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Another example: let $R=k[x,y]$ where $k$ is a field. Then $R$ is a free module over itself, and the ideal $I$ of $R$ generated by $x$ and $y$ is not only not free over $R$, it is not even flat over $ …

If your local rings are Noetherian it's obvious. The Koszul complex consists of finitely generated free modules and the homology modules are subquotients of it so also finitely generated. For non-Noe …

There is a notion of a universal derivation for an algebra. I'll assume everything is commutative for simiplcitity. If $A$ is a $k$-algebra ($k$ a commutative ring) then there is an $A$-module $\Omega …

This reduces to the question: is there an $R$-module injection from $R^{n+1}$ to $R^n$. This is a matrix question: is there a nonzero nullvector for an $n$-by-$n+1$ matrix $M$. Clearly $M$ has a null …

When $R$ is commutative, $K_0(R)$ is a commutative ring with multiplicative unit the class $[R]$ of $R$. The only ring structure with additive group $\mathbb{Z}$ is the familiar one, so every element …

For (b) recall that a prime ideal is the kernel of a map from $G$ into a field. Such a field must be an extension field of $H$ and the image of $G$ is generated by $H$ and the $p^{r_i}$-th roots of ce …

Rather obviously yes. Let $A$ be the algebra over the field $K$ generated by elements $a_1,a_2,\ldots,$ with $a_i$ in dimension $i$ and with $a_ia_j=0$ for all $i$ and $j$. This is an incredibly unin …

A proof I like is that the group of points on the curve is the classgroup of the ring $R=k[x,\sqrt{x^3+Ax+B}]$ where $k$ is the field you're working over. Set $y=\sqrt{x^3+Ax+B}\in R$ and let $K$ deno …

The key word in this context is functor. The point is that homology, homotopy etc. are functors. For example consider homology $H_n$. This is a functor from the category of topological spaces to the c …

The stronger version of the Nullstellensatz states that a maximal ideal $I$ of $R=k[x_1,\ldots,x_n]$ is the kernel of a $k$-homomorphism from $R$ to $L$ where $L/k$ is a finite extension. Let $a_1,\ld …

Wikipedia calls a module over a commutative Noetherian ring with only one associated prime a coprimary module. I don't recall hearing this terminology elsewhere, but it is certainly common to call a s …

How about the algebraic closure of the $p$-adics $\mathbb{Q}_p^{\mathrm{alg}}$. This is not complete under the $p$-adic metric, but it is isomorphic as a field to the complex numbers $\mathbb{C}$ whic …

If every matrix has a Smith normal form, then every finitely generated $R$-submodule $M$ of $R^n$ satisfies $R^n/M$ is a finite direct sum of modules isomorphic to $R/aR$. If $R$ is Noetherian this im …

Well, $\mathbb{Z}[t,t^{-1}]$ is the localization of the polynomial ring $R=\mathbb{Z}[t]$ with respect to the multiplicative set $S$ consisting of the powers of $t$. The ideals of $S^{-1}R=\mathbb{Z}[ …