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

Since $$ x+y\sqrt{p}\in <!--\; \in \; -->\mathbb{R}$$, and it's non-zero, then in $$ \mathbb{R}$$ we have an inverse: $$ 1/(x+y\sqrt{p})$$, we can then rationalize that as follows

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 $$ y\ne <!--\; \ne \; -->0$$ 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

$\[ \frac{x}{x^2-<!--\; -\; -->y^{2}p}-<!--\; -\; -->\frac{y}{x^2-<!--\; -\; -->y^{2}p}\sqrt{p} \]$

$$ ⟸<!--\; ⟸\; -->$$ 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 $$ \stackrel{¯<!--\; ¯\; -->}{a}\in <!--\; \in \; -->\mathbb{Z}/n\mathbb{Z}\setminus <!--\; \setminus \; -->\left\{\stackrel{¯<!--\; ¯\; -->}{0}\right\}$$. Since $$ \stackrel{¯<!--\; ¯\; -->}{a}\ne <!--\; \ne \; -->\stackrel{¯<!--\; ¯\; -->}{0}$$ then $$ a\not\equiv 0(\mathrm{mod}<!--\; \; -->n)$$ therefore $$ n\nmid <!--\; \nmid \; -->a$$ thus $$ gcd(n,a)=1$$ and we must have $$ s,t\in <!--\; \in \; -->\mathbb{Z}$$ such that $$ 1=ns+at$$, taking this equation mod $$ n$$ we see that $$ at\equiv <!--\; \equiv \; -->1(\mathrm{mod}<!--\; \; -->n)$$ in otherwords $$ \stackrel{¯<!--\; ¯\; -->}{a}\stackrel{¯<!--\; ¯\; -->}{t}=\stackrel{¯<!--\; ¯\; -->}{1}$$ so that $$ \stackrel{¯<!--\; ¯\; -->}{t}$$ is $$ \stackrel{¯<!--\; ¯\; -->}{a}$$'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 $$ a,b\in <!--\; \in \; -->1,...n-<!--\; -\; -->1$$ so clearly $$ \stackrel{¯<!--\; ¯\; -->}{a}\ne <!--\; \ne \; -->\stackrel{¯<!--\; ¯\; -->}{0}$$ and $$ \stackrel{¯<!--\; ¯\; -->}{b}\ne <!--\; \ne \; -->\stackrel{¯<!--\; ¯\; -->}{0}$$ but at the same time $$ n|ab$$ therefore $$ \stackrel{¯<!--\; ¯\; -->}{a}\stackrel{¯<!--\; ¯\; -->}{b}=\stackrel{¯<!--\; ¯\; -->}{0}$$, 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