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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.


3

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.


3

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


2

$m_p$ 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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