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First let's note that by the paper, $Q(R)$ is the total ring of fractions of $R$, i.e. it is the localization of $R$ by the subset of non-zero-divisors. The natural map $R \to Q(R),\,r \mapsto \frac{r …

[Sorry for answering my own question] By choosing integers $a_i,b_i$ with $a_ip_i+b_iq_i = 1$ one can define an automorphism on the polynomial ring by mapping $x_i \mapsto x'_i = p_ix_i-q_iy_i,\, y_ …

Is there a nice way to show that $$(p_1x_1-q_1y_1,\ldots ,p_nx_n-q_ny_n) \subseteq \mathbb{Z}[x_1,...,x_n,y_1,...,y_n]$$ is a prime ideal for coprime non-zero integers $p_i,q_i\,(i=1,...,n)$ ? I h …

Picking up a simplified version of Jason Starr's idea: $R = \mathbb{Z}_p \times \mathbb{Z}$ has dimension $1\,\,( \mathbb{Z}_p$ denotes the p-adic integes) and Jacobson radical $(p) \times 0 \not= 0 …

Given free modules $N \le M$ of finite rank over a PID $R$, it's well-known that there is a basis $\{x_1,\ldots,x_n\}$ of $M$ and there are $e_1,\ldots,e_n \in R$ such that $\{e_ix_i\mid e_i \neq 0\} …

These rings are called right uniform rings. Generally, a right ideal $I$ is called (right) uniform if all nonzero right subideals $J,K\subseteq I$ have nonzero intersection: $J\cap K \neq 0$. Note: …

Let $R$ be an integral domain and for $a \in R$ denote by $\text{eval}_a: R[X] \to R$ evaluation at $a$. It's well-known (and easy to see) that $$\ker(\text{eval}_a)=(X-a).$$ The next more complicate …

Two classes of torsion-free abelian groups having the desired property are free abelian groups torsion-free divisible groups (here I use the axiom of choice) By noting that a torsion-free divis …

No. Counterexample: Let $\deg x = \deg y = 1$ and take $$A=\mathbb{F}_2[x,y],\,\,\,B=\mathbb{F}_2[x,y^2],\,\,\,\,\,g(x)=x,\,\,g(y)=x+y$$ As already observed by the OP, since the generators of $B$ hav …

I don't think that analytic power series can be controlled by algebraic power series in an algebraic way, because they are often transcendental over the former. Suppose an automorphism $\phi$ of …