynomial $x^2 + 1$ factors as $(x-\alpha)(x+\alpha)$. Furthermore, in $\mathbb{Q}_p[x]$ the ideals $(x-\alpha)$ and $(x+\alpha)$ are comaximal (i.e. their sum is the unit ideal) and their intersection is $(0)$, so by the Chinese remainder theorem, there is an isomorphism \begin{equation*} \mathbb{Q}_p[x]/(x^2+1) = \mathbb{Q}_p[x]/(x-\alpha)(x+\alpha) \cong \mathbb{Q}_p[x]/(x-\alpha) \times \mathbb{Q}_p[x]/(x+\alpha) \cong \mathbb{Q}_p \times \mathbb{Q}_p \end{equation*}

Answer 2

Let $u$ be a root of $x^2+1=0$, then $x^2+1=(x-u)(x+u)$ and $\Bbb Q_p/(x^2+1)=\Bbb Q_p((x-u)(x+u))$. If $K$ is a field and $a$, $b\in K$ with $a\ne b$, then $$\frac{K[x]}{((x-a)(x-b))}\cong K^2$$ as $K$-algebras. The isomorphism is induced by $f(x)+((x-a)(x-b)) \mapsto (f(a),f(b))$.