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?

  • $\begingroup$ 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? $\endgroup$ – Florian Suess Oct 24 '18 at 0:42
  • $\begingroup$ First step: take out the $(x)$ because it's not really used, and the expressions become easier to read when simplified. $\endgroup$ – CommutativeAlgebraStudent Oct 24 '18 at 0:56
  • $\begingroup$ Oh yes, okay fair enough. $\endgroup$ – 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!

Hope this helps :)

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