🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

We already know that since $F$ is a field then $F$ is a crone, so one way would be just showing that it's a domain through this characterization of a domain

Suppose that $x,y\in F$ and assume that $x\otimes y=0_F$, if $x=0_R$, we've proven the statement, if $x\ne 0_R$, then since $F$ has multiplicative inverses for non-zero elements we know that $x^{-1}\otimes x\otimes y=x^{-1}{0}_R=0_R$, thus after cancellation and multiplication by one we have $y=0_R$ as needed.

Now let's observe the denominator $x^2-y^{2}p$, if this equals zero, then we have the following ${\left(\frac{x}{y}\right)}^2=p$ note that division is justified since $y\ne 0$ from the fact that $x+y\sqrt{p}$ was non-zero. For this equation to make any sense we require that ${\left(\frac{x}{y}\right)}^2$ be an integer, which isn't necessarily the case and if it's not the case we've reached a contradiction.

Perhaps ${\left(\frac{x}{y}\right)}^2$ is an integer (non-zero), in that case we're saying that there exists some integer $a$ such that $a^2=p$, but this is a contradiction as clearly the square of any non-zero integer is not prime, thus our original assumption that $x^2-y^{2}p=0$ is false, and thus we can safely write our multiplicative inverse as

$⟸$ Suppose that $n$ is prime, we recall that $\mathbb{Z}/n\mathbb{Z}$ is a crone. Therefore to show it's a field we have to show it has multiplicative inverses, so let $\bar{a}\in \mathbb{Z}/n\mathbb{Z}⧵\left\{\bar{0}\right\}$. Since $\bar{a}\ne \bar{0}$ then $a\not\equiv 0(modn)$ therefore $n\nmid a$ thus $gcd(n,a)=1$ and we must have $s,t\in \mathbb{Z}$ such that $1=ns+at$, taking this equation mod $n$ we see that $at\equiv 1(modn)$ in otherwords $\bar{a}\bar{t}=\bar{1}$ so that $\bar{t}$ is $\bar{a}$'s multiplicative inverse so that $\mathbb{Z}/n\mathbb{Z}$ is a field.

$⟹$ We use the contrapositive so assume that $n$ is not prime, therefore $n$ is composite, so we have $n=ab$ for some $a,b\in 1,...n-1$ so clearly $\bar{a}\ne \bar{0}$ and $\bar{b}\ne \bar{0}$ but at the same time $n|ab$ therefore $\bar{a}\bar{b}=\bar{0}$, which shows that $\mathbb{Z}/n\mathbb{Z}$ is not a domain, therefore it cannot be a field, as needed for the contrapositive, so the original implication holds true