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I added to field a mention of some other constructive variants of the definition, with a couple more references.
I fixed up the DOI link for Johnstone’s paper (at field and at local ring), since the geniuses at Elsevier use parentheses in their DOIs, which break markdown syntax.
For reference, one needs to replace (nn)
in the DOI url by %28nn%29
.
Thanks! I’ve noticed that before, but always forget about it.
cross-linked the discussion of weakly initial sets in the category of fields with the corresponding example at multi-adjoint
added pointer to:
There appears to be an error in the section on construction notions of a field. Specifically, the claim is that a residue field is discrete iff equality is decidable. But this seems not to be true.
In fact, the statement “A residue field with decidable equality is a Heyting field” is equivalent to excluded middle.
To see this, consider a proposition . Consider the set , which is a subring of . Since is a subset of and has decidable equality, also has decidable equality. And of course in .
I claim that is a residue field iff . For suppose , and consider some . Suppose does not have a multiplicative inverse. Now suppose . Then we see that . If held, we would have . So we know holds. But this is a contradiction. Therefore, must be zero (using decidable equality).
Conversely, suppose is a residue field. Then , so 2 does not fail to have an inverse. That is, is not not in . Then .
I claim that is a Heyting field iff iff is a discrete field. For suppose is a Heyting field. Then either 2 or 3 has a multiplicative inverse, so either or . In either case, we see that holds. If holds, then , which is a discrete field. And if is a discrete field, it is clearly a Heyting field.
With these facts in hand, we see that if every residue field with decidable equality is a Heyting field, then holds for all . So we have full excluded middle.
Of course, assuming excluded middle, it is clear that all residue fields with decidable equality are discrete fields.
I have not yet determined whether all Heyting fields with decidable equality are discrete, but it seems very likely that this also cannot be proven.
If no one objects, I will change the page to say that a residue field is discrete if and only if unithood is decidable.
Added reference
and used the more clear “weak Heyting field” over the ambiguous “residue field” throughout the page.
Anonymouse
I recently came back to this page and, after some thought, have demonstrated we cannot constructively prove all Heyting fields with decidable equality are discrete.
Theorem: if all Heyting fields with decidable equality are discrete, then excluded middle holds.
Proof: Fix any prime number . Suppose is a proposition. Then define by . Then is a prime filter, so is a local ring. Let be the corresponding Heyting field.
Note that in if and only if is not a unit in , if and only if , if and only if . Thus, if has decidable equality, then is decidable. Conversely, suppose is decidable. The ideal of nonunits in is , which is a decidable subset of , so has decidable equality. Thus, has decidable equality if and only if is decidable.
Note that is a unit in if and only if is a unit in , iff , iff . So if is a discrete field, then is decidable. Conversely, if is decidable, then is either or , and thus is a discrete field. So is a discrete field if and only if is decidable.
Now suppose all Heyting fields with decidable equality are discrete fields, and suppose . Then is decidable, so is a Heyting field with decidable equality, so is a discrete field. Because is a discrete field, is decidable. Since is decidable and , we can conclude . We thus have double negation elimination, and hence full excluded middle.
If no one sees a flaw in this argument, then we should delete the portion of the page claiming “A Heyting field is a discrete field if and only if equality is decidable; it is in this sense that a discrete field is ‘discrete’.” Decidable equality is constructively weaker than decidable unithood.
Replaced
A Heyting field is a discrete field if and only if equality is decidable; it is in this sense that a discrete field is ‘discrete’
with
A Heyting field is a discrete field if and only if its apartness relation is a decidable relation.
This exact mistake was made and later corrected on the principle of omniscience page, where editors confused the analytic WLPO (reals have decidable equality) with the analytic LPO (reals have decidable apartness).
Also removed
It is not true that every weak Heyting field with decidable equality is Heyting. See this proof for details.
since decidable equality isn’t really all that important for constructive fields. Locality (in the sense of local ring) and decidable apartness are more important.
changed higher algebra - contents to algebra - contents in context sidebar
Anonymouse
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