I am at a very initial stage of commutative algebra.

The permutations and combinations of $abcd$ taken $3$ at a time are respectively.

Show that the image of ${\mathbb{P}}^n\times <!--\; \times \; -->{\mathbb{P}}^m$ under the Segre embedding $\mathrm{\psi <!--\; \psi \; -->}$ is actually irreducible.

Is function composition commutative?

If $B$ and $C$ are $A$-algebras with ring morphisms $f:A\to <!--\; \to \; -->B$ and $g:<!--\; :\; -->A\to <!--\; \to \; -->C$, and $D=B{\otimes <!--\; \otimes \; -->}_{A}C$ is an $A$-algebra with morphism $a\mapsto <!--\; \mapsto \; -->f(a)\otimes <!--\; \otimes \; -->g(a)$, then $uf=vg$, where $u:B\to <!--\; \to \; -->D$ is $u(b)=b\otimes <!--\; \otimes \; -->1$.The map $v:C\to <!--\; \to \; -->D$ is not defined in the text, but my guess is it's $v(c)=1\otimes <!--\; \otimes \; -->c$.I don't understand why the diagram is commutative though. That would imply $f(a)\otimes <!--\; \otimes \; -->1=1\otimes <!--\; \otimes \; -->g(a)$ for all $a\in <!--\; \in \; -->A$. Is that true, or is v something else?Added: On second thought, does this follow since $f(a)\otimes <!--\; \otimes \; -->1=a\cdot <!--\; \cdot \; -->(1\otimes <!--\; \otimes \; -->1)$ and $1\otimes <!--\; \otimes \; -->g(a)=a\cdot <!--\; \cdot \; -->(1\otimes <!--\; \otimes \; -->1)$ where ⋅ is the $A$-module structure on $D$?
$\begin{array}{ccc}A& \stackrel{f}{\to }& B\\ g\downarrow & \mathrm{\#<!--\; \#\; -->}& \downarrow u& \\ C& \underset{v}{\overset{}{\to }}& D\end{array}$

Let $A\subseteq <!--\; \subseteq \; -->B$ be rings, $B$ integral over $A$; let $\mathfrak{q},{\mathfrak{q}}^{\prime }$ be prime ideals of $B$ such that $\mathfrak{q}\subseteq <!--\; \subseteq \; -->{\mathfrak{q}}^{\prime }$ and ${\mathfrak{q}}^c={\mathfrak{q}}^{\prime c}=\mathfrak{p}$ say. Then $\mathfrak{q}={\mathfrak{q}}^{\prime }$.
Question 1. Why ${\mathfrak{n}}^c={\mathfrak{n}}^{\prime c}=\mathfrak{m}$?
My attempt: Since $\mathfrak{p}\subseteq <!--\; \subseteq \; -->\mathfrak{q}$, we have $\mathfrak{m}=S^{-<!--\; -\; -->1}\mathfrak{p}\subseteq <!--\; \subseteq \; -->S^{-<!--\; -\; -->1}\mathfrak{q}=\mathfrak{n}\subseteq <!--\; \subseteq \; -->B_{\mathfrak{p}}$. But m is maximal in $A_{\mathfrak{p}}$, which is not necessarily maximal in $B_{\mathfrak{p}}$. I can't get $\mathfrak{m}=\mathfrak{n}$ by this.
Question 2. When we use that notation $A_{\mathfrak{p}}$, which means the localization $S^{-<!--\; -\; -->1}A$ of $A$ at the prime ideal $\mathfrak{p}$ of $A$. But in this corollary, $\mathfrak{p}$ doesn't necessarily be a prime ideal of $B$. Why can he write $B_{\mathfrak{p}}$? Should we write $S^{-<!--\; -\; -->1}B$ rigorously?

Proof of commutative property in Boolean algebra
$a\vee <!--\; \vee \; -->b=b\vee <!--\; \vee \; -->a$
$a\wedge <!--\; \wedge \; -->b=b\wedge <!--\; \wedge \; -->a$

The theorem is stated in the context of commutative Banach (unitary) algebras, but the proof seems to show that it is valid for any commutative algebra defined as a linear space where a commutative, associative and distributive (with respect to the addition) multiplication is defined such that $\mathrm{\forall <!--\; \forall \; -->}\mathrm{\alpha <!--\; \alpha \; -->}\in <!--\; \in \; -->\mathbb{K}\mathrm{\alpha <!--\; \alpha \; -->}(xy)=(\mathrm{\alpha <!--\; \alpha \; -->}x)y=x(\mathrm{\alpha <!--\; \alpha \; -->}y)$.
In any case, whether it concerns only commutative Banach unitary algebras or commutative algebras as defined above, I think we must intend contained as properly contained. Am I right?

Let $H$ be a Hilbert space and $\mathcal{A}$ a commutative norm-closed unital $\ast <!--\; \ast \; -->$-subalgebra of $\mathcal{B}(H)$. Let $\mathcal{M}$ be the weak operator closure of $\mathcal{A}$.
Question: For given a projection $P\in <!--\; \in \; -->\mathcal{M}$, is the following true?
$P=inf\{A\in <!--\; \in \; -->\mathcal{A}:P\le <!--\; \le \; -->A\le <!--\; \le \; -->1\}$
It seems that the infimum must exist and is a projection, but I am not able to show that the resulting projection cannot be strictly bigger than $P$. Also, if the above is true, what happens if $\mathcal{A}$ is non-commutative?

Just a simple question. What does Eisenbud mean by $(x:y)$ where $x,y\in <!--\; \in \; -->R$ a ring. An example on this is in the section 17 discussing the homology of the koszul complex. I assume it's something along the lines of $\{r\mid <!--\; \mid \; -->ax=ry\}$ for some $a\in <!--\; \in \; -->R$.

Let $R$ be a commutative $k$-algebra, where $k$ is a field of characteristic zero.
Could one please give an example of such $R$ which is also:
(i) Not affine (= infinitely generated as a $k$-algebra).
and
(ii) Not an integral domain (= has zero divisors).My first thought was $k[x_1,x_2,\dots <!--\; \dots \; -->]$, the polynomial ring over k in infinitely many variables, but unfortunately, it satisfies condition (i) only. It is not difficult to see that it is an integral domain: If $fg=0$ for some $f,g\in <!--\; \in \; -->k[x_1,x_2,\dots <!--\; \dots \; -->]$, then there exists $M\in <!--\; \in \; -->\mathbb{N}$ such that $f,g\in <!--\; \in \; -->k[x_1,\dots <!--\; \dots \; -->,x_M]$, so if we think of $fg=0$ in $k[x_1,\dots <!--\; \dots \; -->,x_M]$, we get that $f=0$ or $g=0$, and we are done.

Let $A$ be a commutative Banach algebra. Let ${\mathrm{\chi <!--\; \chi \; -->}}_1$ and ${\mathrm{\chi <!--\; \chi \; -->}}_2$ be characters of $A$.
I am having some difficulty seeing why the following statement is true:
If $\ker <!--\; \; -->{\mathrm{\chi <!--\; \chi \; -->}}_1=\ker <!--\; \; -->{\mathrm{\chi <!--\; \chi \; -->}}_2$, then since ${\mathrm{\chi <!--\; \chi \; -->}}_1(\mathbf{\text{1}})={\mathrm{\chi <!--\; \chi \; -->}}_2(\mathbf{\text{1}})=\mathbf{\text{1}}$, we have that ${\mathrm{\chi <!--\; \chi \; -->}}_1={\mathrm{\chi <!--\; \chi \; -->}}_2$.

Need to find a functor $T:$ Set $\to <!--\; \to \; -->$ Set such that Alg(T) is concretely isomorphic to the category of commutative binary algebras.
The first idea is that the functor is likely to map object $X\in <!--\; \in \; -->Ob($ to the $X\times <!--\; \times \; -->X$ because then we have to get a binary algebra, i.e., the operation $X\times <!--\; \times \; -->X\to <!--\; \to \; -->X$, which have to be commutative.
So the question (if these thoughts are right) is: how to map $X$ to $X\times <!--\; \times \; -->X$ to get later a commutative binary algebra?

I would like to know, under what condition on the group $G$ (abelian, compact or localement compact ...), the algebra $L^1(G)$ is commutative?

Let $\mathcal{A}$ be a $C^{\ast <!--\; \ast \; -->}$-algebra.
(i) Let $\mathrm{\phi <!--\; \phi \; -->}$ be a state on a $u\in <!--\; \in \; -->\mathcal{A}$-algebra $\mathcal{A}$. Suppose that $|\mathrm{\phi <!--\; \phi \; -->}(u)|=1$ for all unitary elements u∈A. Show that φ is a pure state. [Hint: $\mathrm{span}<!--\; \; -->\mathcal{U}(\mathcal{A})=\mathcal{A}$ ]
(ii) Let $\mathrm{\phi <!--\; \phi \; -->}$ be a multiplicative functional on a $C^{\ast <!--\; \ast \; -->}$-algebra $\mathcal{A}$. Show that $\mathrm{\phi <!--\; \phi \; -->}$ is a pure state on $\mathcal{A}$.
(iii) Show that the pure and multiplicative states coincide for commutative $\mathcal{A}$.
I managed to work out the first two problems but I have no idea about the last one. How to see from being an extreme element in the state space of a commutative $\mathcal{A}$ that the extreme element is multiplicative?

$A\subset <!--\; \subset \; -->B$ is a ring extension. Let $y,z\in <!--\; \in \; -->B$ elements which satisfy quadratic integral dependance $y^2+ay+b=0$ and $z^2+cz+d=0$ over $A$. Find explicit integral dependance relations for $y+z$ and $yz$.