Let $f$ be a homomorphism from commutative ring $A$ to $B$.  Suppose $f$ induces a bijection between $\operatorname{Spec} B$ and $\operatorname{Spec} A$.  Must $\operatorname{Spec} A$ and $\operatorname{Spec} B$ be homeomorphic?  Are there any counterexamples？(The topology is Zariski topology)
121116
The following question occurred to me while doing exercises in Hartshorne. If $A \to B$ is a homomorphism of (commutative, unital) rings and $f : \text{Spec } B \to \text{Spec } A$ is the corresponding morphism on spectra, does $f$ bijective imply that $f$ is a homeomorphism? If not, can anyone provide a counterexample? The reason this seems reasonable to me is because I think that the inverse set map should preserve inclusions of prime ideals, which is the meaning of continuity in the Zariski topology, but I can't make this rigorous.
103896
Let $R$ be a commutative ring with unit and $M$ a right $R$-module. Consider the assignment $\varphi: \operatorname{Hom}_{R} (R,M) \rightarrow M$, $f \mapsto \varphi(f) = f(1)$. Show that $\varphi$ is an isomorphism.  My approach: We have to prove $\varphi$ is a morphism, then $\varphi$ is a injection, surjection, hence $\varphi$ is an isomorphism. $\forall f,g \in \operatorname{Hom}_{R} (R,M)$ $\varphi(fg) = (fg)(1) = f(1)g(1) = \varphi(f) \varphi(g)$ $\Rightarrow \varphi$ is a morphism $\forall f,g \in \operatorname{Hom}_{R} (R,M)$ $\varphi(f) = \varphi(g) \Rightarrow f(1) = g(1) \Rightarrow f = g$ $\Rightarrow \varphi$ is injective $\Rightarrow \varphi$ is a homomorphism  How can I prove $\varphi$ is surjective to conclude $\varphi$ is an isomorphism? Sorry for my poor English.
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.
119925
Let $f:A \rightarrow B$ be a ring homomorphism and $\mathfrak{a}_1,\mathfrak{a_2}$ be ideals of $A$. Let $\mathfrak{a}^e$ denote the extension of an ideal $\mathfrak{a}$ of $A$ in $B$. An exercise shows that $$(\mathfrak{a}_1 \cap \mathfrak{a}_2)^e \subseteq \mathfrak{a}_1^e \cap \mathfrak{a}_2^e.$$  It seems that $(\mathfrak{a}_1 \cap \mathfrak{a}_2)^e =\mathfrak{a}_1^e \cap \mathfrak{a}_2^e$ is wrong in general, but I can not find a counter-example.
112299
Prove or disprove: for any comm. ring $R $,   For three $R $-modules $M $, $N $ and $P $, $M \oplus P \simeq N \oplus P $ implies $M \simeq N $. Let $f:M\to N $ and $g:N\to M $ be two $R $-module homomorphisms such that $gf=1_M $ (where $1_M = \operatorname{id}_M$ means the identity map on $M$). Then, $N\simeq\operatorname{Im}(f)\oplus\ker(g)$.   If $R $ is a PID and the modules are f.g., 1 is true, but I have no idea on general cases.
119197
Let $A \subset B$ be a finite ring extension. Fix a ring homomorphism $\nu : A \to \Omega$, where $\Omega$ is an algebraically closed field.   I want to show there exists a non-zero homomorphism $v : B \to \Omega$ of $A$-modules.  I want show that the space $V$ of all the homomorphisms $v$ in (i) is a non-zero finite dimensional vector space over $\Omega$. Show that the algebra $B$ acts on this space. Show that eigenvectors of this action correspond to homomorphisms of algebras $\lambda : B \to \Omega$ that extend the homomorphism $\nu$. Deduce that the set of such homomorphisms $\lambda : B \to \Omega$ is finite and non-empty.    We can assume that $B = Ab_1 + ... + Ab_n$ where $b_1,...,b_n \in B$. The chose $v(b_i)=\omega_i \in \Omega-\{0\}$ and then define $$ v( a_1 b_1 + ... + a_n b_n ) = \nu(a_1) v(b_1) + ... + \nu(a_n) v_n = \nu(a_1) \omega_1 + ... + \nu(a_n) \omega_n \ .$$ This is an $A$-module homomorphism. Am I correct? For the second part, that $V$ is not-zero follows from the first part. That it is finite dimensional follows from the fact that $v_i(b_j)=\delta_{i,j}$ (Kronecker's delta) with $i=1,...,n$ spans $V$ and hence $\dim V \le n &lt; \infty$. I define the action of $B$ over $V$ to be $$b \bullet v = \left( x \mapsto v(b)v(x) \right) \mbox{  for all } b,x \in B \ .$$ This seems to be the natural action to define. Am I correct? However, I haven't managed to solved the rest (the eigenvector stuff and the relation to $\nu$).
39347
Let $X$ be a subset of $\mathbb{R^n}$. The topology on $X$ is induced by the topology of $\mathbb{R^n}$. If there is an homeomorphism from $X$ onto $\mathbb{R}^n$, is it true that $X$ is open in $\mathbb{R}^n$ ? I think yes, but how can it be shown? (of course, $X$ is open in $X$).  If $x \in X$, $H_{n-1}(\mathbb{R}^n-\{f(x)\})= \mathbb{Z}$ so $H_{n-1}(X-\{x\})= \mathbb{Z}$, but I don't know how to continue.  Thanks in advance.
116610
Prove that $\mathbb{Q}$ isn't homeomorphic to $\mathbb{Z}$  I know that a homeomorphism is a bijection function from two sets where both the function and its inverse are continuos.   So I figure that if I show that there is no continuos invective mapping from $\mathbb{Q}\rightarrow\mathbb{Z}$ then I have done the proof. But I do not know how to do this (if his is the right way). How can I start to show this?
6226
Possible Duplicate:          $f:K\rightarrow R$ is a homomorphism from field to ring. To prove is that it is injective or zero.
119714
I am thinking about the proof of the second isomorphism theorem, and something isn't very clear to me.   Let $R$ be a ring, $S\subset R$ a subring and $I\subset R$ an ideal. We have the natural homomorphism $f:R\rightarrow R/I$. The theorem states that $\mathrm{Im}(S)=(S+I)/I$. My question is why not simply $\mathrm{Im}(S)=S/I$?  I understand that it is not true (for starters, $S/I$ need not be a subring), but I cannot explain that to my self in a convincing way.
80411
On $\mathbb{R}$ with the usual topology, generate an equivalence relation by $x\sim x$ for all $x$ in $\mathbb{R}$ and $x\sim 0$ for all $x$ in $[0,1]$. Prove $\mathbb{R}/\sim$ is homeomorphic to $\mathbb{R}$.
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).
11709
I know a proof of the following theorem using determinants. For some reason, I'd like to know a proof without using them.     Theorem   Let $A$ be a commutative ring.   Let $E$ and $F$ be finite free modules of the same rank over $A$.   Let $f:E → F$ be a surjective $A$-homomorphism.   Then $f$ is an isomorphism.
86406
Let $k$ be a ﬁeld and let $A$ be a $k$-algebra. Denote by $End_A(A)$ the set of all $A$-homomorphisms of the regular $A$-module $A$ into itself. Fix $a ∈ A$, and deﬁne the $A$-module homomorphism $r_a : A → A$ by $r_a(x) = x*a$.  $\\$ Show that $End_A(A)$ = {$r_a$ | $a ∈ A$}  $\\$ Answer: $\\$ Obviously I can see that, for any $a$ we have $r_a$ is in $End_A(A)$, but how do I go about the other direction? Namely why is it that any endomorphism at all has this very simple form? I just don't see it...
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.
56117
In chapter 3 of Atiyah and Macdonalds Intro to Comm Algebra there is the statement  "We also have a ring homomorphism $f:A\rightarrow S^{-1}A$ defined by $f(x)=x/1$. This is not in general injective." (Note: $A$ denotes an integral domain and $S$ a multiplicatively closed subset of $A$.)  How could this possibly not be injective? If $f(x)=f(y)$ then $x/1=y/1$ so $1x=1y$ hence $x=y$. What am I missing here?
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?