Let $k$ be a commutative ring with unit and let $(A,\varepsilon)$ be a (not necessarily commutative) augmented, finitely generated $k$-algebra with augmentation ideal $I$.
If $\mu(A)$ denotes the minimal number of algebra generators of $A$ and $\mu(I)$ the minimal number of generators of $I$ as ideal, we have $\mu(I) \le \mu(A)$.
(For, if $A$ is generated as algebra by $x_1,...,x_n$, then it's also generated by $y_i := x_i - \varepsilon(x_i)\cdot 1\in I$. If $y$ is any element in $I$, write $y = a + f\langle y_1,...,y_n\rangle$ where $a \in k$ and $f$ is a polynomial in the (noncommuting) vaiables $y_1,..,y_n$ having no constant term. Hence $0=\varepsilon(y)=\varepsilon(a)=a$, i.e. $y = f\langle y_1,...,y_n\rangle$ is in the ideal generated by $(y_1,...,y_n)$. This shows $I=(y_1,...,y_n)$ and hence $\mu(I) \le n$)
I wonder what can be said in the case of group algebras:
Question: Does for finitely generated groups the equality $\mu(\mathbb{Z}[G]) = \mu(I)$ hold ?
Note: If $\mu(G)$ denotes the minimal number of generators of the group $G$, we have for finite $G$: $$\mu(I) \le \mu(\mathbb{Z}[G]) \le \mu(G).$$ So the equality would follow from $\mu(I)=\mu(G)$ but I don't know if the latter is true.
