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What is the fraction field $K$ of the domain $\mathbb Z[[X]]$? It is strictly smaller than the field of Laurent series $L=\operatorname {Frac}\mathbb Q[[X]]$, since $\sum_{i\geq 0}\frac {X^i}{i!}\in …

Since this subject is full of misunderstandings (see here, here, here, and here) let us fix a precise terminology. Let $A$ be a commutative ring and $P$ an $A$-module. I) We'll say that $P$ is a lo …

1) About the second definition: $\alpha$) It is not true that for an arbitrary ring a) is equivalent to c): Indeed Bourbaki in Algèbre commutative, Chapitre II, Exercices §5, 12) c) exhibits a ring …

Let $R$ be a domain and $\tilde R$ its integral closure in its fraction field: $R\subset \tilde R\subset Frac(R)$. Is it true that a prime ideal $ \tilde {\mathfrak p} \subset \tilde R$ and its trac …

1) It is sometimes stated that a finitely generated module $M$ over a Noetherian commutative ring $R$ is projective if for all maximal ideals $\mathfrak m\subset R$ the localized module $M_\mathfrak m …

Consider a connected holomorphic manifold $X$ and its ring of holomorphic functions $\mathcal O(X).$ My general question is simply: in which cases is the Krull dimension $\dim \mathcal O(X)$ known? …

The answer is no: for example, $k[[x]]\otimes_k k((x))$ is not noetherian. Indeed, if it were, so would be $ k((x))\otimes_k k((x))$. But this would contradict the following interesting general theore …

In an anwswer to a question on our sister site here I mentioned that a reduced commutative ring $R$ has zero Krull dimension if and only if it is von Neumann regular i.e. if and only if for any $r …

Consider two extension fields $K/k, L/k$ of a field $k$. A frequent question is whether the tensor product ring $K\otimes_k L$ is a field. The answer is "no" and this answer is often justified by so …

Let me give a self-contained proof of the existence of prime ideals of the type you require, for $\mathbb R$ and many other spaces . Given a topological space $X$, let ${\mathfrak m }_x\subset C(X)$ d …

Recall that in a commutative ring $A$ an ordered pair of elements (a,b) is said to form a regular sequence if the ideal $\langle a,b\rangle $ is strictly included in $A$ ,if $a$ is not a zero-divisor …

A simple example is obtained by taking $P$ to mean "has positive dimension". Every local domain of positive dimension $(A,\mathfrak m)$ has $P$ at all maximal ideals (i.e. just at $ \mathfrak m$ !) …

You should look at Atiyah's aricle "On the Krull-Schmidt theorem with applications to sheaves", freely available here. He gives fairly general conditions on an exact category for the Krull-Remak-Sc …

Consider the inclusion $k\subset A$ of the field $k$ in the domain $A$ and the fraction field $K=Frac(A)$ of $A$. Obviously if a family $(a_i)_{i\in I}$ of elements $a_i \in A$ is algebraically indep …

No, here is an example of a morphism $f:X\to Y$ which is not affine although $X$ is affine. Take $X=\mathbb A^2_k$, the affine plane over the field $k$ and for $Y$ the notorious plane with origin do …