I am looking for an example of an injective ring homomorphism $\phi\colon A\to B$ such that the following holds: there exists a prime ideal $\frak q$ of $B$ such that $\phi^{-1}(\frak q) = \frak p$, and the induced map $A_{\frak p} \to B_{\frak q}$ is not injective.  The algebro-geometric picture should be clear: if we put $X = \mathrm{Spec\, A}$ and $Y = \mathrm{Spec\, B}$, then $\phi$ induces a morphism $(\psi,\theta)\colon Y\to X$ of affine schemes with $\psi(\frak q) = \frak p$. The map $A_{\frak p} \to B_{\frak q}$ corresponds to the induced map on the level of stalks at $\frak q$ of the morphism of sheaves $\theta^\sharp \colon \psi^* \mathcal O_X \to \mathcal O_Y$ [I have used the notation from EGA I].
122757
Let $R\subset T$ be two commutative rings, and $T$ is integral over $R$. Let $\mathfrak m\in \operatorname{Max} R,\mathfrak n\in\operatorname{Max}T$ such that $\mathfrak m=\mathfrak n\cap R$. Show that $R_\mathfrak m\subset T_\mathfrak n$, i.e. the canonical map $\{\frac ru\mid r\in R,u\in R\setminus\mathfrak m\}\to\{\frac tv\mid t\in T,v\in T\setminus\mathfrak n\}$ must be an injection.   I feel confused because the exercise 4 from Chapter 5 in the book of Atiyah and Macdonald just assume that the localization is a ring extension. And I don't know if the condition "integral over" is necessary.
62097
Let $\phi\colon R \to R'$ be a ring homomorphism. Prove that if $R$ is a field then either $\phi$ is an isomorphism or $\phi(r) = 0$ for all $r \in R$.   I am stuck on this problem and don't know where to begin. I feel like I'm very weak in writing proofs.
75584
Let me ask Prob.F6 in Pinter's algebra book.  'Let $p$ be a prime and $f$ be a homomorphism from $G$ to $H$ where $G$ and $H$ are groups.  If range of $f$ has an element of order $p$, then $G$ has an element of order $p$.'   If $f$ is injective, life gets easier. But without the injectivity of the homomorphism, i feel lost... Would somebody help please?
13716
Let $\phi:R\to S$ Be a surjective ring homomorphism and $M \subset S$ is maximal, then $\phi^{-1}(M)$ is maximal in $R$. Here is what I wrote (it is known the preimage is an ideal, so prove only maximality)  Assume $\phi^{-1}(M) \subset W \subset R$. Then first $\phi(\phi^{-1}M) = S$ because $\phi$ is surjective. Hence $\phi(\phi^{-1}M) \subset \phi(W) \subset \phi(R) \implies S \subset \phi(W) \subset S \implies \phi(W) = S$. Now let $w \in W$ and $\phi(w) \in \phi(W) = S = \phi(\phi^{-1}M)$, so $w \in \phi^{-1}M$? Or this only shows $x \in \phi^{-1}(M)$ such that $\phi(x) = \phi(w)$?  Is there a way to complete the proof with what I wrote? (I already have witness the solution, basically $\phi^{-1}M \subset R$ is an ideal. Then by isomorphism theorems $S/M \approx R/\ker(\pi \circ \phi)$ where $\pi$ is the quotient map and usually field argument implies $\ker (\pi \circ \phi) = \phi^{-1}M$ because fields contain only two ideals)
115145
Does there exist a surjective homomorphism from $(\mathbb R,+)$ to $(\mathbb Q,+)$ ?  ( I know that there 'is' a   'surjection' , but I don't know whether any surjective homomrophism from $\mathbb R$ to $\mathbb Q$ exist or not . Please help . Thanks in advance )
109739
Question 2.3.20 in Qing Liu's Algebraic Geometry and Arithmetic Curves:  Let $A$ be a ring, $G$ a finite group of automorphisms of $A$. Let  $$    p: \mathrm{Spec}\ A \to \mathrm{Spec}\ A^{G} $$  be the morphism of affine schemes induced by the obvious inclusion.     Show that, for $x_{1}$, $x_{2} \in \mathrm{Spec}\ A$,    $$ p(x_{1}) = p(x_{2}) \iff \exists \ \sigma \in G : \sigma(x_{1}) = x_{2}.$$   The '$\impliedby$' direction is easy, but I'm having trouble with the other direction (although I have no doubt that it is also simple).   My attempts so far have included assuming that there is no $\sigma \in G$ such that $\sigma(x_{1}) = x_{2}$, so that for each $\sigma$ there is $g_{\sigma} \in x_{1}$ with $\sigma(g_{\sigma}) \notin x_{2}$, and then trying to construct an element of $p(x_{1})\backslash p(x_{2})$ out of these $g_{\sigma}$, for example by taking $$   g = \prod_{\sigma \in G}g_{\sigma} $$ and then  $$   h = \prod_{\sigma}\sigma(g) $$ but this appears to fail as I can't ensure that $\sigma(g_{\tau}) \notin p(x_{2})$ for $\sigma \neq \tau$.  Any hints would be much appreciated!
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.
82810
Let $\phi: R \to S$ be a surjective homomorphism. Prove that $\frac{R/ \ker \phi}{(\ker \phi + J)/ \ker \phi} \cong \frac{S}{\phi(J)}$ for an ideal $J$ of $R.$  Obviously, $S \cong R/ \ker \phi$(first isomorphism theorem) and $\phi(J)$ is an ideal of $S$, since $\phi$ is surjective( $\forall s \in \phi(J), s = \phi(j)$ for some $j \in J, \forall s' \in S, s' = \phi(r)$ for some $r \in R$, so $ss' = \phi(j)\phi(r) = \phi(jr) \in \phi(J)$, and $ss' = \phi(r)\phi(j) = \phi(rj) \in \phi(J)$ ).  However, I'm having trouble with going on. If it's important, I mean a unitary ring by a ring.
8621
Let $k$ be a field, n a positive integer.  Vakil's notes, 17.4.B: Show that all the automorphisms of the projective scheme $P_k^n$ correspond to $(n+1)\times(n+1)$ invertible matrices over k, modulo scalars.   His hint is to show that $f^\star \mathcal{O}(1) \cong \mathcal{O}(1).$ (f is the automorphism. I don't if $\mathcal{O}(1)$ is the conventional notation; if it's unclear, it's an invertible sheaf over $P_k^n$) I can show what he wants assuming this, but can someone help me find a clean way to show this?
28976
Let $\mathcal F$ and $\mathcal G$ be sheaves of groups of some topological space $X$ and let $\varphi : \mathcal F \to \mathcal G$ be a surjective morphism. That means that for each $x \in X$ the induced group morphism $\varphi_x : \mathcal F_x \to \mathcal G_x$ is surjective.  Does this imply that for each $x\in X$ there exists an open $U \subseteq X$ with $x \in U$ such that $\varphi_U : \mathcal F(U) \to \mathcal G(U)$ is surjective?  I don't think so but would be glad to see an easy counter example.
34488
I've been doing some old exam problems and I've come across a problem that I've answered, but my gut is telling me that there's something I'm glossing over.     Let $R$ be a commutative ring with identity and let $U$ be an ideal that is maximal among non-finitely generated ideals of $R$. I wish to show that $U$ is a prime ideal.   Assume that $U$ is not prime. Let $x, y\not\in U$ be such that $xy\in U$. $U$ is contained in a maximal ideal $M$ and $xy\in M$, so either $x$ or $y$ is in $M$; assume $x\in M$. The condition $U\subset M$ then implies that there is a ring homomorphism $$\varphi: R/M\to R/U$$  Since $R/M$ is a field, $\varphi$ is injective. Hence, $\varphi(x)\in U$. This is a contradiction, so $U$ must be prime.  The thing that worries me is that I never explicitly used the hypothesis that $U$ was not finitely generated or the result that $M$ must be finitely generated.
15528
I can't figure out why such an epimorphism cannot exist.  I think the fact that a bijection exists between the $\mathbb{Q}$ and $\mathbb{Z}$ throws me off a little. What is it about having a homomorphic surjection that causes problems?
17872
This is question 2.4 in Hartshorne. Let $A$ be a ring and $(X,\mathcal{O}_X)$ a scheme. We have the associated map of sheaves $f^\#: \mathcal{O}_{\text{Spec } A} \rightarrow f_* \mathcal{O}_X$. Taking global sections  we obtain a homomorphism $A \rightarrow \Gamma(X,\mathcal{O}_X)$. Thus there is a natural map $\alpha : \text{Hom}(X,\text{Spec} A) \rightarrow \text{Hom}(A,\Gamma(X,\mathcal{O}_X))$. Show $\alpha$ is bijective.  I figure we need to start off with the fact that we can cover $X$ with affine open $U_i$, and that a homomorphism $A \rightarrow \Gamma(X,\mathcal{O}_X)$ induces a morphism of schemes from each $U_i$ to $\text{Spec} A$ and some how glue them together. But I have no idea how to show that the induced morphisms agree on intersections. How does this work?
63503
If $\phi$ is a homomorphism from the group $G$ with identity $e$ to the group $G'$ with identity $e'$, then $\phi$ is injective if and only if $\ker\phi=$?  I think the $Ker\phi = \{x\in G:\phi(x)=1\}$  because you are looking at an identity then it is all reals $0$ does not work a because if $x=0, e'x=0$  but then i realize it is injective how does that affect the kernel?
34319
The following simple claim is used without proof in Proposition 2.9 of Atiyah and MacDonald (p.23). Although I believe I can prove it with a fairly involved argument, the claim is treated by the authors as a triviality so I feel like there should be a simple argument. Is there such an argument?      Let $M, N$ be $A$-modules. Let $v\colon M \to N$ be a homomorphism. Suppose that for all $A$-modules $P$, the dual maps $v^*\colon\operatorname{Hom}(N,P) \to \operatorname{Hom}(M,P)$ are injective. Then $v$ is surjective.   (Note that the claim is trivial for free modules, but I am looking for a general proof).
88393
Find an explicit isomorphism of the map $h: \Bbb F_5[x]/(x^2+x+2) \to \Bbb F_5[x]/(x^2+4x+2)$  I take the question to mean showing the 2 rings are isomorphic by finding an explicit mapping of each element.   I think $x^2+x+2$ and $x^2+4x+2$ are irreducible over $\Bbb F_5$, so these are both fields.  Also $x^2=4x+3$ in the first case and $x^2=x+3$ in the second.  But beyond these basic facts I'm not sure how I should proceed.
22683
I need an example of a ring homomorphism such that the image of an ideal needs not to be an ideal.   I found that the image is an ideal if the mapping is onto, so all we need to find a mapping that is not onto.