# $\langle f(x)\rangle + \langle g(x)\rangle = \langle\text{gcd}(f,g)(x)\rangle$

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?

• To elaborate a little $\langle\text{gcd}(f,g)(x)\rangle = p(x)\cdot(t(x)f(x) + v(x)g(x)) = p(x)t(x)f(x) + p(x)v(x)g(x)$ and since p(x) is arbitrary, then we are talking about the generators of f(x) and g(x) right? – Florian Suess Oct 24 '18 at 0:42
• First step: take out the $(x)$ because it's not really used, and the expressions become easier to read when simplified. – AlgebraicGeometryStudent Oct 24 '18 at 0:56
• Oh yes, okay fair enough. – Florian Suess Oct 24 '18 at 1:09

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!