Is it true that any finitely generated ring has no infinite minimal generating set? I want to prove that any finitely generated (not necessarilly commutative) ring $R$ is Noetherian. 
Attempt: Suppose not. Then there is an ascending chain of left-ideals $I_i$ of $R$ such that $I_1\subsetneq I_2\subsetneq...$ Let $a_1\in I_1$ and $a_i\in I_{i}-I_{i-1}$ for all $i=2,3,...$ Then we can complete the set $\{a_1,a_2,...\}$ to a generating set of $R$. I claim that any finitely generated ring has no infinite minimal generating set. But I'm not sure whether the claim is true or not. Thanks for your help! 
 A: Fix a finite set $g_1, \ldots, g_n$ generating R.  Let X be an infinite set that also generates R.  Then we can write each $g_i$ as a (non-commutative) polynomial (with integer coefficients) in finitely many elements of X, say $$g_i = f_i (x_{i,1}, \ldots, x_{i,N}).$$  Then the finite subset of X consisting of the $x_{i,j}$ generates R.
As for the ring being noetherian, I think this is not true.  Take $\mathbb{Z}\{x,y\}$ and consider the ideal $(xy^n x | n>1)$.
A: The integers are noetherian. Use Hilbert's basis theorem in the commutative case and that rings are $\Bbb Z$-algebras.
Under the assumption that the ring is not commutative, the claim is false. See this for an explicit example of a non finitely generated ideal in the free algebra of two elements over $\Bbb Z$. 
A: Hint: The easiest examples of finitely generated rings are $\mathbb{Z}[x_1,\ldots,x_n]$, which you should know (or prove) are noetherian. Then, remember that if a ring $R$ is noetherian and $f:R\to S$ is a surjective ring map, then $S$ is noetherian.
You can take this as a direct argument, or adapt this approach by pulling back your work so far to a ring $\mathbb{Z}[x_1,\ldots,x_n]$.
