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While it's certainly true (per Fernando's comment) that this is a special case of the Quillen-Suslin theorem, it was certainly known long before Quillen and Suslin came along. There's a paper of Mu …

Better yet, you can replace $f(x,y)$ with $f(x)$. See the answer to this question. Edited to add: At Martin Brandenburg's request, I'm expanding this to add the details I thought were too obvious t …

Depending on what you consider simple, let $k$ be the complex numbers, or the integers, or the field with two elements (or any other commutative ring you're fond of). Let $R=k[a,b,c,x,y,z]/(ax+by+cz- …

Let $M$ be a finitely generated module over the commutative noetherian ring $R$. Let ${\cal C}(M)$ be the set of all primes $P$ in $R$ such that $R/P$ appears as a quotient in every composition serie …

Others have given examples. It might be worth noting that (in the category of finitely generated abelian groups, or more generally in the category of finitely generated modules over a noetherian ring …

Call your displayed matrix $H_{M,N}$. Theorem The following are equivalent: $u$ is the Taylor series of a rational function. There is a finite sequence $q_0,\ldots, q_N$, not all zero, such that for …

The vanishing of the cohomology group $H^1(Spec(R),GL_n)$ doesn't actually say that all projectives of rank $n$ are free; it says only that all projectives of rank $n$ are isomorphic. Combining this …

For your final question, the answer is that all vector bundles over $U$ are trivial. Sketch of proof: Let $R=k[x_1,\ldots,x_n]$. Let $L_1, L_2,\ldots, L_m$ be the equations of hyperplanes in $R$. Wi …

I don't know what Serre (or Auslander and Buchsbaum?) was thinking, but it would have been natural to observe that $R$ is regular iff its maximal ideal is generated by a regular sequence, which (by wr …

For $\alpha\neq n$, the symmetric polynomial is of the form $f\cdot x_n + g$ where $f,g$ are non-zero elements of $A={\mathbb C}[x_1,...,x_{n-1}]$ with no common factor. Thus $${\mathbb C}[x_1,...,x_ …

The Jacobian conjecture is trivial in this case. Write $M$ for $dF$ and $N$ for its inverse, which is also triangular with 1's on the diagonal. (I assume upper triangular.) Now just look at the con …

1) In the positive direction: If any entry in $e$ is an $(n-1)!$ power, then $e$ is an element of a basis. (More generally, it's enough to have $e=(z_1^{m_1},\ldots z_n^{m_n})$ with $(n-1)!$ dividin …

If you think of the projective line as two affine lines --- namely $Spec(R[t])$ and $Spec(R[t^{-1}])$ --- patched together over $Spec(R[t,t^{-1}])$, then Horrocks's Theorem tells you that any vector …

Pick your favorite bijection $\phi$ from the natural numbers to the rational numbers. For each real number $\alpha$, let $I_\alpha$ be the ideal generated by all $x_n$ with $\phi(n)<\alpha$. Then $I …

Presumably you are looking at the conductor square on the left below, where $\epsilon^2=0$: $$\matrix{k[t^2,t^3]&\rightarrow& k[t]\cr \downarrow&&\downarrow\cr k&\rightarrow &k[\epsilon]\cr}\qquad \ma …