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5

$(0:x)$ is the set of elements $r$ of $R$, such that $rx=0$; called annihilator of $x$. Note that In the complex $K(x): 0\to R\xrightarrow{x} R$, the homomorphism $\xrightarrow{x}$ is multiplication by $x$. So $ker (\xrightarrow{x}) = \{r\in R| rx=0\}=(0:x)$ (and this is the homology). The complex is not an exact sequence unless the homology is ...

3

I think your assumptions are wrong (not that it is important for the issues). Arguably, one of the first books on commutative algebra was written by Zariski and Samuel with the explicit intention of codifying the algebra necessary for their work in algebraic geometry. It still is one of the deepest books in the field, though not easy to read. For example, it ...

3

$I_a = (x,y) \cdot (z,x+a)$ and $(x,y) + (z,x+a) =(x,y,z,x+a) = (1)$, since it contains $a \in k \setminus \{0\}$. By the Chinese remainder theorem $$k[x,y,z]/I_a \cong k[x,y,z]/(x,y) \times k[x,y,z]/(z,x+a) \cong k[z] \times k[y]$$ is the product of two domains, hence reduced.

2

Part 2 the answer is no, and it can be seen in the simplest example where $k = \mathbb{C}$ and $K = \overline{\mathbb{C}(t)}$, and $n=1$. Let $(a,b) = (t,\sqrt{t}) \in K$. Then $F(X,Y)= X - Y^2$ vanishes at $(a,b)$ but if we specialize $t$ to any complex number $a'$ then $F(a',b) = a'-t \ne 0$. Edit: Part 1 the answer is also no, and can again be seen in ...

2

First, the assumption $(B/I)_t = (B/J)_t$ has to be made precise: I guess it's meant to say that the kernels of the projections $B_t \to (B/I)_t$ and $B_t \to (B/J)_t$ are equal, i.e. $IB_t = JB_t$. Now, suppose $i\in I$. Considering it as an element of $IB_t=JB_t$, there exist $j\in J$ and $k> 0$ such that $i = j/t^k$, i.e. $t^{k+l} i = t^l j$ in $B$ ...

2

Let $p \in \mathbb{Z}$ be a prime integer and consider $\mathbb{Z}_p$, the ring of $p$-adic integers. Then $R$ has prime ideals $(0)$ and $(p)$, and hence is a discrete valuation ring. It is easy to check that the localization of $\mathbb{Z}_p$ at each prime ideal is a regular local ring, verifying that $\mathbb{Z}_p$ is a regular commutative ring. Since ...

1

Your question is equivalent to the following statement: Suppose deg $f$ $\geq$ deg $g$, $f$, $g$ are homogeneous then $f|_{x_i=1}=g|_{x_i=1}$ if and only if $f=x_i^kg$ for some $k\in \mathbb{N}$ Since we can divide $f$ and $g$ by $x_i$ if necessary, we can assume both $f$ and $g$ are not multiples of $x_i$ Then the statement can be proved by noticing ...

1

A module in which every submodule is an essential submodule is called a uniform module. A ring $R$ for which $R_R$ is a uniform module is called a right uniform ring. They were notably used in A. W. Goldie's theory of uniform dimension. You can find in Lam's Lectures on modules and rings a chapter devoted to this.

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What about Eisenbud's Commutative algebra with a view toward algebraic geometry? Now, I haven't read much of the thing, and I'm hardly knowledgeable about much algebraic geometry, but I enjoyed what I read and the title seems to be exactly what you're looking for.

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Consider the exact sequence $0\to A\cap B\to Q\to Q/A\oplus Q/B$. Tensor this sequence with $F$ to get an exact sequence due to flatness, your statement follows from this resulting exact sequence. Here we are just assuming that $A$, $B$ are $R$ submodules of the $R$ module $Q$. ADDED LATER: after tensoring with $F$, the rightmost term will be $Q/A\otimes ... 1 If$A, B$are submodules of$Q$(it is not important that$Q$is the fraction field of$R$) we have an inclusion$0\to Q/A\cap B\to Q/A\oplus Q/B$. Tensoring with$F$, using flatness, we have$0\to F\otimes Q/A\cap B\to F\otimes (Q/A\oplus Q/B)$to be exact. Now, one easily checks that$F\otimes Q/A\cap B=F\otimes Q/F\otimes (A\cap B)$and similarly, ... 1 1) There is such a great book (of 120 pp.) which is Very easy to read but it ofcourse it covers less material (than Eisenbud's book) in its 120 pages. -- The book is: Miles Reid, Undergraduate Commutative Algebra. --With immediate applications to Algebraic Geometry with many examples and Pictures. Moreover, it is written for begining grad. students. 2) ... 1 In such questions the Proposition 2.4 from Atiyah and Macdonald, Introduction to Commutative Algebra, is a kind of universal tool. (See also here.) In the notation of the book, set$M=I^n/y$,$\mathfrak a= R$. Note that$M^2=M$within the field of fractions of$R$. Now let$x\in M$and define$\phi:M\to M$by$\phi(a)=ax$. Then$\phi(M)\subseteq M$, and ... 1 If I understand your definitions correctly, any finite product of fields, including the zero ring, has this property. 1 The answer to your question depends on your definition of multivariate division algorithm. A good algorithm in any case might be the following: Let$(g_1,\ldots,g_n)\in R^n$be a fixed tuple of polynomials. Consider the following replacement rules:$\operatorname{LM}(g_i) \mapsto (\operatorname{LM}(g_i)-g_i)$The algorithm now does the following: Start ... 1 Set$R=k[x_1,\ldots,x_n,y_1,\ldots,y_m]$. Denote the generators of$I$by$g_i = y_i - \psi_i$. Then, in the lexicographic order you chose, the leading monomial of$g_i$is$\operatorname{LM}(g_i) = y_i$. Because these monomials are pairwise coprime, the tuple$(g_1, \ldots, g_m)\$ is a Grobner basis.

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