# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorTodd_Trimble
• CommentTimeJul 5th 2015

Created splitting field.

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeJul 5th 2015

Made improvements to splitting field, noting in particular that existence and uniqueness up to isomorphism of splitting fields of arbitrary sets of polynomials doesn’t require the full axiom of choice, but only the ultrafilter principle. This applies in particular to algebraic closures.

• CommentRowNumber3.
• CommentAuthorDavidRoberts
• CommentTimeJul 6th 2015

Is it worth pointing out that for a countable field (and so a countable set $S$ of polynomials) one doesn’t need such a strong axiom to get algebraic closure etc? Or better, that for a set of polynomials with bounded cardinality, there much be an ultrafilter principle (for sets up to a certain size) that implies the existence of splitting fields.

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeJul 6th 2015

Sure, there are a number of such statements one could make: for finite fields no choice is required, and for any field (such as $\mathbb{Q}$) which sits canonically inside $\mathbb{C}$ you don’t need choice, in addition to the statements you gave. Please feel free to add such, or I might do so sometime later.

• CommentRowNumber5.
• CommentAuthorZhen Lin
• CommentTimeJul 6th 2015

Isn’t there some subtlety even in the countable case? I vaguely recall something terrible like, a countable field has a unique countable algebraic closure, but there may also be uncountable algebraic closures.

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeJul 6th 2015
• (edited Jul 6th 2015)

Right, there was this thread on FOM started by Timothy Chow about uniqueness of algebraic closure of even $\mathbb{Q}$ if one doesn’t assume some amount of choice or choice-consequence, perhaps such as the statement that the union of countably many finite sets is countable. I haven’t looked at the cited paper, but it sounds like Lauchli has constructed these monsters that Zhen Lin is referring to. I didn’t trace through to the end of the discussion there, which proceeds by fits and (false) starts. (Sorting by thread is highly recommended.)

• CommentRowNumber7.
• CommentAuthorDavidRoberts
• CommentTimeJul 7th 2015
• (edited Jul 7th 2015)

Let me point out for completeness that the thread continues, at times, in the following month. No conclusion seems to be reached!

Note that the contention is mostly around the definition of algebraic closure.