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3

The answer is no. Why did you expect that? For example, consider, say $A$ is a domain and $B=A[x]$, $I=(x-a)B$ for some $0\neq a\in A$. Then the natural map $A\to B/I$ is an isomorphism and in particular flat. If $J=I+xB$, clearly $A\to B/J=A/aA$ is not flat.

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This is not a definition, it is a property of free objects. Every object surjecting onto a free object satisfies this too, and these do not have to be free. (Consider the group $S_n \times \mathbb{Z}$ for instance.) I advise you to read any text / book on universal algebra. There, you can find concise definitions and constructions of free objects.

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This is almost Exercise 2.2.24(b) of Bruns-Herzog. And other parts of Exercise complete this. Hint. Localization is flat. $\left(R_{(p)}\right)_{pR_{(p)}}=R_p.$

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$m_p$ is the maximal (irrelevant) ideal of $K[X_1,\dots,X_n]/I(V)$, that is, the ideal generated by the images of $X_1,\dots,X_n$ hence $m_p=M_P/I(V)$. Then $m_p^2=(M_P^2+I(V))/I(V)$. Now it's clear why $$m_p/m_p^2=M_P/(M_P^2+I(V)).$$

2

For $s\in U$ and $x\in M$ we have $sx=0\implies sx\in Q$, where $Q$ is a primary submodule of $M$ which appears in a primary decomposition of $(0)$ and $r_M(Q)\cap U=\emptyset$. Conclude that $x\in Q$. For the converse, $x\in\cap Q_j$ with $r_M(Q_j)\cap U=\emptyset$. On the other side, for some $Q_i$ such that $r_M(Q_i)\cap U\ne\emptyset$ we get an $s_i\in ... 1 Let$k$be a field, and consider the domain$A = k[x,y]/(x^2-y^3)\cong k[t^2,t^3]$(the isomorphism is given by$x\mapsto t^3$and$y\mapsto t^2$). Note that$t\notin A$. The field of fractions of$k[t^2,t^3]$contains$t = t^3/t^2$, but$t$is integral over$k[t^2,t^3]$, since it satisfies the monic polynomial$z^2 - t^2$. Hence$A$is not integrally closed ... 1 Let$A$be a commutative ring and$S$a multiplicative set. Then the family of rings$\left\{A_s \right\}_{s \in S}$forms a directed family. To see this, first we define a partial order on$S$by$s \le t$if$t = u s$for some$u \in S$. Next for$s \le t$with$t = u s$, there exists a ring homomorphism$f_{s,t}: A_s \rightarrow A_t$, which is defined by ... 1 This is kind of useless at this point, but since I was able to get a copy of one of Olivier's original articles on weakly étale/absolutely flat morphisms, I thought I'd share what I found. The article Ferrand, Daniel. "Epimorphismes d'anneaux et algèbres séparables." C. R. Acad. Sci. Paris Sér. A-B 265 1967 A411–A414. MR0244313 (39 #5628) is cited as ... 1 Your answer is not right. The ideal quotient$I \colon J$is the set of all$r \in R$such that$rJ \subseteq I$, for$I,J$ideals of a commutative ring$R$. Observe that$\langle xyz \rangle \subset \langle x \rangle$already, so for any$r \in k[x,y,z,o]$, we have that$r\langle xyz \rangle \subseteq \langle xyz \rangle \subseteq \langle x \rangle$and ... 1 The inverse limit$L$of the inverse system$(A_i,f_{ij}:A_i \to A_j)$is defined by the exact sequence $$0 \to L \to \prod_i A_i \xrightarrow{g} \prod_{jk} A_{jk}$$ where$A_{jk} = A_j$and where$g$is defined by$g((a_i))_{jk} = a_j - f_{kj}(a_k)$. It is possible, in principle, to implement this directly for example in Macaulay2, but if$(I, <)$gets ... 1 For the forward direction: If$I\cap{}\hat{P}=(0)$, then since$f\in{}I$, we have$f\notin{}\hat{P}$, since$f\neq{}0$and$f\notin{}(0)$(since P is an integral domain). And backwards: If$f\notin{}\hat{P}$, then note that every element of$I$is of the form$h=\sum_{i=1}^ng_if=gf$where the$g_i\in{}P$and$g=\sum_{i=1}^ng_i$. Since$f\notin{}\hat{P}$... 1 If$P=(a_1,\dots,a_n)$, then$M_P=(x_1-a_1,\dots,x_n-a_n)/I(V)$; see Proving that kernels of evaluation maps are generated by the$x_i - a_i$(the proof given there works for any field). Then$\overline K[V]/M_P\simeq\overline K[x_1,\dots,x_n]/(x_1-a_1,\dots,x_n-a_n)\simeq\overline K$; for the last isomorphism see Maximal ideals in$K[X_1,\dots,X_n]\$.

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