Commutative Algebra Explained: In-Depth Examples, Equations, and Expert Insights

Let $R$ be a commutative ring (with identity) and let $R\mathbf{A}\mathbf{l}\mathbf{g}$ denote the category of $R$-algebras.
Is there a suitable notation for the full subcategory of commutative $R$-algebras?
I would like to know this in order to classify the polynomial ring $R\left[S\right]$ - where $S$ is a set - as an object of category ? free over set $S$.
In the special case $R=\mathbb{Z}$ the notation $\mathbf{C}\mathbf{R}\mathbf{i}\mathbf{n}\mathbf{g}$ will do, since rings can be recognized as $\mathbb{Z}$-algebras. But what in the general case?

Let $A,B$ be commutative unital Banach algebras and let $\mathrm{\phi <!--\; \phi \; -->}:A\to <!--\; \to \; -->B$ be a continuous unital map such that
$\stackrel{¯<!--\; ¯\; -->}{\mathrm{\phi <!--\; \phi \; -->}(A)}=B$
Let
${\mathrm{\phi <!--\; \phi \; -->}}^{\ast <!--\; \ast \; -->}:\text{Max}(B)\to <!--\; \to \; -->\text{Max}(A)$
${\mathrm{\phi <!--\; \phi \; -->}}^{\ast <!--\; \ast \; -->}(m)=m(\mathrm{\phi <!--\; \phi \; -->})$
be the map from the space of maximal ideals of $B$ to the space of maximal ideals of $A$ induced by $\mathrm{\phi <!--\; \phi \; -->}$.
How to prove that ${\mathrm{\phi <!--\; \phi \; -->}}^{\ast <!--\; \ast \; -->}$ is a topologically injective map?
(Recall that an operator $T:X\to <!--\; \to \; -->Y$ is called topologically injective if $T:X\to <!--\; \to \; -->\text{im}(T)$ is a homeomorphism)
My progress on the problem is the following: first of all, the space of maximal ideals of a commutative Banach algebra $A$ can be identified with the space of continuous functionals of the form $m:A\to <!--\; \to \; -->\mathbb{C}$. Clearly the map above is continuous, since the pointwise convergence of a net $(n_i)$ in $\text{Max}(B)$ implies the convergence of the net $m_i\circ <!--\; \circ \; -->\mathrm{\phi <!--\; \phi \; -->})$)
(the space of continuous linear functional is endowed with the weak* topology)
If we assume for the moment that the map is bijective, then the fact that a continuous bijective map between compact Hausdorff spaces is a homeomorphism, yields the result.
(here the space of maximal ideals is compact in weak* topology since the algebra is unital)
The suggested proposition looks like a relaxation of the aformentioned reasoning above though i cannot figure out an easy way to modify it to make it work. Are there any hints?

Define the $\ast <!--\; \ast \; -->$ on $l^1(\mathbb{Z})$ by
$(x\ast <!--\; \ast \; -->y)(n)=\underset{k=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}x(n-<!--\; -\; -->k)y(k),\text{with }n\in <!--\; \in \; -->\mathbb{Z}\text{ and }x,y\in <!--\; \in \; -->l^1(\mathbb{Z})$
I want to show that this defines a commutative Banach algebra.
Firstly I must show that for $x,y,z\in <!--\; \in \; -->l^1(\mathbb{Z})$,
$(x\ast <!--\; \ast \; -->(y\ast <!--\; \ast \; -->z))(n)=((x\ast <!--\; \ast \; -->y)\ast <!--\; \ast \; -->z)(n)$
However, I am struggling to derive the expression for (x∗(y∗z))(n). I came up with something like
$(x\ast <!--\; \ast \; -->(y\ast <!--\; \ast \; -->z))(n)$
$(x\ast <!--\; \ast \; -->(y\ast <!--\; \ast \; -->z))(n)=\underset{k=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}x(n-<!--\; -\; -->k)(y(n-<!--\; -\; -->k)z(k))$
But I am sure that is wrong.

Suppose $k$ is a field and $A$ is a $k$-algebra of dimension no larger than 3. If $A$ is semi-simple, then $A$ can be written as a direct sum of simple $k$-algebras. Further one can find $A$ is commutative by exhausting all the cases.
Without the semi-simple condition, what can we say about $A$? Is it still commutative?

Does the category of representations over an augmented algebra over a commutative ring have enough projectives?
I think it would be easier to begin with my motivation. Let $R$ be a commutative ring and $\mathfrak{g}$ a Lie algebra over $R$. In Weibel's Homological algebra he states that the category of $\mathfrak{g}$-representations (which he calls g-modules; they are $R$-modules with a $\mathfrak{g}$ action compatible with the $[,]$ of $\mathfrak{g}$) has enough projectives and injectives. To prove this, he shows that the category of representations over g and the category of representations over $U\mathfrak{g}$ are naturally isomorphic, where Ug is the universal enveloping algebra.
He claims this is enough, but I don't understand why. I turned to Cartan & Eilenberg, where they make the argument that, because $U\mathfrak{g}$ is an augmented $R$-algebra, we can compute for example $To{r}_n^{U\mathfrak{g}}(A,R)$ for any $A$ a left $U\mathfrak{g}$ module by using a projective resolution of $R$ as a $U\mathfrak{g}$ module. This is only useful for defining homology as such. But does it show that the category of $U\mathfrak{g}$-representations has enough projective objects (which is required for Weibel's definition, since he uses another derived functor)?

Form the free algebra on a set which is in bijection with $S$
$A⟨<!--\; ⟨\; -->i_s:<!--\; :\; -->s\in <!--\; \in \; -->S⟩<!--\; ⟩\; -->$
and impose the relation that $i_s$ is a two-sided inverse of $s\in <!--\; \in \; -->S$ for each $s\in <!--\; \in \; -->S$:
$S^{-<!--\; -\; -->1}A:=\frac{A⟨<!--\; ⟨\; -->i_s:<!--\; :\; -->s\in <!--\; \in \; -->S⟩<!--\; ⟩\; -->}{⟨<!--\; ⟨\; -->s{i}_s-<!--\; -\; -->1,i_{s}s-<!--\; -\; -->1:<!--\; :\; -->s\in <!--\; \in \; -->S⟩<!--\; ⟩\; -->}.$
Then define $\mathrm{\varphi <!--\; \varphi \; -->}A\to <!--\; \to \; -->S^{-<!--\; -\; -->1}A$ by letting $\mathrm{\varphi <!--\; \varphi \; -->}(a)$ be the image of $a$ in $S^{-<!--\; -\; -->1}A$. By definition $\mathrm{\varphi <!--\; \varphi \; -->}(S)$ consists of the units of $S^{-<!--\; -\; -->1}A$.
What I am confused about is what is precisely meant here by
Form the free algebra on a set which is in bijection with $A⟨<!--\; ⟨\; -->i_s:<!--\; :\; -->s\in <!--\; \in \; -->S⟩<!--\; ⟩\; -->$.
What is the definition of free algebra over a non-commutative ring $A$?

Let $A$ be a commutative unital Banach algebra. We know that for any two elements of $A$ $\mathrm{\sigma <!--\; \sigma \; -->}(a+b)\subset <!--\; \subset \; -->\mathrm{\sigma <!--\; \sigma \; -->}(a)+\mathrm{\sigma <!--\; \sigma \; -->}(b)$ and $\mathrm{\sigma <!--\; \sigma \; -->}(ab)\subset <!--\; \subset \; -->\mathrm{\sigma <!--\; \sigma \; -->}(a)\mathrm{\sigma <!--\; \sigma \; -->}(b)$.
I know that if we consider $2\times <!--\; \times \; -->2$ matrices, then we get examples of a non-commutative unital banach algebra such that $\mathrm{\sigma <!--\; \sigma \; -->}(a)+\mathrm{\sigma <!--\; \sigma \; -->}(b)\not\subset \mathrm{\sigma <!--\; \sigma \; -->}(a+b)$ and $\mathrm{\sigma <!--\; \sigma \; -->}(a)\mathrm{\sigma <!--\; \sigma \; -->}(b)\not\subset \mathrm{\sigma <!--\; \sigma \; -->}(ab)$.
I wonder if there are examples of $a$ and $b$ in a non-unital commutative $A$ such that $\mathrm{\sigma <!--\; \sigma \; -->}(a)+\mathrm{\sigma <!--\; \sigma \; -->}(b)\not\subset \mathrm{\sigma <!--\; \sigma \; -->}(a+b)$ and $\mathrm{\sigma <!--\; \sigma \; -->}(a)\mathrm{\sigma <!--\; \sigma \; -->}(b)\not\subset \mathrm{\sigma <!--\; \sigma \; -->}(ab)$. (I can't think of any at the moment)

Let $k$ be any field and let $A=k[X,Y,Z]/(X^2-<!--\; -\; -->Y^3-<!--\; -\; -->1,XZ-<!--\; -\; -->1)$.
How can I find $\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}\in <!--\; \in \; -->k$ such that $A$ is integral over $B=k[X+\mathrm{\alpha <!--\; \alpha \; -->}Y+\mathrm{\beta <!--\; \beta \; -->}Z]$?
For these values of $\mathrm{\alpha <!--\; \alpha \; -->}$ and $\mathrm{\beta <!--\; \beta \; -->}$, how can I find concrete generators for $A$ as a $B$-module?

Is every finite ring a matrix algebra over a commutative ring?
- Can every finite ring $R$ be written as a subring of ${\text{Mat}}_{n\times <!--\; \times \; -->n}A$ for some commutative ring $A$?
- If not, then what is/are the smallest ring(s) that cannot be?

If $R$ is a commutative associative ring with neutral element. Then if $R|P$ is an integral domain $ab\in <!--\; \in \; -->R|P⟹<!--\; ⟹\; -->a=0orb=0$ , then $1\notin <!--\; \notin \; -->$ ? This is the only thing I can deduct from another proof, is this correct , or must I search for another reason?

Brown's Matrices over Commutative Rings book discusses the theory of eigenvalues, eigenvectors, and diagonalizing matrices over commutative rings, but unless I've missed something, nothing like generalized eigenvectors is even mentioned. Has the notion of generalized eigenvectors ever been studied over commutative rings in any context more general than working over a field?
My interest stems from trying to find a closed form expression for the matrix exponential $e^{zA}$, where $z$ is an element of a finite dimensional real associative commutative algebra, and $A$ is a matrix built of elements of said algebra, which in the real case relies on the theory of generalized eigenvectors.

For a finite abelian $p$-group $G$ we have that
$G\simeq <!--\; \simeq \; -->\mathbf{Z}/(p{)}^{{\mathrm{\lambda <!--\; \lambda \; -->}}_1}\oplus <!--\; \oplus \; -->\cdots <!--\; \cdots \; -->\oplus <!--\; \oplus \; -->\mathbf{Z}/(p{)}^{{\mathrm{\lambda <!--\; \lambda \; -->}}_r}$
for some positive integers ${\mathrm{\lambda <!--\; \lambda \; -->}}_1\ge <!--\; \ge \; -->\cdots <!--\; \cdots \; -->\ge <!--\; \ge \; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_r$. Note that $G$ is uniquely determined by p and this partition $\mathrm{\lambda <!--\; \lambda \; -->}=({\mathrm{\lambda <!--\; \lambda \; -->}}_1,\dots <!--\; \dots \; -->,{\mathrm{\lambda <!--\; \lambda \; -->}}_r)$, so let's call λ the type of $G$. For types $\mathrm{\lambda <!--\; \lambda \; -->}$, $\mathrm{\mu <!--\; \mu \; -->}$, and $\mathrm{\nu <!--\; \nu \; -->}$, define the Hall number $g_{\mathrm{\mu <!--\; \mu \; -->},\mathrm{\nu <!--\; \nu \; -->}}^{\mathrm{\lambda <!--\; \lambda \; -->}}(p)$ to be the number of normal subgroups $N◃<!--\; ◃\; -->G$ of type $\mathrm{\nu <!--\; \nu \; -->}$ such that $G/N$ has type $\mathrm{\mu <!--\; \mu \; -->}$. These Hall numbers serve as the structure constants of an associative algebra called the Hall algebra.
It turns out that this algebra is commutative, i.e. $g_{\mathrm{\mu <!--\; \mu \; -->},\mathrm{\nu <!--\; \nu \; -->}}^{\mathrm{\lambda <!--\; \lambda \; -->}}(p)=g_{\mathrm{\nu <!--\; \nu \; -->},\mathrm{\mu <!--\; \mu \; -->}}^{\mathrm{\lambda <!--\; \lambda \; -->}}(p)$ю The proof of this that I'm looking at, following the more general theory in MacDonald's Symmetric Functions and Hall Polynomials, goes like this: You realize that we're looking at the category of finite-length modules over ${\mathbf{Z}}_p$, the $p$-adic integers. The Prüfer $p$-group $\mathbf{Z}(p^{\mathrm{\infty <!--\; \infty \; -->}})$ is the injective hull of $\mathit{k}=\mathbf{Z}/(p)$ in this category, and the functor $\mathrm{H}\mathrm{o}\mathrm{m}(-<!--\; -\; -->,\mathbf{Z}(p^{\mathrm{\infty <!--\; \infty \; -->}}))$, via Matlis duality, gives you a bijection of the short exact sequences in question, so $g_{\mathrm{\mu <!--\; \mu \; -->},\mathrm{\nu <!--\; \nu \; -->}}^{\mathrm{\lambda <!--\; \lambda \; -->}}(p)=g_{\mathrm{\nu <!--\; \nu \; -->},\mathrm{\mu <!--\; \mu \; -->}}^{\mathrm{\lambda <!--\; \lambda \; -->}}(p)$.
Proving this can also be approached by developing the theory of characters of finite abelian groups, section 3 in particular. But this is really the same approach in a different language: $\mathbf{Z}(p^{\mathrm{\infty <!--\; \infty \; -->}})$ plays the role of $S_1$ in this context. But in either approach, we're introducing some heavy stuff just to prove a fact about $p$-groups and partitions. Is there a elementary way to prove that $g_{\mathrm{\mu <!--\; \mu \; -->},\mathrm{\nu <!--\; \nu \; -->}}^{\mathrm{\lambda <!--\; \lambda \; -->}}(p)=g_{\mathrm{\nu <!--\; \nu \; -->},\mathrm{\mu <!--\; \mu \; -->}}^{\mathrm{\lambda <!--\; \lambda \; -->}}(p)$ in the case of finite abelian $p$-groups?

Let $(A,||\cdot <!--\; \cdot \; -->||)$ be a commutative Banach algebra over $\mathbb{C}$. Consider a formal power series $f(z):=\underset{n=0}{\sum <!--\; \sum \; -->}{a}_n{z}^n\in <!--\; \in \; -->A[[z]]$ and let
$r:=\frac{1}{\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim sup}||a_n|{|}^{1/n}}$
For $b\in <!--\; \in \; -->A$ denote by $\mathrm{\rho <!--\; \rho \; -->}(b)$ its spectral radius.
Is it generally true that $f(b)$ is divergent (in the $||\cdot <!--\; \cdot \; -->||$-topology) for any $b\in <!--\; \in \; -->A$ with $\mathrm{\rho <!--\; \rho \; -->}(b)>r$?

Determining a differential on an exterior algebra
Let $V$ be a graded vector space (over $\mathbb{N}$). Determine a derivation on the free, graded-commutative algebra $\mathrm{\Lambda <!--\; \Lambda \; -->}V$ it suffices to define it on $V$. I don't understand why this is so. Presumably it's referring to the fact that $\mathrm{\Lambda <!--\; \Lambda \; -->}$ is left adjoint to the forgetful functor from commutative graded algebras to graded vector spaces, that's ok, but the problem is that the differential on a graded vector space is not supposed to be a morphism of graded commutative algebras, but a derivation!
What's going on?

Let $X$ be a compact hausdorff topological space with more than one element.Then prove that the ring $C(X)$ of complex valued continuous functions on $X$ is not an integral domain. Thanks for any help. Actually this question arose when I was trying to prove that any commutative $C\ast <!--\; \ast \; -->$-algebra which is also an integral domain must be isomorphic to $C$ and I think this statement is correct. The problem I am having is with the case when $X$ is connected.

I.e. the bilinear map/product is not only associative, but also commutative. I am looking for examples of unital associative algebras, so they should be a vector space and a ring, not a vector space and a rng.One example, inspired by When is matrix multiplication commutative?, is the set of all diagonal matrices. Generalizing this, we could look at all $n\times <!--\; \times \; -->n$ matrices over $\mathbb{R}$ that share a common eigenbasis (in the case of all diagonal matrices, this eigenbasis forms $I_n$), though they need not have the same eigenvalues $\in <!--\; \in \; -->\mathbb{R}$. Are there other examples?

suppose $A$ is a $C^{\ast <!--\; \ast \; -->}$-algebra and $x$ is a normal element in $A$. $C^{\ast <!--\; \ast \; -->}$-algebra generated by $x$ denote by $A[x]$. then
1) $A[x]$ is commutative.
2) $A[x]$ is the clouser of polynomials of two variable $x$ and $x^{\ast <!--\; \ast \; -->}$.

Consider $M$, an $A$-module where $A$ is a ring. It defines ${\mathrm{\mu <!--\; \mu \; -->}}_f:M\to <!--\; \to \; -->M$ for the map $m\mapsto <!--\; \mapsto \; -->fm$, where $f\in <!--\; \in \; -->A$. Then the text claims that $f\mapsto <!--\; \mapsto \; -->{\mathrm{\mu <!--\; \mu \; -->}}_f$ is a ring homomorphism $A\to <!--\; \to \; -->\mathrm{End}<!--\; \; -->(M)$ from $A$ to the noncommutative ring of endomorphisms of $M$.
So, am I correct to think that in this case $\mathrm{End}<!--\; \; -->(M)$ is a noncommutative ring because $A$ is not commutative?
$\mathrm{End}<!--\; \; -->(M)$ seems to commutative if $A$ is commutative.

Let $A:=C^{(n)}([0,1])$ be the set consisting of the n-times continuously differentiable complex-valued functions. Consider $A$ with the norm
$\Vert <!--\; \Vert \; -->f\Vert <!--\; \Vert \; -->:=\underset{0\le <!--\; \le \; -->t\le <!--\; \le \; -->1}{max}\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}{\displaystyle \frac{|f^{(k)}(t)|}{k!}}.$
I want to show that $A$ is a commutative Banach algebra and find its maximal ideal space.
I know that the following holds:
- for all $f,g\in <!--\; \in \; -->A$: $(fg)(x)=f(x)g(x)=g(x)f(x)=(gf)(x)$;
- for all $f,g,h\in <!--\; \in \; -->A$:
$((fg)h)(x)=(fg)(x)h(x)=g(x)f(x)h(x)=f(x)(gh)(x)=(f(gh))(x)$
- for all $f,g,h\in <!--\; \in \; -->A$:
$(f(g+h))(x)=f(x)(g+h)(x)=f(x)(g(x)+h(x))=f(x)g(x)+f(x)h(x)=(fg)(x)+(fh)(x)=(fg+fh)(x)$
- hence $(g+h)f=gf+hf$;
- for all $f,g\in <!--\; \in \; -->A,\mathrm{\alpha <!--\; \alpha \; -->}\in <!--\; \in \; -->\mathbb{R}$:
$(\mathrm{\alpha <!--\; \alpha \; -->}(fg))(x)=\mathrm{\alpha <!--\; \alpha \; -->}(fg)(x)=\mathrm{\alpha <!--\; \alpha \; -->}f(x)g(x)=(\mathrm{\alpha <!--\; \alpha \; -->}f)(x)g(x)=((\mathrm{\alpha <!--\; \alpha \; -->}f)g)(x)=f(x)\mathrm{\alpha <!--\; \alpha \; -->}g(x)=(f(\mathrm{\alpha <!--\; \alpha \; -->}g))(x)$
Now, let $f,g\in <!--\; \in \; -->A$, then we see that
$\begin{array}{rl}\Vert <!--\; \Vert \; -->fg\Vert <!--\; \Vert \; -->& =\underset{0\le <!--\; \le \; -->t\le <!--\; \le \; -->1}{max}\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}{\displaystyle \frac{|(fg{)}^{(k)}(t)|}{k!}}\\ & =\underset{0\le <!--\; \le \; -->t\le <!--\; \le \; -->1}{max}\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}{\displaystyle \frac{|\underset{j=0}{\overset{k}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{k}{j})\cdot <!--\; \cdot \; -->f^{(k-<!--\; -\; -->j)}(t)\cdot <!--\; \cdot \; -->g^{(j)}(t)|}{k!}}\\ & =\underset{0\le <!--\; \le \; -->t\le <!--\; \le \; -->1}{max}\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}{\displaystyle \frac{|\underset{j=0}{\overset{k}{\sum <!--\; \sum \; -->}}{\displaystyle \frac{k!}{j!(k-<!--\; -\; -->j)!}}\cdot <!--\; \cdot \; -->f^{(k-<!--\; -\; -->j)}(t)\cdot <!--\; \cdot \; -->g^{(j)}(t)|}{k!}}\\ & \le <!--\; \le \; -->\underset{0\le <!--\; \le \; -->t\le <!--\; \le \; -->1}{max}\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}\underset{j=0}{\overset{k}{\sum <!--\; \sum \; -->}}{\displaystyle \frac{|f^{k-<!--\; -\; -->j}(t)g^k(t)|}{j!(k-<!--\; -\; -->j)!}}\\ & =\underset{0\le <!--\; \le \; -->t\le <!--\; \le \; -->1}{max}\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}\underset{j=0}{\overset{k}{\sum <!--\; \sum \; -->}}{\displaystyle \frac{|f^{k-<!--\; -\; -->j}(t)}{(k-<!--\; -\; -->j)!}}{\displaystyle \frac{g^k(t)|}{j!}}\\ & =?\end{array}$
Obviously, I want to show that $\Vert <!--\; \Vert \; -->fg\Vert <!--\; \Vert \; -->\le <!--\; \le \; -->\Vert <!--\; \Vert \; -->f\Vert <!--\; \Vert \; -->\Vert <!--\; \Vert \; -->g\Vert <!--\; \Vert \; -->$, but I dont know how to do this. Any hints?