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Since the flatness is a local property one can assume that $A$ is local, so $A$ is a DVR. Let $\mathfrak m$ be the maximal ideal of $A$, and $a\in A$ such that $\mathfrak m=aA$. For DVRs flatness is equivalent with torsion free. Then it's enough to prove that $B/fB$ is a torsion free $A$-module. (Note that this is equivalent to $a$-torsion free.) Let $b\in ... 2 In fact, you are looking for a minimal primary decomposition of$I=(X^2Z,YZ,Z-XY)$in$k[X,Y,Z]$. As it is noticed in the comments,$k[X,Y,Z]/(Z-XY)\simeq k[X,Y]$. The image of$I/(Z-XY)$under this isomorphism is$(X^3Y,XY^2)$and a primary decomposition of this ideal is given by$(X)\cap(Y)\cap (X^3,Y^2)$. The inverse image of this decomposition is ... 2 For a finitely generated module over a local ring being flat is the same as being free. Can the fraction field of a domain have two elements linearly independent over the domain? Try this for$\mathbf{Z}$, for example. 2$S^{-1}R$is noetherian, reduced ($N(S^{-1}R)=0$), and every element of$S^{-1}R$is a zerodivisor or invertible. Now your question follows from the following result: Let$A$be a reduced noetherian ring with the property that every element of$A$is a zerodivisor or invertible. Then$A$is artinian. Let$\{\mathfrak p_1,\dots,\mathfrak p_n\}$be the ... 2 Actually, you can view the code, http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.7/share/doc/Macaulay2/Macaulay2Doc/html/_code.html, and I believe that it used the variable of ambient ring as the maximal ideal. My experience is that sometimes its output is a negative number which is not possible. But most of the time, instead computing the dimension of ... 1 No. The ideal generated by 3,5 and 6 is$\mathbb{Z}$. Thus$(3,5,6) \cdot M =M$for every$\mathbb{Z}$-module$M$. 1 Although this is the usual property that defines the faithfully flatness, (one of the) Liu's definition(s) which is helpful here is the following: Let$M$be a flat$A$-module. Then$M$is faithfully flat iff for any$A$-module$N$such that$M\otimes_AN=0$we have$N=0$. Recall that$N'\stackrel{u}\to N\stackrel{v}\to N''$is exact iff ... 1 First of all, the correspondence between$K$-algebra homomorphisms and morphisms of affine algebraic varieties is contravariant, so$f$being surjective does not imply that$f'$is surjective - it does imply that$f'$is a monomorphism. The monomorphisms in the category of affine varieties are the injective morphisms. I mention this explicitly because the ... 1 Let$M$be a maximal ideal of$A[X,Y]/(Y^2-aX)$. Then$M=\mathfrak m/(Y^2-aX)$, where$\mathfrak m$is a maximal ideal in$A[X,Y]$containing$Y^2-aX$. We have $$\left(\dfrac{A[X,Y]}{(Y^2-aX)}\right)_M=\dfrac{A[X,Y]_{\mathfrak m}}{(Y^2-aX)}.$$ (I) If$a=0$, then$Y\in \mathfrak m$, so$Y^2\in \mathfrak m^2$, hence$A[X,Y]_{\mathfrak m}/(Y^2)$isn't regular. ... 1 I don't get Liu's hint, but the following it's useful and not hard to prove: Let$M$be a flat$A$-module. Then if tensor a short exact sequence of$A$-modules$0\to M_1\to M_2\to M\to 0$by any$A$-module it remains exact. Now consider$0\to J\to B\to B/J\to 0$a short exact sequence of$A$-modules. By tensoring this with$A/I$we get$0\to ...

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I don't think the answer is true in general. Let $(A,m)$ be a Noetherian local domain which is not Cohen-Macaulay. Let $R$ be the localization of $A[x]/(x^2)$ at $(m, x)$. Then $(x)$ is the unique minimal prime with $\dim R = \dim R/(x)$, but $R/(x) \cong A$ is not Cohen-Macaulay. Let $I$ be an $R$-ideal such that $\dim R/I = \dim R$. Write $p = (x)$. Then ...

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Let $R = \{f(1,1) = f(-1,-1)\} \subset k[x,y]$. So $\mathrm{Spec} R$ looks like a surface passing back through itself at an isolated point. The singular point is not Cohen-Macaulay since it looks like two planes meeting at a point (analytically). Now let $S = R \oplus R$, so $\mathrm{Spec} S$ is the disjoint union of two copies of of this surface. There are ...

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Here is another approach; it is more involved than user26857's, but may also help make things clearer (or maybe not; I've left several details to check): The ring $R$ is reduced (i.e. has trivial nil-radical) and is Noetherian, so has finitely many minimal prime ideals, say $\mathfrak p_1, \dots, \mathfrak p_n$. Combining these facts, we find that ...

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Unless I'm missing something here, this is not too complicated. I'll leave it as a nice exercise in using definitions that in a Noetherian domain, every nonzero prime ideal contains an irreducible element. So there exists $q\in Q$ irreducible. Then $q\in P$, so $p\mid q$; since $q$ is irreducible and $p$ is not a unit, $q$ is an associate of $p$, so ...

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for your first Question; the book Steps in Commutative Algebra (bySharp) Theorem 15.13: Let $R$ be a commutative Noetherian ring and let $P \in Spec(R)$; suppose that $htP = n$. Then there exists an ideal $I$ of $R$ which can be generated by $n$ elements, has $ht\ I = n$, and is such that $I \subseteq P$ note that $ht\ I=ht\ P$, so $P$ is minimal prime ...

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First of all you can suppose that $A$ and $B$ are local ring by with $A$ neotherian, by localization, as $A$ is a Dedekind domain. Now, See Bourbaki, Algèbre Commutative, paragraph 5, number 2, theorem 1, and the equivalence of the first and third assertions of the theorem : More details here. But as you will see, whatever you do/write, you will write ...

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You can go to various website, such as US News & World Reports here, that drills down rankings by department, but I never saw any ranking that drills down to one particular subject in algebra. Hope this helps.

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