Timeline for Invariance of $\mathbb{Z}[x]$ under a self-equivalence of the category of commutative rings with 1
Current License: CC BY-SA 4.0
17 events
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S Dec 24, 2019 at 3:41 | history | suggested | Emily | CC BY-SA 4.0 |
Corrected typo + removed abbreviation, since question was already on front page
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Dec 24, 2019 at 2:38 | review | Suggested edits | |||
S Dec 24, 2019 at 3:41 | |||||
Sep 12, 2013 at 6:59 | comment | added | Buschi Sergio | Martin Brandenburg: If I don't wrong (...), the ring structure of $(\mathbb{Z}[x], R)$ (natural in $R$) is equivalent to a coring struture on $\mathbb{Z}[x]$ (only observe that the tensor product is the sum in the commutative ring category, then apply Yoneda lemma). The some for more variables $x_1, \ldots x_n$. Is this fact used elsewhere? | |
May 20, 2011 at 8:32 | answer | added | Martin Brandenburg | timeline score: 12 | |
Mar 20, 2011 at 20:31 | vote | accept | Nico Bellic | ||
Feb 21, 2011 at 15:35 | answer | added | Martin Brandenburg | timeline score: 23 | |
Feb 19, 2011 at 15:19 | comment | added | Martin Brandenburg | I just want to remark that I've proven, among several other things of which I don't know if they are useful, that the whole thing works if we replace (rings) to ($k$-algebras) for an algebraically closed field $k$. | |
Feb 19, 2011 at 13:19 | comment | added | Martin Brandenburg | For every integral domain $R$ we have that $A \otimes R$ is also an integral domain, but $\mathbb{Z}[2x,x^2] \cong \mathbb{Z}[a,b]/(a^2=4b)$ is not integral after tensoring with $\mathbb{F}_2$. | |
Feb 19, 2011 at 10:09 | comment | added | Kevin Ventullo | Well, it's not quite a solution since I haven't ruled out, say $Z[2x,x^2]$, so I didn't post it as an answer. Universal is probably the wrong word to use, but basically $Q(x)$ is the only transcendental extension of $Q$ such that any other transcendental extension can be factored through it (though not uniquely). | |
Feb 19, 2011 at 9:54 | comment | added | Martin Brandenburg | This works: First, a purely transc. extension is a transc. extension which contains no nontrivial algebraic subextension. For a purely transc. extension the transc. degree is the length of the longest chain of purely transc. extensions in this extension. Thus $\mathbb{Q}(x)$ is preserved under $F$. | |
Feb 19, 2011 at 9:45 | comment | added | Martin Brandenburg | @Kevin: You should post this as an answer. Small correction: Fields are the nonterminal rings such that every morphism into a nonterminal ring is mono. In the end I don't know what you mean by saying that $\mathbb{Q}(x)$ is a universal transcendental extension. I think that we need to recover the transcendence degree ... | |
Feb 19, 2011 at 4:12 | answer | added | Charles Rezk | timeline score: 8 | |
Feb 19, 2011 at 4:06 | comment | added | Kevin Ventullo | of finite extensions). So $F(\mathbb{Z}[x])$ is a domain whose field of fractions is $Q(x)$. Moreover, it admits morphisms into any field, so its intersection with $Q$ is just $Z$. | |
Feb 19, 2011 at 4:04 | comment | added | Kevin Ventullo | Well, since mono=injective in rings, you know you can pick out the subcategory of fields in a categorical manner; these are the precisely the rings such that all morphisms out are monomorphisms. Hence, you can make sense of domains (they admit monos into fields), hence the field of fractions of a domain, since this is a universal property. $Q$ is the field of fractions of the initial object, and $Q(x)$ is a universal transcendental extension of $Q$ (finite algebraic extensions are those which have only finitely many morphisms into another field; arbitrary algebraic extensions are colimits... | |
Feb 19, 2011 at 3:01 | comment | added | Martin Brandenburg | ... but the question $A:=F(\mathbb{Z}[x]) \cong \mathbb{Z}[x]$ seems to be hard to decide. I've already proven many things about $A$, but I don't know if these properties actually characterize $\mathbb{Z}[x]$ ... | |
Feb 19, 2011 at 1:35 | comment | added | Martin Brandenburg | $F(\mathbb{Z}[x]) \cong \mathbb{Z}[x]$ already implies $F \cong \text{id}$. Proof: If $|R|$ is the set underlying $R$, then canonically $|R| \cong Hom(\mathbb{Z}[x],R)$. Apply $F$ on the right and get $|R| \cong |F(R)|$. This map is a ring homomorphism: To see this, use naturality to reduce to the case $R = \mathbb{Z}[x]$ or $R = \mathbb{Z}[x,y]$. The first works by definiton, and the second by the description of $\mathbb{Z}[x,y]$ as the coproduct of $\mathbb{Z}[x]$ and $\mathbb{Z}[y]$. | |
Feb 19, 2011 at 0:02 | history | asked | Nico Bellic | CC BY-SA 2.5 |