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**SlipEternal** You can generate every element of $\displaystyle L = F(u+v)$ by all products of the form $\displaystyle a+b(u+v)$. This is obvious since $\displaystyle [0+1(u+v)]^k$ gives the basis you mentioned. So, to show that $\displaystyle L(u) = F(u,v)$, you only need to show that for any $\displaystyle a + bu \in L(u)$ with $\displaystyle a,b \in L$, there exists $\displaystyle x \in F(u,v)$ such that $\displaystyle x = a+bu$. Then, for any $\displaystyle c + du + ev \in F(u,v)$ with $\displaystyle c,d,e \in F$, show that there exists $\displaystyle y \in L(u)$ such that $\displaystyle c+du+ev = y$.

SlipEternal, HallsofIvy, thanks so much for your help on this Example!!

peter

So, since $\displaystyle (u+v) \in L, u \in L(u)$, we know $\displaystyle v \in L(u)$. Since $\displaystyle F \subseteq F(u+v) = L \subseteq L(u)$, we know $\displaystyle c,d,e \in L(u)$. Hence, $\displaystyle c+du+ev \in L(u)$.