True or False: For n xx n matrices A and B, define A ox B = AB − BA. The operator ox is not associative or commutative.

In general, we have functors ${\mathcal{S}\mathcal{C}\mathcal{R}}_{R/}\stackrel{\varphi }{\to }{\mathcal{D}\mathcal{G}\mathcal{A}}_R\stackrel{\psi }{\to }{\mathcal{E}\mathcal{I}}_{R/}$. If $R$ is a $\mathbf{Q}$-algebra, then $\psi$ is an equivalence of $\mathrm{\infty }$-categories, $\varphi$ is fully faithful, and the essential image of $\varphi$ consists of the connective objects of ${\mathcal{D}\mathcal{G}\mathcal{A}}_R\simeq {\mathcal{E}\mathcal{I}}_{R/}$ (that is, those algebras $A$ having ${\pi }_{i}A=0$ for $i<0$).
What is the explicit functor $\varphi :{\mathcal{S}\mathcal{C}\mathcal{R}}_{R/}\to {\mathcal{D}\mathcal{G}\mathcal{A}}_R$? I suppose that the natural thing would be to take a simplicial $R$-algebra $A$ and assign it to
$\begin{array}{r}\varphi (A)=\underset{i=0}{\overset{\mathrm{\infty }}{⨁}}{\pi }_{i}A,\end{array}$
and take a map $f:A\to B$ and assign it to
$\begin{array}{r}\varphi (f)=\underset{i=0}{\overset{\mathrm{\infty }}{⨁}}(f_i:{\pi }_{i}A\to {\pi }_{i}B),\end{array}$
but as far as I could find this isn't stated explicitly in DAG. Is this the case, and if so, do you have a source or proof? And how does one show that $⨁{\pi }_{i}A$ is a differential graded algebra?

Show that a commutative ring with the cancellation property (under multiplication) has no zero-divisors.

Let $\times$ be a binary operation on set of rational number ${\displaystyle \mathbb{Q}}$ defined as follows: $a\cdot b=a+b+2ab$, where ${\displaystyle a,b\in \mathbb{Q}}$
a) Prove that $\times$ is commutative, associate algebraic operation on ${\displaystyle \mathbb{Q}}$

Determining Lie algebras from commutative diagrams of exact sequences.
Suppose that we have the following commutative diagram of graded Lie algebras.
$\begin{array}{}0& ⟶& C_n& ⟶& A_{n+1}& ⟶& A_n& ⟶& 0\\ & & \downarrow & & \downarrow & & \downarrow \\ 0& ⟶& D_n& \stackrel{}{⟶}& B_{n+1}& \stackrel{}{⟶}& B_n& ⟶& 0& \end{array}$
for all $n\in {\mathbb{Z}}^{+}$, where the both rows are split exact sequences and every vertical map is onto.
My question is, suppose that we know $A_n$, $C_n$, $D_n$ for all $n\in {\mathbb{Z}}^{+}$ and $B_1$, do we have enough information to uniquely determine up to isomorphism every $B_n$ (likely by induction)?

Let x not be quasinilpotent, so $li{m}_{n\to \mathrm{\infty }}||x^n|{|}^{1/n}=\lambda \ne 0$. Let $\pi (x)=\overline{x}$, where $\pi (x):A\to A/rad(A)$ is the canonical quotient map. Suppose that $||{\overline{x}}^n|{|}^{1/n}=0$. Then it's spectrum $\sigma (\overline{x})=0$, so $\overline{x}$ is not invertible in $A/rad(A)$, and thus generates a proper ideal in $A/rad(A)$. So then ${\pi }^{-1}(\overline{x})$ generates a proper ideal in $A$ containing $rad(A)$.
From here, if $rad(A)$ is maximal I think I'd have a contradiction, but I don't know if that's true. If not, does anyone have another strategy I could try?

Say a function is commutative if it remains unchanged under any permutation of its arguments. E.g. ${\displaystyle f(0,1)=f(1,0)}$. (Alternatively we could describe these as functions over multi-sets, or say that they are reflective about any hyperplane ${\displaystyle x_i=x_j}$). Some examples are sum, product and average.
1) Is there a name for these functions? Google searches for commutative and reflective functions dont

Let $A$ be a commutative Banach algebra such that set of all characters is infinite. I want to prove that there exists an element in $A$ such that its spectrum is infinite.
Why is there an element with infinite spectrum in a commutative Banach algebra with infinitely many characters?