@fresh_42 I am posting the quoted text above without enclosing it in quotes:
"We shall now undertake a slight generalization of the concept of a quotient ring. We consider a homomorphism ##f## of ##R## into a ring ##S## such that ##f(m)## is a unit for every ##m\in M##, where ##M## is a given multiplicative system in ##R##. If ##x## is an element of ##R## such that ##mx=0## for some ##m## in ##M##, we have ##0=f(xm)=f(x)f(m)##, and since ##f(m)## is a unit in ##S##, this implies ##f(x)=0##. In other words, the kernel of ##f## must contain the set ##\mathfrak{n}## of all elements ##x## in ##R## for which there exists an element ##m## in ##M## such that ##mx=0##. Since ##M## is multiplicatively closed, this set ##\mathfrak{n}## is an
ideal in ##R##, as is readily verified; and since ##0\not\in M##, we have ##1\not\in \mathfrak{n}## and ##\mathfrak{n}\neq R##. Thus, the image ##f(R)## of ##R## in ##S## is isomorphic to a residue class ring of ##R/\mathfrak{n}##, and ##f## defines a homomorphism ##f'## of ##R/\mathfrak{n}## into ##S##. Now, the canonical image ##\bar{M}=\frac{M+\mathfrak{n}}{\mathfrak{n}}## of ##M## in ##R/\mathfrak{n}## is obviously closed under multiplication. Further, ##\bar{M}
## does not contain any zero divisor: for, if ##\bar{x}\cdot\bar{m}=0 \;(\bar{x}\in R/\mathfrak{n},\bar{m}\in\bar{M})## and ##x,m## are representatives of ##\bar{x},\bar{m}## in ##R## and ##M##, then ##xm\in \mathfrak{n},xmm'=0## for a suitable element ##m'\in M##, and since ##mm'\in M##, we deduce that ##x\in \mathfrak{n}## and ##\bar{x}=0##. Thus ##\bar{M}## is a regular multiplicative system in ##R/\mathfrak{n}##, and we can construct the ordinary quotient ring ##(R/\mathfrak{n})_\bar{M}##. Since every element of ##f'(\bar{M})\;(=f(M))## is a unit in ##S##, the homomorphism ##f'## may be extended to a homomorphism (still denoted by ##f'##) of ##(R/\mathfrak{n})_\bar{M}## into ##S## by setting ##f'(\bar{x}/\bar{m})=f'(\bar{x})/f'(\bar{m})## (the fact that ##f'## is single valued and is a homomorphism is easily proved. The ring ##(R/\mathfrak{n})_\bar{M}## is called the
quotient ring of ##R## with respect to the multiplicative system ##M## and is denoted by ##R_M##. We notice that if ##M## is regular, we have ##\mathfrak{n}=(0), R/\mathfrak{n}=R, \bar{M}=M##, and the new terminology and notation is consistent with the old one."
I tried to figure out what the term 'extended' mean and how the canonical map ##\bar{M}## is defined and its associated homomorphic map.
A straight reading allows me to deduced the following:
We have ##R/\mathfrak{n}## is the quotient ring of ##R## by the ideal ##\mathfrak{n},\\\\##
##S## is a ring such that ##f(m)## is a unit in ##S## for all ##m \in M\\\\##
The ring ##(R/\mathfrak{n})_{\bar{M}}## is the quotient ring (or localization) of ##R/\mathfrak{n}## with respect to the multiplicative system ##\bar{M} \subseteq R/\mathfrak{n}\\\\##
We have the quotient set ##(R/\mathfrak{n})_{\bar{M}}=\{[(\bar{x}, \bar{m})]\mid \bar{x} \in R/\mathfrak{n},\bar{m} \in \bar{M}\}\{[\frac{\bar{x}}{\bar{m}}]\mid \bar{x} \in R/\mathfrak{n},\bar{m} \in \bar{M}\}## under the equivalence relation: ##(\bar{x}, \bar{m}) \sim (\bar{y}, \bar{n})## if there exists ##\bar{t} \in \bar{M}## such that ##\bar{t}(\bar{n}\bar{x} - \bar{m}\bar{y}) = 0\\\\##
The term: ''extended'' seem to mean one can define a new homomorphism, using the notation: ##f', f':(R/\mathfrak{n})_{\bar{M}}\to S##,since ##(R/\mathfrak{n})_{\bar{M}}## is a larger ring of ##S##, so ##S\subset (R/\mathfrak{n})_{\bar{M}}\to S##, and ##f'=f'|_{R/\mathfrak{n}}.## Then ##\\\\##
##f'(\bar{x}/\bar{m}) = \{\frac{f'(\bar{x})}{f'(\bar{m})}\in S\mid \bar{x}/\bar{m} \in (R/\mathfrak{n})_{\bar{M}},f'(\bar{m}) = f(m) \text{ is a unit in }S \text{ for all } \bar{m} \in \bar{M}, m\in M\}\\\\##
The canonnical image of ##\bar{M}.\\\\##
The set ##\bar{M}## is the canonical image of the multiplicative system ##M \subseteq R## in the quotient ring ##R/\mathfrak{n}##, where ##\\\\##
##\mathfrak{n} = \{ x \in R \mid \exists m \in M \text{ such that } mx = 0 \}## is an ideal in ##R.\\\\##
We have the canonical projection ##\pi: R \to R/\mathfrak{n}## is defined by ##\pi(x) = x + \mathfrak{n}.\\\\##
The set ##\bar{M}## consists of the images of elements of ##M## under ##\pi:\\\\##
##\bar{M} = \frac{M + \mathfrak{n}}{\mathfrak{n}} = \{ m + \mathfrak{n} \mid m \in M \}= \{\bar{m} \in R/\mathfrak{n} \mid \exists m \in M \text{ such that } \bar{m} = m + \mathfrak{n} \}.\\\\##
The second equality is written using set builder notation.##\\\\##
the mapping associated with canonical image of ##\bar{M}## is the homomorphic mapping ##f' : R/\mathfrak{n} \to S##, which is created from the original homomorphism ##f : R \to S## defined by ##f'(x + \mathfrak{n}) = f(x), \bar{m} = m + \mathfrak{n} \in \bar{M};## which gives:##\\\\##
##f'(\bar{m}) = f'(m + \mathfrak{n}) = f(m).\\\\##
Because ##f(m)## is a unit in ##S## for all ##m \in M##, the set ##f'(\bar{M}) = \{ f(m) \mid m \in M \}## consists of units in ##S##.
However from an MSE post, I got:
Explainations from an answer on MSE:##\\\\##
There are a lot of misunderstandings. Let me rewrite some of the text:##\\\\##
Consider a ring homomorphism ##f \colon R \to S## and a multiplicative system ##M## in ##R.\\\\##
Suppose that each element of ##f(M)## is a unit in ##S.\\\\##
Then it can be shown that the set##\\\\##
##\mathfrak n := \{x \in R : mx=0 \text{ for some } m \in M\}\\\\##
is a proper ideal of ##R## such that ##\mathfrak n \subset \ker f.\\\\##
##\textbf{Note 1:}\\\\##
We don't know if ##\mathfrak n = \ker f.\\\\##
Thus, ##f## defines a ring homomorphism ##f' \colon R/\mathfrak n \to S.\\\\##
##\textbf{Note 2:}\\\\##
It is not correct to say that ##f'=f \colon R/\mathfrak n \to S\\\\##.
Here we are using the ##\textit{universal property of the quotient ring}##: If ##f \colon R \to S## is a ring homomorphism and ##\mathfrak n## is an ideal of ##R## such that ##\mathfrak n \subset \ker f##, then there exists a (unique) ring homomorphism ##f' \colon R/\mathfrak n \to S## such that ##
f'(x+\mathfrak n) = f(x) \quad \text{for all } x \in R.\\\\##
The condition ##\mathfrak n \subset \ker f## is used to show that ##f'## is well-defined with the above formula.##\\\\##
Next, it can be shown that the quotient##\\\\##
##\bar M := (M+\mathfrak n)/\mathfrak n\\\\##
is a regular multiplicative system in ##R/\mathfrak n.\\\\##
##\textbf{Note 3:}\\\\##
As you observe, ##\bar M = \{m+\mathfrak n : m \in M\}##, that is, ##\bar M## is the image of ##M## under the canonical map ##\pi \colon R \to R/\mathfrak n## (this is why ##\bar M## is called the canonical image of ##M## in ##R/\mathfrak n).\\\\##
However, it is incorrect to write ##\bar M## as: ##\\\\##
- ##M/\mathfrak n##, since we don't know if ##\mathfrak n \subset M##. (The notation ##A/B## only makes sense when ##B \subset A.)\\\\##
- ##M+\mathfrak n##, since this sum denotes the subset ##\{m+n : m \in M, n \in \mathfrak n\}## of ##R.\\\\##
Hence, as each element of ##f'(\bar M) = \{f'(m+\mathfrak n) : m \in M\} = \{f(m) : m \in M\} = f(M)## is a unit in ##S##, the homomorphism ##f' \colon R/\mathfrak n \to S## can be extended to a homomorphism ##f'' \colon (R/\mathfrak n)_{\bar M} \to S## by defining##\\\\##
##f''(\bar x/\bar m) = f'(\bar x)/f'(\bar m) \quad \text{for all } \bar x \in R/\mathfrak n \text{ and } \bar m \in \bar M.\\\\##
By simplicity, ##f''## is still denoted by ##f'.\\\\##
##\textbf{Note 4:}\\\\##
Since ##R/\mathfrak n## is a subset of ##(R/\mathfrak n)_{\bar M}##, ##f''## is indeed an extension of ##f'##, meaning that the restriction of ##f''## to ##R/\mathfrak n## is equal to ##f'##:##\\\\##
For each ##\bar x \in R/\mathfrak n,\\\\##
##f''(\bar x) = f''(\bar x/\bar 1_R) = f'(\bar x)/f'(\bar 1_R) = f'(\bar x)/1_S = f'(\bar x).\\\\##
How did Zariski and Samuel hid so much information in so few a sentences? I am hoping you can help with relating or help with tying all the various homormophic mappings together and how each set (quontient rings, ideals,etc) are related to their respective larger sets,
