Let $A, B$ be commutative rings with identity elements and let $\mathfrak{p} \subseteq B$ be a prime ideal. Let $\varphi: A \to B$ be an injective ring homomorphism.  I want to show that the induced homomorphism $\varphi_{\mathfrak{p}} : A_{\varphi^{-1}(\mathfrak{p})} \to B_{\mathfrak{p}}$ is also injective. Moreover: Why is the induced morphism of schemes $f: \text{Spec}(B) \to \text{Spec}(A)$ dominant?  The motivation of the first question: To show this is sufficient for showing that the induced morphism of sheaves $\mathcal{O}_{\text{Spec}(A)} \to f_{*}\mathcal{O}_{\text{Spec}(B)}$ is injective.
12703
Let $\alpha:A\to B$ be a ring homomorphism, $Q\subset B$ a prime ideal, $P=\alpha^{-1}(Q)\subset A$ a prime ideal. Consider the natural map $\alpha_Q:A_P\to B_Q$ defined by $\alpha_Q(a/b)=\alpha(a)/\alpha(b)$. Suppose that $\alpha$ is injective. Then is $\alpha_Q$ always injective?    I think so, but I'm clearly being too dense to prove it! My argument goes as follows.  Let $\alpha(a)/\alpha(b)=0$. Then $\exists c \in B\setminus Q$ s.t. $c\alpha(a)=0$. If $B$ is a domain we are done. If not we must exhibit some $d\in A\setminus P$ s.t. $da=0$. Obviously this is true if $c =\alpha(d)$. But I don't see how I have any information to prove this!  Am I wrong and this is actually false? If so could someone show me the trivial counterexample I must be missing?   Many thanks!
60234
The following is the first part of Proposition 2.9 in "Introduction to Commutative Algebra" by Atiyah &amp; Macdonald.     Let $A$ be a commutative ring with $1$. Let $$M' \overset{u}{\longrightarrow}M\overset{v}{\longrightarrow}M''\longrightarrow  0\tag{1} $$ be sequence of $A$-modules and homomorphisms. Then the sequence (1) is exact if and only if for all $A$-modules $N$, the   sequence  $$0\longrightarrow \operatorname{Hom}(M'', N)  \overset{\overline{v}}{\longrightarrow}\operatorname{Hom}(M,  N)\overset{\overline{u}}{\longrightarrow}\operatorname{Hom}(M', N)  \tag{2} $$ is exact.   Here, $\overline{v}$ is the map defined by $\overline{v}(f)=f\circ v$ for every $f\in\operatorname{Hom}(M'', N)$ and $\overline{u}$ is defined likewise.   The proof one of direction, namely $(2)\Rightarrow (1)$ is given in the book, which I am having some trouble understanding. So assuming (2) is exact sequence, the authors remark that "since $\overline{v}$ is injective for all $N$, it follows that $v$ is surjective". Could someone explain why this follows?  Given that $\overline{v}$ is injective, we know that whenever $f(v(x))=g(v(x))$ for all $x\in M$, we have $f=g$. I am not sure how we conclude from this surjectivity of $v$.  Thanks!
71544
I am reading the proof of proposition 5.3.1 of , and I have a problem with the last sentence : "If $g' = g''/f^n$ ($g''\in A$) then $\textrm{Spec}((A_f)_{g'}) = \textrm{Spec} (A_{fg''})$, and we're done."  Noting $V = \textrm{Spec} (B)$ and $V' = \textrm{Spec} (B_g)$, and noting $D_Z (h)$ the distinguished affine open subset of an affine scheme $Z$ associated to a section $h\in\Gamma(Z,\mathcal{O}_Z)$, I understand perfectly that if we note $U' = \textrm{Spec} (A_f)$ then the inclusion $U'\subset V$ induces a morphism $f : \Gamma(V',\mathcal{O}_X)\to \Gamma(U',\mathcal{O}_X)$ and that if we note $\varphi$ the associated morphism of affine schemes then $\varphi^{-1}(V') = D_{U'}(g')$ where $g'$ is the image of $g$ by $f$, but as $\varphi$ is the the inclusion $V'\subset U'$, we have $\varphi^{-1}(V') = V'$ because $V'\subset U'$ so finally $V' = D_{U'}(g')$ and $V'$ is distinguished in $U'$. I guess that the last sentence of the proof is supposed to show that this $V'$ is also distinguished in $U$, but I don't understand why.
111066
The problem is stated as follows:  "Let $R$ be a Noetherian ring and $\theta$ be a ring homomorphism from $R$ to $R$. Show that if $\theta$ is surjective then it is also injective."  Regardless of the right solution, I don't understand why is the following wrong:     We have $\theta: R\to R$. By the isomorphism theorem $R/\ker\theta\cong\operatorname{Im}\theta$. Since $\operatorname{Im}\theta = R$, it follows $\ker\theta =0$, so it's injective.   Harsh criticism will be appreciated. Thanks.
36387
Let $S$ be a scheme and let $f:X\to Y$ be a morphism of separated $S$-schemes.  shows that, when $X$ is reduced, the obvious morphism $g_f:\Gamma_f \to X\times_S Y\to X$ is an isomorphism, where $\Gamma_f$ is the scheme-theoretic image of $\Delta_f:X\to X\times_s Y$.     Can we show that $g_f$ without assuming that $X$ is reduced?   I see how the reduced induced structure is being used there, but I do not see if $X$ being reduced affects the statament, $X$ cannot have double/triple/... points if $\Gamma_f$ does not! So, I wonder if there is a probably more elaborate proof that shows $g_f$ to be an isomorphism.   The only assumption I have is that $S$ is Noetherian.   A hint or a reference would be appreciated.
20206
Let $A$ be a commutative ring with 1, set $V(\frak a)=$ the set of prime ideals of $A$ that contains $\mathfrak a$, and write $\operatorname{Supp}(M)$ for the support of the $A$-module $M$. We always have $\operatorname{Supp}(M)\subseteq V(\operatorname{Ann} M)$; if $M$ is finitely generated then the other inclusion also holds. Assume now $M$ is finitely generated. It is ok to show that if $M_{\frak p}=0$ then $\operatorname{Ann} M\cap A\setminus\frak p\ne\varnothing$, I was just wondering if the following procedure for showing this inclusion is flawed:  Suppose $\frak p$ contains $\operatorname{Ann} M$. Since $M$ is finitely generated we have the equality $(\operatorname{Ann} M)_{\frak p}=\operatorname{ Ann} (M_p)$, hence the annihilator of $M_{\frak p}$ is contained in maximal ideal of $A_{\frak p}$ and is therefore different from (1), so $M_{\frak p}\ne0$.
34692
Let $R$ be a commutative ring with unit, $I$ an ideal of $R$ and consider the following three constructions.   The  $R_I$ of $R$ at $I$ (i.e. the localization of $R$ at the multiplicative system $R\setminus I$) gives a morphism $$ f_1:\operatorname{Spec}(R_I)\to \operatorname{Spec}(R)=:X $$ The  $\widehat{R}_I$ of $R$ at $I$ gives a morphism $$ f_2:\operatorname{Spec}(\widehat{R}_I)\to X $$ For the special case $I=(a)$, the localization $R_a$ (i.e. the localization of $R$ at the multiplicative system $\{1,a,a^2,\ldots\}$) gives a morphism $$ f_3:\operatorname{Spec}(R_a)\to X $$   My question is:     What is the geometric meaning of all three constructions and how are they related?   This is what I ''know'' already or describes at least the style of answer I would appreciate. As remarked in the comments, $I$ has to be a prime ideal.   $R\to R_I$ is injective iff $R\setminus I$ contains no zero divisors which is the case if $I$ is a prime ideal. The scheme $\operatorname{Spec}(R_I)$ is the intersection of all the neightbourhoods of $I$ in $X$. This is a little contra-intuitive for me since the last statement sounds like ''$f_1$ is injective'' but the $\operatorname{Spec}$-operator should turn 'injective'' and ''surjective'' around somehow (I know this is literally not true but I only want to get a feeling like ''is contained'', ''is bigger'', ''is smaller'', etc.). $\operatorname{Spec}(R_a)$ is somehow the opposite of $\operatorname{Spec}(R_{(a)})$ (= the intersection of all the neightbourhoods of $(a)$ in $X$) because $\operatorname{Spec}(R_a)$ seems to be something like the union of all the open sets of $X$ not containing the point $(a)$. $R\to\widehat{R}_I$ is injective iff $\cap I^n=(0)$ and this holds very often (e.g. for $R$ noetherian and either an integral domain or a local ring). Hence (this is probably false intuition as remarked before), $f_2$ should be something like a ''projection'' (from something ''big'' into something ''small''). But what is geometrically the difference between localization and completion. I don't have a geometric idea of completion at all.   As remarked by Qiaochu Yuan in the comments below, one should not think of $\operatorname{Spec}$ as injective-surjective switching.
54678
I'd like to see a proof why $\varphi \in \operatorname{Hom}{(S^1, S^1)}$ looks like $z^n$ for an integer $n$.  At first I thought I could argue that if I have a homomorphism that maps $e^{ix}$ to some $e^{iy}$ for $x,y \in \mathbb R$ then $z = rx$ for some real number $r$. But on second thought I'm not sure why I can't have $\varphi (e^{ix} ) = e^{g(x)i}$ for a $g$ other than $g(x) = \lambda x$.   I'm also interested in seeing different proofs. I'm sure there are several ways to prove this.  Thanks for your help.
82048
This is Exercise 2.4.E. from .     Show that a morphism of shaves of sets is an isomorphism if and only if it induces an isomorphism of all stalks.    My failed attempts:  I found some similar questions  and . But I still couldn't figure out the details.   I have shown the "only if" part. I have also shown that if the morphism induces an isomorphism of stalks, it is injective.   The surjective part confused me a lot. Here is a commutative diagram:   To show surjectivity, suppose $g\in \mathscr{G}(U)$. Its image in the stalks at $p$ is $(g,U)_p$. Since $\phi_p$ is an isomorphism for all $p$, we have $(f_p,U_p)\in \mathscr{F}_p$, such that $\phi_p(f_p,U_p)=(g,U)_p$ for all $p\in U$. Now I don't know how to glue the $f_p$ together. Since even we have the pairs $(f_q,U_q)\in \mathscr{F}_q$, we cannot show that their restrictions on $U_p\cap U_q$ are equal. We have to find a small enough open set to glue them.   I also tried the open cover argument from the right hand side:  Since $\prod(g,U)_p$ consists of compatible germs, there exists an open cover $U_i$, and section $g_i$ in $\mathscr{G}(U_i)$, such that $(g,U)_p=(g_i,U_i)$ for all $p\in U_i$. But this doesn't help much. Since $\phi_p$ are different maps, we still cannot conclude anything about the $(f_p,U_p)$.    Any help will be greatly appreciated!
59613
I need help to solve the following exercise:   If $K$ is an algebraic extension of $\mathbb{Q}$ (finite or infinite), then we let $O_{K}$ denote the ring of algebraic integers in $K.$ If $\frak{P}$ is a prime ideal of $O_{K}$ then $\frak{P}∩\mathbb{Z}$ is a prime ideal of $\mathbb{Z}.$ Moreover, if $\frak{P}$ is non-zero, then so is $\frak{P}∩\mathbb{Z}$ . Thus $\frak{P}∩\mathbb{Z}=$$ p\mathbb{Z}$ for some prime $p.$ Show that $O_{K}/\frak{P}$  is a field, and is an algebraic extension of $\mathbb{F}_{p}$; then show that it is Galois over $\mathbb{F}_{p}$ with abelian Galois group.  Now suppose that $K/F$ is a (finite or infinite) Galois extension. Let $\frak{P},$  $O_{K}$ be as before and put $\frak{p}=\frak{P} \cap $$O_{F}$, a prime ideal of $O_{F}.$  Show that $\mathrm{Gal}(K/F)$ acts transitively on the set of prime ideals of $O_{K}$ above $\frak{p}.$   Thanks in advance
98026
Let $A$ be a commutative ring and $S$ a multiplicative subset of $A$ generated by $s\in A$ which is not a zero-divisor. Consider the polynomial ring $A[x]$.      Given a polynomial $f\in A[x]$, suppose that $f(1/s)=0$ for $s\in S$. Does it follow by some application of the first isomorphism theorem that $f(x)=(sx-1)g(x)$, with $g(x)\in A[x]$?   On the one hand I think this can be proven by constructing an isomorphism from $A[x]$ modulo the relation $x=1/s$ (really the replacement of $x$ with $1/s$). This is evidently an isomorphism from $A[x]/(x=1/s)$ into $S^{-1}A$. It is tempting to conclude (though I'm not certain that this should work), that there must be some surjective homomorphism $\phi: A[X]\to S^{-1}A$ such that $\ker(\phi)=(sx-1)$. [Since generally $(A/B\cong A/C) \nRightarrow (B\cong C)$, it is not sufficient to conclude that, say, the kernel of $\psi: f\to f(1/s)$ is $(sx-1)$.]  I am especially interested in answers as to whether, say, $\psi$ has kernel $(sx-1)$. But for anyone who can follow my own attempt at a proof, I'm also interested in whether my argument breaks down or can be made to work rigorously.
49526
For part a) of this exercise we need to show that for a scheme $X$ and an open affine subset $U=$ Spec $B$ that $U \cap X_f = D(\bar{f})$ where $\bar{f}$ is the restriction of $f \in \mathcal O_X(X)$ to $\mathcal O_X(U)$ and $X_f= \{x \in X :$ the stalk of $f$ at $x$ $(f_x)$ is not contained in the maximal ideal $(m_x)$  of $\mathcal O_x \}$.  Now I believe the proof goes something along the lines of using the fact that this maximal ideal in the affine scheme is just $xB_x$ but I can't see how $f_x \not\in m_x$ implies $\bar{f}_x \not\in xB_x$. Like how can we go from the local ring $\mathcal O_x$ to $B_x$?
13732
I am solving an exercise from Dummit, Foote. Let $R$ be a ring with $1$. Then the following are equivalent:   Every $R$-module is projective. Every $R$-module is injective.   Proof:  (1) implies that for any R-mod $A$ we have $Ext^i(-,A)$ is a zero functor. But this means that for any R-mod $Ext^i(L,A) = 0$ for all $A$ so $L$ is injective. I show $(2) \Rightarrow (1)$ exactly the same way.  It is quite easy so I am afraid I missed something. Is my proof correct?
61985
Let $(A,\mathfrak{m})$ and $(B,\mathfrak{n})$ be local Noetherian rings. Suppose that $\phi : A\rightarrow B$ is a map such that $\phi(\mathfrak{m}) \subset \mathfrak{n}$ and suppose    $A/\mathfrak{m} \cong B/\mathfrak{n}$; $\mathfrak{m} \rightarrow \mathfrak{n/n^2}$ is surjective; $B$ is a finitely generated $A$ module.   Show that $\phi$ is surjective.  Here is my approach so far: Consider $0\rightarrow \mathfrak{m} \rightarrow A \rightarrow A/\mathfrak{m}\rightarrow 0$ and the exact sequence for $B$. Then we can establish an commutative diagram of the two exact sequences. (Sorry i do not know how to type that.) Then apply the snake's lemma, the isomorphism in condition 1 yields an isomophism $\mathfrak{n}/\phi(\mathfrak{m})\cong B/\phi(A)$. To see $\phi$ is surjective, it suffices to show $\phi(\mathfrak{m})=\mathfrak n$. Since $\mathfrak{n}$ is the maximal ideal of $B$, if $\phi(\mathfrak{m})+\mathfrak{n}^2 = \mathfrak{n}$, Nakayama will give the desired result. Now somehow I want to conclude $\phi(\mathfrak{m})+\mathfrak{n}^2 = \mathfrak{n}$ from conditions 2) and 3), which seems to be true but cannot really proceed.
115141
Let $R$ be a simple ring and $T$ be a non-zero ring. Let $f\colon R \rightarrow T$ be a ring homomorphism. Show $f$ is injective.  Proof: $\ker f \lhd R$, so $\ker f=R$ or $\ker f=\{0\}$. If $\ker f=R$ then $R/R \cong \{0\}$, but $T \neq \{0\}$, which is a contradiction. I can't see why does the last part means contradiction.   Thanks for any assistance!
41618
Let $R$ be a commutative ring with $1$ and $I$, $J$ ideals in $R$.   For a prime ideal $P$, let $I_P=(R-P)^{-1}I$ be the localization of $I$ at $P$.   Question:     If $I_P=J_P$ for all prime ideals $P$ then is $I=J$?   My first attempt at a solution: take $x\in I$ then for each prime $P$ write $x/1=y/s$ for some $y\in J$ and $s\in R-P$. Then for some $t\in R-P$, $xst=yt\in J$ where $y,s,t$ all depend on $P$. From this can I deduce that $x\in J$?   Is there a better way to show this?  Thank you!
61393
I have a question which should be pretty basic commutative algebra, but I can't find a reference and I'm stuck on proving a result myself, so here it goes:  Let $\varphi: S \to R$ be a morphism of commutative rings and let $M$ be a finitely generated $S$-module. Then, we can "extend the scalars of $M$" by considering $M \otimes_S R$, which is naturally an $R$-module. I would like to get a description of $\text{Supp}(M \otimes_S R)$ (as an $R$-module) involving $\text{Supp}(M)$ and $\varphi$.  Maybe this helps: by Eisenbud's Commutative algebra with a view toward algebraic geometry, Corollary 2.7, the problem can probably be solved by finding a description of $\text{Ann}(M \otimes_S R)$ and it is not difficult to see that $\varphi(\text{Ann}(M))R \subset \text{Ann}(M \otimes_S R)$, but I fail to prove the other inclusion.