0

For the following post on One-to-one correspondence of ideals in the quotient also extends to prime ideals? Bill refer to the proof by Arturo given in his answer to theorem asked from the author of that post which is

For a comm. ring $A$ and an ideal $I$ of $A$, does the one-to-one correspondence between ideals of the quotient $A/I$ and ideals of $A$ containing $I$ extends to a correspondence of prime ideals? $(*)$

Bill states:

Yes, this is a standard result. The proof you give is the same as that in Zariski and Samuel, Commutative Algebra I, S.3.8, Theorem 11. See also the slightly more general presentation on extension and contraction, p.9 in Atiyah and MacDonald, Introduction to Commutative Algebra. $\ $ I had no problem locating it by web searches, e.g. the Lemma at the bottom of p.8 Boocher's notes.

I looked up Zariski and Samuel vol 1 of commutative algebra chapter 3 section 8 theorem 11 and all I found was:

Theorem 11. Let $T$ be a homomorphism of a ring $R$ onto a ring $R'$ with kernel $N$. If $\mathfrak{N}$ is an ideal in $R$ containing $N$, then $\mathfrak{N}$ is respectively prime or maximal if and only if $\mathfrak{N}T$ is prime or maximal. If $\mathfrak{N'}$ is an ideal in $R'$, then $\mathfrak{N'}$ is respectively prime or maximal if and only if $\mathfrak{N'}T^{-1}$ is prime or maximal.

I am not clear on the meaning of Zariski and Samuel's notatation for $\mathfrak{N}T\mathfrak{N'}T^{-1}$ Also, he doesn't the notion of define completely prime ideal. Am I suppose to do some notation matching with what Theorem 11 says with that of $(*)$? I also looked at Atiyah and MacDonald's text for page 9 on contraction and extension of ideals. I was not able to find anything that matches what the theorem in $(*)$ states. I also consulted the referenced p.8 Boocher's notes., I am not seeing how what is stated at the bottom there matches what $(*)$ states.

Thank you advance.

Seth
  • 4,043
  • 1
    Please have a look at Bourbakis Algebra. Or Aluffis Algebra. Or basically any standard book on that topic guess. – Martin Brandenburg Feb 25 '25 at 01:30
  • Zarsiki-Samuel of course talk about image and preimage. Just read it from right to left. – Martin Brandenburg Feb 25 '25 at 01:32
  • @MartinBrandenburg oh Zariski and Samuel were referring to images and pre-images. ok ok, thank you for the clear up – Seth Feb 25 '25 at 02:02
  • Re: completely prime: the linked question is about commutative rings, for which a "completely prime ideal" is equivalent to a "prime ideal", as Arturo mentions there. – Bill Dubuque Feb 25 '25 at 02:12
  • 4
    Again, as has been mentioned in many of your prior questions, you are having trouble with basic matters because you are trying to run before you have learned to walk. When studying mathematics it is essential to have competence in the listed prerequisite fields. – Bill Dubuque Feb 25 '25 at 02:19
  • This book is an old one, from a time when functions were conventionally written opposite to how we write them now. So, if you have a function $f\colon X\to Y$, then for a subset $A\subseteq X$, the notation $Af$ means $f(A)$ and for $B\subseteq Y$, the notation $Bf^{-1}$ is $f^{-1}(B)$. – ultralegend5385 Feb 25 '25 at 07:05
  • @ultralegend5385 thank you for the clear up also. I wasn't sure since Zarski and Samuel also uses the $f(x)$ notations, transformation as mappings $xT$ notations and the product of two ideals $AB$ notations. The thing is like about them is there is a lot more exposition as if you are there hearing them lecture or reading Richard Courant's Calculus texts or Hardy's analysis' text. There is none of the "readers are suppose to have certain prerequisite". Besides, according to Abel, it is always good to be reading and learning from the masters. – Seth Feb 25 '25 at 09:04

0 Answers0