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The answer is yes. (In what follows, I write $f(x^2,y)$ for your $f(x,y)$.) 1) By symmetry, if $h(x,y)$ is an irreducible factor of $g(x^2,y)$ then so is $h(-x,y)$. 2) In particular, if $h(x,y) …

Start with the integers. Adjoin variables $X_{in}$ and $Y_{in}$ for all $1\le i \le n$. Mod out by all relations of the form $$\sum_{i=1}^nX_{in}Y_{in}=1$$ Call the resulting ring $R$. Then, by co …

If $M=R/(p^n)$ with $p$ prime, the result is clear. Since an arbitrary $M$ is a direct sum of such modules, the result is still clear.

For what it's worth, it suffices for $(r)$ to be an interesection of minimal primes; in fact, more generally if $J$ is any intersection of minimal primes then $J=Annih(Annih(J))$. (This requires only …

Hints: 1) First prove that $I\otimes(R/J)=I/IJ$ . 2) If $I+J=R$, write $1=i+j$ and use the fact that $x=1x$.

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 …

The implication does not hold. Let $K$ be the real numbers, $F$ the complex numbers, $A=K[X,Y,Z]/(X^2+Y^2+Z^2-1)$, and $M$ the kernel of the map $A^3\rightarrow A$ given by the unimodular row $(X,Y,Z …

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 …

I won't be surprised if you already know this, but here is a proof for $A$ noetherian and/or $B$ finite over $A$: 0) We can replace $A$ with its image in $B$ and assume $A\rightarrow B$ is injective …

The two sides of your "that is" are different questions: 1) Does $C$ necessarily come from $D(A_f)$ ? 2) Is there some complex $C'$ of $A_f$ modules and a quasi-isomorphism $C'\rightarrow C$ of co …

Let $v$ be any map at all from $R$ to $S\approx R^{n-1}$. Then $v=qu$ where $u:R\rightarrow R^n$ is given by $r\mapsto (vr,r)$ and $q:R^n\rightarrow S$ is the obvious projection. Therefore your ma …

I did this by hand and got $$ coker(f)=(R/J_{xv}\oplus R/J_{yv})/(xv,yv)$$, where $J_t$ is the ideal generated by all quadratic monomials except for $t$. (Sorry for the garbled version of this I br …

Put $B=k[x,y]/(xy,y^2)$ and $A=B[z]/(yz-x)$. Then $(y)$ is an associated (in fact minimal!) prime of $B$. But any prime of $A$ that contains $(y)$ must also contain $x$, and so cannot contract to ju …

It is a theorem of Pierre Samuel that if $R$ is normal and of finite type over a field $k$, then the following are equivalent: 1) The map taking $x$ to ($1$ plus the $k$-dimension of $R/(x)$) is a d …

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 …