Let $I$ be a proper ideal of the commutative ring $R$. Then $IR[\alpha_1, ... , \alpha_n]$ is a proper ideal of $R[\alpha_1, ... , \alpha_n]$
I thought of using the fact that an ideal of any ring $A$ is the same thing as an $A$-module. Basically, I was trying to show that $IR[\alpha_1, ... , \alpha_n]$ is an $R[\alpha_1, ... , \alpha_n]$-module.
Obviously $IR[\alpha_1, ... , \alpha_n]$ is an $R$ module. Since $R[\alpha_1, ... ,\alpha_n]$ is a free module it is isomorphic to a direct sum of copies of $R$. So I need to show that $I(\oplus_{i=1}^n R)$ is a $\oplus_{i=1}^n R$ module. But this is obvious since $IR$ is an $R$ module, and we can use the same operation coordinate wise.
Is my answer correct?
The statement is true for a polynomial ring $R[X_1,\dots,X_n]$. To see this suppose $IR[X_1,\dots,X_n]$ is not proper. Then $1\in IR[X_1,\dots,X_n]$. It then follows that $1\in I$, which implies that $I$ is not proper.
It is not necessarily true that $IR[a_1,\dots,a_n]$ is a proper ideal. Consider $R[X_1,\dots,X_n]/M$ where M is a maximal ideal. Then $R[X_1,\dots,X_n]/M$ is a field and is of the form $R[a_1,\dots,a_n]$ where $a_i$ is the images of $X_i$ in $R[X_1,\dots,X_n]/M$. Then for any nonzero ideal of $R$, $IR[a_1,\dots,a_n]$ is not proper (since the only proper ideal of a field is $\langle 0\rangle$).