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To add to the answer above, I'd like to advertise the so-called sumless Sweedler notation here. This notation works as follows: let $C$ be a coalgebra; then if $c \in C$ then we write $\Delta(c) = c …

Instead of updating my previous answer, I've decided to add a new answer in order to keep it short(ish). In the comments following his original question, TonyS added the extra assumption that $R$ is …

Since $A$ is regular local commutative ring, it's an integral domain. Let $S = A - 0$ so that the localisation $A_S$ is the field of fractions $F$ of $A$. Assuming the algebra $A$ is central in $R$, …

No. The problem is that $\Lambda$ might have too few units. Here is an example that illustrates this point. Let $\Lambda$ be the ring of Hurwitz quaternions. This is the subring of the usual quaterni …

This is an algebraic elaboration on Emil's answer. Let $A = K \langle x_1,\ldots,x_n \rangle$ and let $\hat{A} = K \langle\langle x_1,\ldots, x_n \rangle \rangle$. Since $A$ is a subring of $\hat{A}$ …

The natural map $\mathbb{Z} \to \mathbb{Z}G$ has central image and therefore induces a map between prime spectra $Spec(\mathbb{Z}G) \to Spec(\mathbb{Z})$. The preimage of the ideal generated by $(p)$ …

Just for the record, here is an example of a (necessarily infinite dimensional) $k$-algebra $A$ where the Jacobson radical is not equal to the intersection of all maximal two-sided ideals. Let $k$ be …

Yes. Let $J(\mathfrak{g}) := (Z(\mathfrak{g}) \cap \mathfrak{g} U( \mathfrak{g} ) ) \cdot U(\mathfrak{g})$ denote the ideal of $U(\mathfrak{g})$ you're interested in. First note that if $\mathfrak{g} …

Suppose that $R$ satisfies the following properties: $R$ is a simple ring : it has no non-trivial two-sided ideals, $R$ has left Krull dimension equal to $1$. This means that $R$ has at least one in …