Of Algebra and Diagrams...

Prologue~
“Algebra is generous; she often gives more than is asked of her.”
— Jean le Rond d'Alembert

The aim of this post is to discuss a few theorems based of ring homomorphisms, while giving an intuition of diagram chasing, since our magic tool here would be commutative diagrams. We also encounter what a natural map would be in such a setting.

Verse I: The Epimorphism Theorem~

Let $f : R \to S$ be a ring homomorphism then there exists an isomorphism from $R/ Ker\ f$ to $S$, iff $f$ is an epimorphism.

Fig. 1

Now if you consider the adjacent figure, then clearly $f$ has been rightfully depicted as a map from $R$ to $S$.

We were supposed to find the map between the quotient ring and $S$, which here is represented as $\tilde{f}$.

Notice that the map denoted by $\eta$ here is a natural (or canonical) surjection, and it is always a surjective ring homomorphism. 

The meaning of surjective here is simply onto, i.e., the map $\eta$ takes an element of $R$ and lands it in $R/Ker \ f$, such that every element of $R/Ker\ f$ has a pre-image in $R$.

Now we must define $\tilde{f}$ to complete the proof. So what our $\tilde{f}$ does here is takes an element from $R/Ker \ f$ and maps it to the image of $R$ under $f.$

Hence, the above paragraph(s), for all elements $x \in R$ can be concluded as, 

$f(x)= \tilde{f} \circ \eta (x)$ 

and we say that the diagram in Fig. 1 commutes!

Here $\circ$ implies pre-right composition.

Verse II: Quotient of a Quotient Theorem~

If $I \subseteq J$ are both 2-sided ideals in $R$, then $(R/I)/(J/I)$ is naturally isomorphic to $R/J.$

Fig. 2

Let us dissect the adjacent figure which, one can claim that gives a precise and swift proof of the theorem stated above.

Here again, we have two quotient rings $R/I$ and $R/J$, where our $I$ and $J$ are ideals. And we have been given that $I \subseteq J$. (In)formally, our $I$ sits in $J$ and hence from what we did in the previous proof, we can say that the map $\eta_{IJ}$ forms a natural (or canonical) surjection, which again means that for every element of $R/I$, there exists a pre-image in $R$. 

Similar applies for $\eta_J$ and the map here is (again) a canonical surjection.

Interestingly, $\eta_{R/J}$ and $\eta_{IJ}$ are also canonical surjections from their respective domains to their codomains.

Fig. 2.1

Now, talking about the map which actually concerns us, because everything so far arose naturally from the ideal and quotient ring relations. 

We are interested in $\tilde{\eta}$, which is a map from $(R/I)/(J/I)$ to $R/J$, and since $I \subseteq J$, this again turns out to be another natural surjection, the only difference being that it's slightly more demanding than the others.

What this map does is take an element of the form $x+I+J/I$, where $x \in R$ and maps it to an element in $R/J$ of the form $x+J$, which is obvious, since we started with $x+I+J/I= x+I + j+I , \  \forall j \in J$

and since $I \subseteq J$, $j+I, \ \forall j \in J,$ belongs to $J$. Therefore, we can concisely write the $  J+I = J$.

Hence, $x+I+J/I = x+I + j+I= x+I+J= x+J $.

And our Fig. 2 commutes, which implies that $  \eta_I \circ \eta_{IJ}= \eta_J $ and $\eta_{IJ}= \tilde{\eta} \circ \eta_{R/ J}$

One may verify these.

Also see that we can combine the other two relations and write 

$ \eta_I \circ (\tilde{\eta} \circ \eta_{R/ J} )= \eta_J $

Verse III: Extension and Ideal Theorem~

If $I$ is a 2-sided ideal in $R$, then $I[X]$ is a 2-sided ideal of $R[X]$ and the quotient ring $R[X]/I[X]$ is naturally isomorphic to $(R/I)[X]$.

Fig. 3

The first part of the theorem can be proved by considering $I$ to be the 2-sided ideal and then we can proceed by writing $I[X]$ as $$ I[X]=\{  a_0 +a_1 X + a_2 X^2 + \cdots + a_r X^r , \ \forall \ a_i \in I \} $$

And, $I[X]$ is a 2-sided ideal in $R[X]$.

Now, consider the adjacent figure. 

Here, the maps $\gamma_{R/I}$ and $\gamma_{I[X]}$ are both natural surjections.

And our $\gamma_{R/I}$ is defined by 

$a_0 +a_1 X + a_2 X^2 + \cdots + a_r X^r  = \overline{a}_0 + \overline{a}_1 X + \overline{a}_2 X^2 + \cdots + \overline{a}_r X^r  $

where $\overline{a}= a+I, \ \forall \  a \in R$.

So, we are now concerned with our map $\tilde{\gamma}$ which takes objects from $R[X]/I[X] $ to an object in $(R/I)[X]$. Clearly, the objects of the domain are of the form 

$\tilde{\gamma}(a(X) + I[X]) = (a_0 + I) + (a_1 + I)X + \cdots + (a_n + I)X^n$

 and clearly, $(a_0 + I) + (a_1 + I)X + \cdots + (a_n + I)X^n \ \in (R/I)[X]$ and it's a surjection. every element of $(R/I)[X]$ has a pre-image in $R[X]/I[X] $.

Verse IV: Quotient maps~

Fig. 4

In all the above commutative diagrams, we obtain a figure of the adjacent form, and if all the maps in the figure, i.e., $\alpha, \beta$ and $\delta$ are onto, then the map $\alpha$ is said to factor through the object $C$ such that $ \alpha = \delta \circ \beta$ and hence in all the above figures, we obtain this universal property.

In general, a universal property describes how a particular object (such as a quotient ring, product, coproduct, or free object) is characterized by its relationships to other objects via maps that uniquely factor through it. 

Note that the diagram in Fig. 4 commutes.

Epilogue~

Algebra's real generosity is in its ability to unify results.

Many fundamental theorems, such as the isomorphism theorems for ring homomorphisms, become clear and accessible when approached through the lens of commutative diagrams and natural maps. These tools reveal underlying connections and provide a systematic framework for understanding why these results hold and how they fit together.

Reference- C. Musili, Rings and Modules.