How is multiplication defined in the field extension $\mathbb Z_p(\sqrt3+5i)$

Question

Consider the field extension $\mathbb Z_p(\sqrt3 +5i)$ where $\mathbb Z_p$ is a field ($p$ is prime).

Every element in this extension will be of the form $a+b(\sqrt3 +5i)$ with $a,b \in Z_p$. Therefore multiplication between two elements is defined as follows: $$(a+b(\sqrt3 +5i))(c+d(\sqrt3+5i)) = ac+ad(\sqrt3+5i)+bc(\sqrt3+5i)+ bd(\sqrt3+5i)^2.$$

This is where I am having trouble grouping elements together so they are of the form $c+d(\sqrt3+5i)$. I would appreciate any help with grouping the elements.

Answer 1

It works just fine if you make the (in my humble opinion reasonable) assumption that $i$ stands for an element satisfying the equation $i^2+1=0$, and similarly $\sqrt3$ is a solution of $x^2=3$. The catch is that the choice of solutions will have an impact on the answer! As will the choice of $p$. Furthermore, either $x^2+1=0$ or $x^2=3$, or both may have solutions already in $\Bbb{Z}_p$. The formulas will bifurcate accordingly.

Listing representative examples so that you see what's going on. Case D is the trickiest, please read at least that example carefully.

A) If $p\equiv1\pmod{12}$, then both $i$ and $\sqrt3$ belong to $\Bbb{Z}_p$. For example when $p=13$, we have $5^2\equiv-1$ and $4^2\equiv3$. Therefore we can use either $5$ or $8=-5$ as $i$ and either $4$ or $9=-4$ as $\sqrt3$. The field we get is just $\Bbb{Z}_{13}$. The four combinations give rise to four different looking multiplication formulas.

B) If $p\equiv5\pmod{12}$, then $i$ belongs to $\Bbb{Z}_p$, but $x^2=3$ has no solutions in $\Bbb{Z}_p$, so we need to extend the field by adjoining $\sqrt3$. For example when $p=17$ we have $4^2\equiv-1$, so either $4$ or $-4=13$ can take the duties of $i$. Assuming $i=4$ we get $$a+b(5i+\sqrt3)=(a+20b)+b\sqrt3.$$ I'm sure you can figure out how to write the product of two such beasts in the same form!

C) If $p\equiv11\pmod{12}$, then $x^2+1=0$ has no solutions in $\Bbb{Z}_p$ but $x^2=3$ has. For example when $p=11$ we see that $5^2=3$. We can thus use $5$ in place of $\sqrt3$, but need to adjoin the element $i$ (and end up with a field $\Bbb{Z}_{11}(i)$ with $11^2=121$ elements). If we use $5$ as $\sqrt3$, then $$ a+b(\sqrt3+5i)=(a+5b)+5bi. $$ But if use $-5=6$ as $\sqrt3$, then we get $$ a+b(\sqrt3+5i)=(a+6b)+5bi $$ instead. Again, I trust you to figure out how this choice affects the product rule.

D) Finally, if $p\equiv7\pmod{12}$ then neither $x^2+1=0$ not $x^2=3$ have solutions in $\Bbb{Z}_p$. This does not mean that we need to adjoin both of them, though. This is in sharp contrast with a biquadratic extension like $\Bbb{Q}(i,\sqrt3)/\Bbb{Q}$. Things go differently in the world of finite fields. It turns out that in this case $i$ and $\sqrt3$ are ALWAYS linearly dependent over $\Bbb{Z}_p$. As an example consider the case $p=19$. If we adjoin $i$, we then have the element $4i$, and, behold, $$(4i)^2=4^2i^2=-16\equiv3,$$ so $4i$ (or $-4i=15i$) can play the role of $\sqrt3$ in $\Bbb{Z}_{19}(i)$. With the choice $\sqrt3=4i$ you thus have $$ a+b(\sqrt3+5i)=a+9bi, $$ another case you can surely handle.

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Concluding remarks:

• In case you wonder why the cases are split according to the residue class of $p$ modulo $12$, you should study the law of quadratic reciprocity to get the explanation.
• Lack of general formula comes from the fact that there is no catch all quadratic polynomial in $\Bbb{Z}_p[x]$ that would have $\sqrt3+5i$ as its zero. The quartic you can find easily (with roots $\pm\sqrt3\pm5i$ over $\Bbb{Q}$), $f(x)=x^4+44x^2+784$, would work, but you need a quadratic to get your product rule. Even though $f(x)$ is irreducible over $\Bbb{Q}$, it is NEVER irreducible over $\Bbb{Z}_p$ (basically we already saw that). But we don't know which factor of $f(x)$ will be needed for which $p$.
• In the case D we used the fact (study those quadratic residues), that the ratio of two quadratic non-residues is always a quadratic residue. Here neither $-1$ nor $3$ was a quadratic residue, $-3\equiv4^2\pmod{19}$ was, saving the day.
• The conclusion is that you never need to adjoin two square roots to the field $\Bbb{Z}_p$. Adjoining at most one will do. This is just as well, as it is a basic fact about finite fields that up to isomorphism there is just one field with $p^2$ elements.