I am looking at the following exercise:
If $A$ is a ring with maximal Ideal $\mathfrak{m}$, define $\tilde{A}:= A\oplus\mathfrak{m}\oplus\mathfrak{m}^{2}\oplus...$. Assume $\mathfrak{m}$ is generated by elements $x_{1},...,x_{n}\in A$. Consider the map $\phi: A[T_{1},...,T_{n}]\longrightarrow\tilde{A}$, defined by $T_{i}\mapsto x_{i}\in\mathfrak{m}$. Show that this map induces a closed immersion $\operatorname{Proj}(\tilde{A})\longrightarrow\operatorname{Spec}(A[T_{1},...,T_{n}])$.
Here comes my question: It is easy to see that $\phi$ is a surjection, so it induces a closed immersion of schemes $\operatorname{Proj}(\tilde{A})\longrightarrow\operatorname{Proj}(A[T_{1},...,T_{n}])$, but I do not see how there is a closed immersion $\operatorname{Proj}(\tilde{A})\longrightarrow\operatorname{Spec}(A[T_{1},...,T_{n}])$.
So I think there should be a closed immersion $\operatorname{Proj}(A[T_{1},...,T_{n}])\longrightarrow\operatorname{Spec}(A[T_{1},...,T_{n}])$.
I have little experience in algebraic geometry (but I know the basic definitions), and I do not even know where to start tackling this question. Please do not be too harsh on me.
This is wrong as stated (probably a typo on your professor's part). Your comment about a closed immersion $\operatorname{Proj} \widetilde{A}\to\operatorname{Proj} A[T_1,\cdots,T_n]$ is what's intended here.
The reason this is wrong is if $X\to Y$ is a closed immersion with $Y$ affine, then $X$ must be affine as well. But the only $A$-schemes which are both affine and projective are finite over $A$. So unless $\operatorname{Proj} \widetilde{A}$ is always finite over $A$ (it's not) you'll have no chance at a closed immersion into any affine scheme.