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A commutative ring $A$ has the property that every non-unit is a zero divisor if and only if the canonical map $A \to T(A)$ is an isomorphism, where $T(A)$ denotes the total ring of fractions of $A$. …

Perhaps the "Rabinowitz trick" is more clear if one writes down the proof backwards in the following way: Let $I \subseteq k[x_1,\dotsc,x_n]$ be an ideal and $f \in I(V(I))$, we want to prove $f \in …

Theorem 1: Let $R$ be an integral domain with field of fractions $K$, and $R \to A$ a homomorphism. Then $Spec(A) \to Spec(R)$ is an open immersion if and only if $A=0$ or $R \to K$ factors through $R …

Let us call a Noetherian commutative ring $R$ an $\mathcal{N}$-position if the next player has a winning strategy; otherwise the previous player has a winning strategy and we will call it a $\mathcal{ …

There are various forms of the Nakayama lemma. Here is a rather general one; note that it does not involve maximal ideals and is a constructive theorem (Atiyah-MacDonald, Commutative Algebra, Prop. 2. …

Sometimes these algebras are fiber products along surjective homomorphisms, and Karl Schwede's paper on Gluing Schemes tells us how the spectrum looks like. For example: 1) The spectrum of $k[x,xy,xy …

Take any compact totally disconnected Hausdorff space $X$ (for example the Cantor set, or the one-point compactification of $\mathbb{N}$). Then $\mathcal{C}(X,\mathbb{F}_2)$ is a ring whose spectrum i …

Every commutative ring is the directed colimit of its subrings that are finitely generated as $\mathbb{Z}$-algebras. The Hilbert Basis Theorem implies that these subrings are noetherian. Actually this …

Bjorn's answer shows that the product of two ideals cannot be defined by means of the partially ordered set of ideals. However, there is a category-theoretic definition of the product of two ideals if …

The category $\mbox{CRing}$ of commutative rings is rigid, i.e. every equivalence $\mbox{CRing} \to \mbox{CRing}$ is isomorphic to the identity, Proof: Let $F : \mbox{CRing} \to \mbox{CRing}$ be an …

Some examples: A. A noetherian commutative ring has only finitely many minimal prime ideals. This is just a corollary of the easy observation that a noetherian space has only finitely many irreducibl …

The free magma on $X$ consists of finite sequences of length $1$ or $2$, which consist of finite sequences of length $1$ or $2$, etc., of elements of $X$; the magma operation is just concatenation, i. …

More generally, the Theorem of Eilenberg-Watts says the following: The category of cocontinuous functors $\mathrm{Mod}(R) \to \mathrm{Mod}(S)$ is equivalent to the category of $(R,S)$-bimodules. A bim …

Yes, it is possible to write down an equational proof for every $n$. This is covered in the preprint Equational proofs of Jacobson's Theorem, arXiv:2310.05301 [math.RA] The rough idea is to prove th …

When $R$ is noetherian, yes: By Baer's criterion it suffices to prove that the map $\hom_{S^{-1} R}(S^{-1} R,S^{-1} M) \to \hom_{S^{-1} R}(J,S^{-1} M)$ is surjective for every ideal $J \subseteq S^{ …