Consider ideals of the ring $\frac{F[x]}{x^n-1}$ where $F$ is an arbitrary finite field.
Suppose $g(x) | (x^n-1)$ and $f(x) | (x^n-1)$ . Also suppose $A = \left <g(x)\right>$ be the ideal that is generated by $g(x)$ and $B = \left<f(x)\right>$ be the ideal that is generated by $f(x)$ and $C =\left <g(x)\right> + \left<f(x)\right>$. Now find $h(x)$ such that $C= \left<h(x)\right>$.
hint = I think $h(x) = f(x) +g(x)+f(x)g(x)$ and I proved $\left<f(x) +g(x)+f(x)g(x)\right> \subseteq C$
If $h(x) = gcd (f(x),g(x))$ and $c \in < f(x) > +< g(x) > $ we have $c \in < h(x) > $ because $$c= a(x)f(x)+b(x)g(x)$$ for some $a(x)$ and $b(x)$ and then exist a $k(x)$ such that $c= k(x)h(x)$. And also $< h(x) > \subset < f(x) > +< g(x) >$ because exist a linear combination of $ f(x), g(x) $ such that $h(x)= t(x)f(x) +s(x)g(x) $ so if $c \in h(x)$ then exist $l(x)$ such that $c= l(x)t(x)f(x)+l(x)s(x)g(x)$ and so we have $$< h(x) > = < f(x) > +< g(x) >$$