Prove two Ideal are equal 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 $A  = \left &lt;g(x)\right&gt;$ be the ideal that is generated by $g(x)$ and $B  =  \left&lt;f(x)\right&gt;$ be the ideal that is generated by $f(x)$ and $C =\left &lt;g(x)\right&gt; + \left&lt;f(x)\right&gt;$. Now find $h(x)$ such that $C= \left&lt;h(x)\right&gt;$.

hint = I think $h(x) = f(x) +g(x)+f(x)g(x)$ and I proved $\left&lt;f(x) +g(x)+f(x)g(x)\right&gt; \subseteq C$
If $h(x) = gcd (f(x),g(x))$ and $c \in &lt; f(x) &gt; +&lt; g(x) &gt; $ we have $c \in &lt; h(x) &gt; $ 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 $&lt; h(x) &gt; \subset &lt; f(x) &gt; +&lt; g(x) &gt;$ 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 $$&lt; h(x) &gt; = &lt; f(x) &gt; +&lt; g(x) &gt;$$
If $h(x) = gcd (f(x),g(x))$ and $c \in &lt; f(x) &gt; +&lt; g(x) &gt; $ we have $c \in &lt; h(x) &gt; $ 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 $&lt; h(x) &gt; \subset &lt; f(x) &gt; +&lt; g(x) &gt;$ 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 $$&lt; h(x) &gt; = &lt; f(x) &gt; +&lt; g(x) &gt;$$