$R_d$ be space of polynomials in $k[x,y]$ of degree $\le d$ , $W_d=R_{d-n_1}f_1+R_{d-n_2}f_2$ ; is $R_d \cap (f_1,f_2) = W_d$ under conditions? Let $k$ be an algebraically closed field , let $f_1 , f_2 \in k[x,y]$ be of degree $n_1,n_2$ respectively . For $d \ge 0$ , let $R_d$ be the vector space of polynomials in $k[x,y]$ of degree $\le d$ . For $d \ge n_1+n_2$ , define $W_d=R_{d-n_1}f_1+R_{d-n_2}f_2$ . Then it is clear that $W_d \subseteq R_d \cap (f_1,f_2) $ . 
Now suppose for $f\in k[x,y]$   of degree $d$ , $f^*=\sum_{i+j=d}a_{ij} x^i y^j$ be the homogenous part of highest degree in $f$ . If $f_1^* , f_2^*$ are relatively prime and $d\ge n_1+n_2$ , then is it true that $R_d \cap (f_1,f_2) \subseteq W_d$ ? 
[motivation : this is needed as a part of a proof of Bezout's theorem for plane curves ]
 A: Let $d \geq n_1+n_2$, let $g \in R_d \cap (f_1,f_2)$. So, $\deg g \leq d$ and $\exists \lambda_1, \lambda_2 \in k[x,y]$ such that $g= \lambda_1 f_1 + \lambda_2 f_2$. We can choose $\lambda_1$ and $\lambda_2$ such that $\max ( \deg \lambda_1 f_1, \deg \lambda_2 f_2)$ is minimum.
If $\lambda_1 \notin R_{d-n_1}$ or $\lambda_2 \notin R_{d-n_2}$, then $\deg \lambda_1 f_1 > (d-n_1)+n_1=d$ or $\deg \lambda_2 f_2 > (d-n_2)+n_2=d$.
Let $n= \max ( \deg \lambda_1 f_1, \deg \lambda_2 f_2)$. 
If $h$ is a polynomial, let $h_n$ be the homogenous part of $h$ of degree $n$.
Then $g_n=0$ because $n>d$.
So $(-\lambda_1 f_1)_n= (\lambda_2 f_2)_n$.
$n= \max ( \deg \lambda_1 f_1, \deg \lambda_2 f_2)$.
So $-\lambda_1 ^*~ f_1^*= \lambda_2^*~ f_2^*$.
$f_1^*$ and $f_2^*$ are relatively prime. So $f_2^*$ divides $\lambda_1^*$ and $f_1^*$ divides $\lambda_2^*$.
$\exists u \in k[x,y]$ such that $\lambda_1^*=-u f_2^*$, and $\lambda_2^*= u f_1^*$ with $u$ homogenous.
Then $g=( \lambda_1+uf_2)f_1 +(\lambda_2-uf_1)f_2$.
And $\deg (\lambda_1+uf_2)= \deg( \lambda_1^*+ uf_2^*) < n$.
And $\deg (\lambda_2-uf_1)= \deg( \lambda_2^*- uf_1^*) < n$.
So $\max ( \deg \lambda_1 f_1, \deg \lambda_2 f_2)$ was not minimum. Contradiction.
So we can choose $\lambda_1 \in R_{d-n_1}$ and $\lambda_2 \in R_{d-n_2}$.
So $R_d \cap (f_1,f_2) \subseteq W_d$ if $d \geq n_1+n_2$.
