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Fernando and Pierre-Yves in the comments are right; $R$ has this property (the version where the canonical map is an isomorphism, as YCor says in the comments) iff it is a solid ring, meaning the mult …

No.

Some commentary on David's very nice answer that might provide some useful context. If $D$ is a subring of $\mathbb{R}$ then it must be an integral domain and its field of fractions $\operatorname{Fra …

More or less by definition this is the case precisely when the coefficients of $f$ satisfies a homogeneous linear recurrence with coefficients in $k$. Equivalently, the coefficients of $f$ are a line …

Here is a less algebraic and more topological answer: it's known that any compact (Hausdorff) topological ring must be totally disconnected. In particular, there's no hope for either $\mathbb{RP}^1$ o …

Let me propose a more natural version of your question that can be asked about more algebras: What can we say about an element $r \in R$ of a commutative algebra $R$ over a field $k$ given that $ …

An algebra is Morita equivalent to a commutative algebra iff it's Morita equivalent to its center, since the center is Morita invariant. So any representative of a nontrivial class in the Brauer group …

No, e.g. $A$ could be $\mathbb{C}^{\mathbb{N}}$ where $t$ is the sequence $(1, 2, 3, ...)$. This is not even Noetherian or countable-dimensional over $\mathbb{C}$. A slight modification of this constr …

The closest analogue is to consider polynomials over finite fields $\mathbb{F}_q[x]$. Here the counting is straightforward: there are $q^n$ monic polynomials of degree $n$ over $\mathbb{F}_q$. If $p$ …

Small comment: certainly there's a natural map $\text{Hom}_R(X, R) \otimes_R S \to \text{Hom}_S(X, S)$ given by composing with $\varphi$ and extending by linearity. We can't hope for this map to be an …

Not an answer, but: It shouldn't be surprising that this product isn't commutative: your definition has no obvious symmetry between $\xi$ and $\eta$, and so to me it's quite unclear that it's "the co …

If $f(x, y)$ is a formal group law then so is $g(f(g^{-1}(x), g^{-1}(y))$ where $g$ is an invertible (under composition) formal power series. This suggests a strategy for writing down $n$-buds, namely …

I believe so. Dirichlet series are a special case of a construction called the reduced incidence algebra of a poset, and the poset of integers under division is precisely a product of chains, one for …

I'll echo Ben Webster's comment from the other thread, but in the other direction: find some natural commuting operators whose simultaneous eigenspaces are all one-dimensional. This is how, for examp …

No, in the sense that this statement is false in a ring without unique factorization. For example, the element $2 + \sqrt{-5}$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$, and $9 \in (2 + \sqrt{-5})$ bu …