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 $m\times n$ constant matrix. Let $I',J'$ be ideals in $k[y_1,...,y_n]$.
Show that $\alpha_A^{-1}(I'+J')=\alpha_A^{-1}(I')+\alpha_A^{-1}(J')$ if $\alpha_A$ is onto.
This exercise is a continuation of my other question.
My attempt: The reverse direction is very easy. Let $f\in \alpha_A^{-1}(I')+\alpha_A^{-1}(J')$. Then $f=g+h$, such that $\alpha_A(g)\in I'$ and $\alpha_A(h)\in J'$. So $\alpha_A(f)=\alpha_A(g)+\alpha_A(h)\in I'+J'$. Thus $f\in \alpha_A^{-1}(I'+J')$, which shows $\alpha_A^{-1}(I'+J')\supset \alpha_A^{-1}(I')+\alpha_A^{-1}(J')$.
I am stuck at the other direction.
Let $f\in \alpha_A^{-1}(I'+J')$. Then $\alpha_A(f)\in I'+J'$. So we can write $\alpha_A(f)=ag'+bh'$ such that $g'\in I', h'\in J'$, and $a,b\in k[y_1,...,y_n]$. I know in general, since the map might not be surjective, this direction is not true. But maybe the matrix $A$ serves some role here to make up that.
Any help will be appreciated!
Thanks to user26857's help, here is my answer for the direction $\subset$:
Let $f\in \alpha_A^{-1}(I'+J')$. Then $\alpha_A(f)=g'+h'$ for some $g'\in I', h'\in J'$.
Since $\alpha_A$ is onto, there exists $g\in k[x_1,...,x_n]$ such that $\alpha_A(g)=g'$.
We then have $\alpha_A(f-g)=\alpha_A(f)-\alpha_A(g)=h'\in J'$. Hence $f=g+(f-g)\in \alpha_A^{-1}(I')+\alpha_A^{-1}(J')$.