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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeNov 26th 2017

I added to field a mention of some other constructive variants of the definition, with a couple more references.

• CommentRowNumber2.
• CommentAuthorDavidRoberts
• CommentTimeNov 26th 2017

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.

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeNov 26th 2017

Thanks! I’ve noticed that before, but always forget about it.

1. Clarify a sentence to say that the discrete field condition implies that $0\neq 1$.

Egbert Rijke

• CommentRowNumber5.
• CommentAuthorUlrik
• CommentTimeJan 8th 2021

Change definition of denial field to actually match MRR87

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeMar 10th 2021

cross-linked the discussion of weakly initial sets in the category of fields with the corresponding example at multi-adjoint

2. one pair of authors defined fields to be possibly trivial.

Anonymous

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeMay 26th 2022
• (edited May 26th 2022)

I have taken the liberty of turning this comment from a sub-section into a remark (here). Also tried to adjust the wording a little, for readability.

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeJun 25th 2022

• CommentRowNumber10.
• CommentAuthorMarkSaving
• CommentTimeJul 3rd 2022

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 $p$. Consider the set $R_p = \mathbb{Z} \cup \{x \in \mathbb{Q} \mid p\}$, which is a subring of $\mathbb{Q}$. Since $R_p$ is a subset of $\mathbb{Q}$ and $\mathbb{Q}$ has decidable equality, $R_p$ also has decidable equality. And of course $0 \neq 1$ in $R_p$.

I claim that $R_p$ is a residue field iff $\neg \neg p$. For suppose $\neg \neg p$, and consider some $x \in R_p$. Suppose $x$ does not have a multiplicative inverse. Now suppose $x \neq 0$. Then we see that $x^{-1} \notin R_p$. If $p$ held, we would have $x^{-1} \in R_p$. So we know $\neg p$ holds. But this is a contradiction. Therefore, $x$ must be zero (using decidable equality).

Conversely, suppose $R$ is a residue field. Then $2 \neq 0$, so 2 does not fail to have an inverse. That is, $2^{-1}$ is not not in $R_p$. Then $\neg \neg p$.

I claim that $R_p$ is a Heyting field iff $p$ iff $R_p$ is a discrete field. For suppose $R$ is a Heyting field. Then either 2 or 3 has a multiplicative inverse, so either $2^{-1}$ or $3^{-1} \in R_p$. In either case, we see that $p$ holds. If $p$ holds, then $R_p = \mathbb{Q}$, which is a discrete field. And if $R_p$ 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 $p \iff \neg \neg p$ holds for all $p$. So we have full excluded middle.

Of course, assuming excluded middle, it is clear that all residue fields with decidable equality are discrete fields. $\square$

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.

3. added Mark Saying’s proof that not every residue field with decidable equality is Heyting.

Anonymous