$\langle f(x)\rangle + \langle g(x)\rangle = \langle\text{gcd}(f,g)(x)\rangle$ My attempt; by extending the Euclidean algorithm we have some polynomials $t,v$ such that $\text{gcd}(f,g) = tf + vg$. So for arbitrary polynomial $p$ we say that $$\langle\text{gcd}(f,g)\rangle = p\cdot(tf + vg) =_{\text{expanding}} \langle f\rangle + \langle g\rangle$$
I'm sorry for putting the entire alphabet in here, but does that make sense?
 A: Since you're using the Euclidean Algorithm and gcd's, I assume that the context of this question is the following: $f$ and $g$ are polynomials with coefficients in some field $F$, $\left< f \right>$, $\left< g \right>$ and $\left< \gcd(f,g) \right>$ are the ideals generated by $f$, $g$, and $\gcd(f,g)$, respectively.
In this case, we can show the result by direct element chasing, using the extended Euclidean Algorithm, as you mentioned.  First, let $h \in \left< f \right> + \left< g \right>$.  Then $q = rf + sg$ for some $r, s \in F[x]$.  Since $\gcd(f,g)$ divides $f$ and $g$, $h = \gcd(f,g)(rq_1 + sq_2) \in \left< \gcd(f,g) \right>$, where $f = \gcd(f,g)q_1$ and $g = \gcd(f,g)q_2$.  Next, let $k \in \left< \gcd(f,g) \right>$.  Then $k = \gcd(f,g)q_3$ for some $q_3 \in F[x]$.  By the extended Euclidean algorithm we have $r$ and $s$ such that $\gcd(f,g) = rf + sg$.  Thus, $k = rfq_3 + sgq_3 \in \left< f \right> + \left< g \right>$.
The reason I emphasize that the coefficients are in a field is that the Euclidean Algorithm isn't available over all rings.  Try writing $3x$ as $q(2x) + r$, where $q, r \in \mathbb{Z}[x]$ and $\deg r < 1$, and you'll start to see why.  Context is important for this problem!
Hope this helps :)
