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As explained in this answerthis answer, the property $V=(V^\perp)^\perp$ holds for submodules $V$ of $(\mathbb Z/q\mathbb Z)^n$ because you can use Smith normal form to find a suitable generating matrix for $V^\perp$. If you just want a reference for the result, I can only suggest the following paper:

D. Wilding, M. Johnson, M. Kambites. Exact rings and semirings. Journal of Algebra, 388 (2013), 324–337; arXiv:1212.5358.

When writing this paper we faced the same problem as you. That $(V^\perp)^\perp=V$ follows quite easily from known results, but we could not find it written down explicitly in this formulation.

As explained in this answer, the property $V=(V^\perp)^\perp$ holds for submodules $V$ of $(\mathbb Z/q\mathbb Z)^n$ because you can use Smith normal form to find a suitable generating matrix for $V^\perp$. If you just want a reference for the result, I can only suggest the following paper:

D. Wilding, M. Johnson, M. Kambites. Exact rings and semirings. Journal of Algebra, 388 (2013), 324–337; arXiv:1212.5358.

When writing this paper we faced the same problem as you. That $(V^\perp)^\perp=V$ follows quite easily from known results, but we could not find it written down explicitly in this formulation.

As explained in this answer, the property $V=(V^\perp)^\perp$ holds for submodules $V$ of $(\mathbb Z/q\mathbb Z)^n$ because you can use Smith normal form to find a suitable generating matrix for $V^\perp$. If you just want a reference for the result, I can only suggest the following paper:

D. Wilding, M. Johnson, M. Kambites. Exact rings and semirings. Journal of Algebra, 388 (2013), 324–337; arXiv:1212.5358.

When writing this paper we faced the same problem as you. That $(V^\perp)^\perp=V$ follows quite easily from known results, but we could not find it written down explicitly in this formulation.

Removed the claim that the result is not written anywhere.
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David Wilding
David Wilding
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As explained in this answer, the property $V=(V^\perp)^\perp$ holds for submodules $V$ of $(\mathbb Z/q\mathbb Z)^n$ because you can use Smith normal form to find a suitable generating matrix for $V^\perp$. If you just want a reference for the result, I can only suggest the following paper:

D. Wilding, M. Johnson, M. Kambites. Exact rings and semirings. Journal of Algebra, 388 (2013), 324–337; arXiv:1212.5358.

When writing this paper we faced the same problem as you. That $(V^\perp)^\perp=V$ follows quite easily from known results, but doeswe could not appear to be explicitlyfind it written anywheredown explicitly in this formulation.

As explained in this answer, the property $V=(V^\perp)^\perp$ holds for submodules $V$ of $(\mathbb Z/q\mathbb Z)^n$ because you can use Smith normal form to find a suitable generating matrix for $V^\perp$. If you just want a reference for the result, I can only suggest the following paper:

D. Wilding, M. Johnson, M. Kambites. Exact rings and semirings. Journal of Algebra, 388 (2013), 324–337; arXiv:1212.5358.

When writing this paper we faced the same problem as you. That $(V^\perp)^\perp=V$ follows quite easily from known results, but does not appear to be explicitly written anywhere.

As explained in this answer, the property $V=(V^\perp)^\perp$ holds for submodules $V$ of $(\mathbb Z/q\mathbb Z)^n$ because you can use Smith normal form to find a suitable generating matrix for $V^\perp$. If you just want a reference for the result, I can only suggest the following paper:

D. Wilding, M. Johnson, M. Kambites. Exact rings and semirings. Journal of Algebra, 388 (2013), 324–337; arXiv:1212.5358.

When writing this paper we faced the same problem as you. That $(V^\perp)^\perp=V$ follows quite easily from known results, but we could not find it written down explicitly in this formulation.

Source Link
David Wilding
David Wilding
  • 66
  • 2

As explained in this answer, the property $V=(V^\perp)^\perp$ holds for submodules $V$ of $(\mathbb Z/q\mathbb Z)^n$ because you can use Smith normal form to find a suitable generating matrix for $V^\perp$. If you just want a reference for the result, I can only suggest the following paper:

D. Wilding, M. Johnson, M. Kambites. Exact rings and semirings. Journal of Algebra, 388 (2013), 324–337; arXiv:1212.5358.

When writing this paper we faced the same problem as you. That $(V^\perp)^\perp=V$ follows quite easily from known results, but does not appear to be explicitly written anywhere.