Timeline for Can transcedence degree be defined for arbitrary ring homomorphism?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 7 at 0:58 | comment | added | Junyan Xu | Does anyone know whether there could exist an injective R-algebra homomomorphism $R[X,Y]\to R[X]$ between polynomial algebras, for some commutative ring R? It is easy to reduce to the case that R is a f.g. $\mathbb{Z}$-algebra; does dimension theory somehow prohibit this in this case? | |
| Oct 13, 2013 at 10:12 | comment | added | Sasha Anan'in | As to me, commutative algebra is enough. | |
| Oct 13, 2013 at 3:38 | comment | added | user39380 | Another question is (may be not a good one):can we also change algebraically independent into integrally independent over $f(A)$ in the tag? | |
| Oct 13, 2013 at 3:34 | comment | added | user39380 | I realized there're two "algebraic over" in my tag and you referred to change only the second one to be "integral over".Thanks for your help! | |
| Oct 12, 2013 at 15:05 | comment | added | Sasha Anan'in | $MC$ consist of all polynomials with coefficients in $M$ (remember, $C=A[b_1,\dots,b_n]$ is the ring of polynomials in $b_1,\dots,b_n$). So, $C/MC\cong(A/M)[b_1,\dots,b_n]$. As $A/M$ is a domain, a polynomial ring over it is also a domain, implying that $MC$ is prime. | |
| Oct 12, 2013 at 14:51 | comment | added | user39380 | A question is I don't know why $MC$ is a prime ideal of $C$. I can show the quotient ring is a domain if ${b_1,.,b_n}$ are algebraic independent over A. But does it still hold if they are just integral independent? Could you explain it a little bit? | |
| Oct 12, 2013 at 14:44 | comment | added | user39380 | You're right the exchange axiom just hold for the algebraic case. Originally I was unable to check axiom 3 for algebraic case.. | |
| Oct 12, 2013 at 10:52 | history | edited | Sasha Anan'in | CC BY-SA 3.0 |
a couple of TeX symbos and a better wording at the end
|
| Oct 12, 2013 at 0:45 | comment | added | Sasha Anan'in | @mqx I am afraid of that the property 4 (Steinitz exchange axiom) does not hold for integral dependence in spite of the fact that $the\ best\ dependence\ is\ the\ integral\ one$ (used to say so to students). | |
| Oct 11, 2013 at 23:32 | comment | added | user39380 | Thanks for your clarification on the definition of algebraic.I think this is the point. Also I think we may use Jacobson's Proof with your clarification. The use of krull dim is unexpected. And thank you for your nice example! | |
| Oct 11, 2013 at 22:40 | vote | accept | CommunityBot | ||
| Oct 11, 2013 at 22:13 | history | edited | darij grinberg | CC BY-SA 3.0 |
latex messed up
|
| Oct 11, 2013 at 19:22 | history | edited | Sasha Anan'in | CC BY-SA 3.0 |
added 10 characters in body
|
| Oct 11, 2013 at 15:14 | history | edited | Sasha Anan'in | CC BY-SA 3.0 |
deleted 5 characters in body
|
| Oct 11, 2013 at 15:08 | history | answered | Sasha Anan'in | CC BY-SA 3.0 |