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We know that for $p \equiv 3 (\text{mod }4)$, there is no solution to $x^2 + 1 \equiv 0 (\text{mod }p^k)$ for $k = 1, 2, \ldots$, by quadratic reciprocity. But can I embed the ring $\mathbb{Z} / p^k \mathbb{Z}$ in some bigger ring so that it does have a solution? What's the simplest way to do that?

As a related question, if I identify $\mathbb{Z} / p \mathbb{Z} \cong F_p$ the finite field of $p$ elments, then it seems there is only one quadratic extension field of $F_p$ that contains square roots of all its elements. Are there good references to this result?

The question was inspired by Example 1.6 of Opera de Cribro.

John Jiang
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    In general, if you have a ring $R$ and you wish to "extend" it by adding roots for a polynomial $P \in R[X]$, the correct ring to consider is $R[X]/(P)$ (the ring $R[X]$ quotiented by the principal ideal $(P)$). However, this notion is more or less well behaved depending on the original ring $R$ and the polynomial $P$. – François Gatine Jul 13 '24 at 15:50
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    For a specific $k$, the most natural way is to consider the quotient $(\mathbb{Z}/p^k\mathbb{Z})[x]/(x^2+1)$. Since $x^2+1$ is monic, $\mathbb{Z}/p^k\mathbb{Z}$ can indeed be seen as a subring. – Mark Jul 13 '24 at 15:52

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A simple way to define your ring is to start with the ring of Gaussian integers and take the congruence modulo $p^k$ on this ring. Or directly, consider the set $$ R = \bigl\{x + iy \mid x, y \in {\Bbb Z}/p^k{\Bbb Z}\bigr\} $$ equipped with the addition $$ (x + iy) + (x' + iy') = (x+x') + i (y+y') $$ and the multiplication $$ (x + iy) (x' + iy') = (xx' -yy') +i(xy' + x'y) $$ This is of course equivalent to the constructions proposed in the comments.

J.-E. Pin
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