Using norms to prove no zero-divisors exist

Question

I was reading a post on a specific problem and there was an equation $(a+b\sqrt{k})(c+d\sqrt{k})\equiv0 \mod p$ where $a,b,c,d,k$ are integers in the integers $\mod p$, $p$ is prime. Supposedly this equation is equivalent to $(a+b\sqrt{k})(a-b\sqrt{k})(c+d\sqrt{k})(c-d\sqrt{k})\equiv 0 \mod p$, why is this true, why can we multiply the first two factors by their conjugate in the first equation, and have the same equation?

Answer 1

It's closely connected to your prior question. If $\,k\,$ is not a square $\bmod p\,$ then a null norm yields

$$ 0\equiv \bar w w\equiv (a-b\sqrt k)(a+b\sqrt k ) \equiv a^2-kb^2\Rightarrow k\equiv (a/b)^2,\ \text{contra hypothesis}\qquad$$

when $\,b\not\equiv 0\ $ (and if $\,b\equiv 0\,$ then $\,w\bar w\equiv a^2\equiv 0\iff 0\equiv a\equiv w,\,$ by $\,p\,$ prime $\Rightarrow \Bbb Z_p\,$ a field.

Thus norms $\bar w w\,$ of nonzero $w$ are nonzero so invertible (being in a field $\,\Bbb Z_p)\,$ so its factor $w$ is also invertible (explicitly $\, \dfrac{1}{w} \equiv \dfrac{\bar w}{w\bar w}$ by rationalizing the denominator (well-defined by $\bar w w\not\equiv 0)$

Hence $\,w\equiv 0\iff N(w) = \bar w w\equiv 0.\,$ Applied to a product using norm multiplicativity we get

$$wv\equiv 0 \iff 0\equiv N(wv) \equiv N(w)N(v) \iff N(w)\equiv 0\ \, {\rm or}\ \, N(v)\equiv 0\qquad$$

which is the sought equivalence.

Remark $ $ The idea generalizes to higher degree algebraic extensions. by using norms, minimal polynomials or the Euclidean algorithm to compute the inverse in a similar way (the linked answer also explains how to view this as a special case of the method of simpler multiples).

The standard way to construct this ring is as a quotient $\,R = \Bbb Z_p[\sqrt k] := \Bbb Z[x]/(x^2-k)$ or, equivalently, by congruences, i.e. $\Bbb Z[x]\bmod (p,\,x^2-k)$. By here or here the quotient ring is a field $\!\iff\! f(x) = x^2-k\,$ is irreducible or $\,\Bbb Z_p\!\iff\! f\,$ has no root in $\,\Bbb Z_p\!\iff\! k\,$ is nomsquare in $\,\Bbb Z_p.\,$

On the other hand, if $\,k\,$ is a square, say $\,0\not\equiv k\equiv j^2,\,$ and $\,p\neq 2\,$ then by CRT we have$\,R = \Bbb Z_p[x]/(x-j)(x+j)\cong \Bbb Z_p[x]/(x-j)\times \Bbb Z_p[x]/(x+j)\cong \Bbb Z_p^2,\,$ and such a product of rings is never a field since it has idempotent zero-divisors $(1,0)(0,1)=(0,0);\,$ equivalently $\,(x-j)(x+j)\equiv 0\,$ yields zero-divisors in $R.\,$ Each component of the product may be viewed as a "possible world" where the sqrt is the negative of the value in the other world, e.g. if $k=1$ then in effect we create a ring by adjoining $\,x = \pm1$. Thus the pairs in the product ring keep track of all the possible values arising from both possible "signs" (branches) of the sqrt, i.e. the arithmetic in the product ring does the computation in parallel in both possible worlds - one world where $\,x=1\,$ and another where $\,x = -1$.

And, of course, if $\,k\equiv 0\,$ then $\,R = \Bbb Z_p[x]/(x^2)$ is also not a field since here $\,x\,$ is zero-divisor - being nilpotent $\,x^2\equiv 0.\,$ This is known as the ring of dual numbers (it proves handy for constructing algebraic models of derivatives and tangent spaces).