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Oct 24, 2010 at 16:13 answer added Greg Kuperberg timeline score: 1
Oct 10, 2010 at 21:56 comment added felix Kevin, this is indeed true. $v_1, \dots, v_n \in \mathcal{R}^n$ give a basis of $\mathcal{R}^n$ iff $\det (v_1, \dots, v_n) \in \mathcal{R}^\ast$, and this is the case iff $\det (v_1, \dots, v_n) \neq 0$ modulo every maximal ideal of $\mathcal{R}$. Hence, the probability that a random $n \times n$-matrix is invertible is the product of the probabilities for random $n \times n$-matrices to be invertible over all residue fields of $\mathcal{R}$.
Oct 10, 2010 at 8:55 comment added Kevin Buzzard I've not thought too hard about this, but: a general finite ring is semi-local so a product of local rings, so that reduces the question to local rings. And then via a Nakayama argument my gut feeling is that you'll be able to reduce to the case of the residue field, which you've already done.
Oct 10, 2010 at 6:29 answer added Aaron Meyerowitz timeline score: 0
Oct 10, 2010 at 4:42 history asked Sarah CC BY-SA 2.5