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This is insanely frustrating to me, as I know I must be described somewhere, but I just cannot seem to find any references. I know that simplicial homology (or thus Betti numbers) is commonly computed through the Smith normal form using the following theorem.

Let $A$ be an $m\times n$ integer matrix, and $B$ be an $l\times m$ integer matrix such that $BA=0$. Then \begin{equation} \mathrm{Ker}(B)/\mathrm{Im}(A)\cong\bigoplus^r_{i=1}\mathbb{Z}/(\alpha_i)\oplus\mathbb{Z}^{m-r-s}, \end{equation} Where $r=\mathrm{rank}(A)$, $s=\mathrm{rank}(B)$, and $\alpha_1, \ldots, \alpha_r$ are the non-zero elements on the diagonal of the Smith normal form of $A$.

It seems to me that almost everywhere they use this result to describe how simplicial homology is computed for integer coefficients. Yet in practice, homology is often computed for coefficients over finite fields. Is there a similar result as above but for finite fields, that explains how simplicial homology is obtained in practice for these? Or is it obtained from the result above with integer coefficients and then transformed somehow (say using the Universal coefficient theorem)?

In summary, my question could be interpreted as "how are Betti numbers computed over finite fields? Is there a more general theorem that allows us to obtain it through the Smith normal form (which exists for fields as well) as above?"

I was really hoping someone could point me to the appropriate references...

rvdaele
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    The Smith Normal form works over any PID, which includes all fields. Is that what you want? – Thorgott Mar 14 '22 at 11:10
  • Thanks, but I am aware that the Smith normal form works over any PID. This is also the reason that I suspect a result such as the isomorphism above should exist more generally, and not only for integer coefficients. – rvdaele Mar 14 '22 at 12:06
  • If you merely want the Betti numbers, that's just simple linear algebra and doesn't need any normal form. All you need is the rank-nullity theorem. – Thorgott Mar 14 '22 at 13:08
  • Thanks, this more in-line with what I am looking for. I guess I can avoid talking about the Smith normal form at all. – rvdaele Mar 14 '22 at 13:10
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    It's not clear whether you are asking about theory or computation. If theory, the comment of @Thorgott applies, and Smith normal form is (mostly) irrelevant. If computation, use the Smith normal form. Of course, in the end you probably need both. But it's important to keep things sorted out. – Lee Mosher Mar 14 '22 at 14:34
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    In terms of computation, SageMath uses Smith Normal Form when computing over the integers. Over fields it just uses rank computations. – John Palmieri Mar 14 '22 at 17:11
  • Thanks all. I was under the (apparently false) impression that the Smith normal form was also used for computation over fields, thanks for the clarification. – rvdaele Mar 15 '22 at 15:08

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