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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.

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The paper The presentation rank of a direct product of finite groups by Cossey, Gruenberg and Kovacs shows that the difference between the minimal number of generators of $G$ and $I$ can be as big as you like for finite groups.

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  • $\begingroup$ Great! Thank you. The paper also let's construct simple explicit examples: Let $H\neq 1$ be a finite solvable group and let $G = H^k$ be the direct product. Then $\mu(I_H)=\mu(H)$ (p. 597 and Prop. 1) and $\mu(I_G)=\mu(I_H)$ (Theorem 2). Hence $\mu(\mathbb{Z}[G]) = k\cdot \mu(\mathbb{Z}[H])=k\cdot \mu(I_H) = k\cdot \mu(I_G) > \mu(I_G)$ for $k> 1$. $\endgroup$ – Todd Leason Sep 16 '15 at 22:51
  • $\begingroup$ BTW: The paper gives a cohomological formula for $\mu(I)$. Do you know if there is also some kind of formula/characterization for $\mu(\mathbb{Z}[G])$ ? $\endgroup$ – Todd Leason Sep 16 '15 at 23:34
  • $\begingroup$ I don't think there is any such formula $\endgroup$ – Benjamin Steinberg Sep 16 '15 at 23:45

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