Let $\alpha_A: k[x_1,...,x_m]\rightarrow k[y_1,...,y_n]$ be a map defined by $\alpha_A(f)(y)=f(Ay)$ where $A$ is an $m\times n$ matrix.
Show that $\left\langle\alpha_A(I\cap J)\right\rangle \subset \left\langle\alpha_A(I)\right\rangle \cap \left\langle\alpha_A(J)\right\rangle $, with equality if $I\supset K$ or $J\supset K$ and $\alpha_A$ is onto, where $K=\ker(\alpha_A)$.
My attempt:
Let $f'\in \left\langle\alpha_A(I\cap J)\right\rangle$, then $$f'=\sum a_i\alpha_A(f_i)$$ where $f_i\in I\cap J$ and $a_i\in k[y_1,...,y_n]$. This is clearly in $\left\langle\alpha_A(I)\right\rangle \cap \left\langle\alpha_A(J)\right\rangle$ because by this equation $f'$ is in both $\left\langle\alpha_A(I)\right\rangle$ and $\left\langle\alpha_A(J)\right\rangle$.
The equality is where I have trouble with.
Suppose now $I\supset K$. Let $g'\in \left\langle\alpha_A(I)\right\rangle \cap \left\langle\alpha_A(J)\right\rangle $ $$g'=\sum a_i\alpha_A(g_i)=\sum b_j\alpha_A(h_j)$$ where $g_i\in I$, $h_j\in J$.
My thought is since $I\supset K$, I should use the first express, and try to write $g_i$ as $g_i+f_i\in J$, such that $f_i$ is in $K$. Then $g'$ would be in $\left\langle\alpha_A(I\cap J)\right\rangle$. But I couldn't find these $f_i$'s, and couldn't think of any other way to go around this.
Any help will be appreciated!
We prove "$\supseteq$".
Let $b\in I^e\cap J^e$. Then $b=\sum_{i=1}^r b_i\alpha(a_i)=\sum_{i=1}^rb_i'\alpha(a_i')$ with $b_i,b_i'\in B$, $a_i\in I$, and $a_i'\in J$. Since $\alpha$ is surjective there exist $u_i,u_i'\in A$ such that $\alpha(u_i)=b_i$ and $\alpha(u_i')=b_i'$. Thus $b=\sum_{i=1}^r\alpha(u_i)\alpha(a_i)=\sum_{i=1}^r\alpha(u_i')\alpha(a_i')$. We then have $\sum_{i=1}^ru_ia_i-\sum_{i=1}^ru_i'a_i'\in K$. Suppose $J\supseteq K$. Then $\sum_{i=1}^ru_ia_i-\sum_{i=1}^ru_i'a_i'\in J$, and therefore $\sum_{i=1}^ru_ia_i\in I\cap J$. Now conclude that $b\in (I\cap J)^e$.