I was trying to prove that a surjective Endomorphism $f:A \to A$ of a noetherian ring is also injective. I would like to know why this argument is not correct? $A/\rm{Ker}f \cong \rm{Im}f=A \Rightarrow Kerf=\{0\}$
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.
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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.
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Suppose $f: A \to B$ and $g: B \to C$ are both surjections.  Since $f$ is surjective, then for every $b \in B$ there exists $a \in A$ such that $f(a)= b$.  Since $g$ is surjective, for every $c \in C$, there's  $b \in B$ such that $g(b) = c$. Then $g(f(a)) = g(b) = c$.   Would this work?  edit; I didn't realize there are similar questions on this site. Don't mind if this post is deleted.
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Let $R$ be an integral domain, and let $a \in R, a \neq 0$. Let $f_a: R \rightarrow R$ be defined by $f_a(r)=ar$. Prove that $f$ is injective.   Then prove that every finite integral domain is a field.   My ideas and concept:   Can someone help articulate my understandings into a proper solution?
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Say I have some function $f:R \rightarrow S$ such that $f$ is a ring homomorphism and $J$ is an ideal of $f$. $I = f^{-1}(J)$ is an ideal of $R$, but I don't really understand why. $J$ being an ideal means that $I$ is non-empty, that much makes sense.  But how can I show that addition/absorption hold if $f$ isn't surjective, nor do I know if $f^{-1}$ is even a ring homomorphism? I've seen a very similar question has been asked, but the only response appeared to just prove that these things held in $J$.  For instance, if I want to prove addition holds, I could take $i_1, i_2 \in I$ so that for $j_1, j_2 \in J$, $i_1 = f^{-1}(j_1)$, $i_2 = f^{-1}(j_2)$. But then I can't really add these elements together because I don't know if addition is preserved in $f^{-1}$. What gives?
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Please help me to prove the following result:  Let   $R$  be  a ring with $1$ and $M$ a uniserial left $R-$module and $\alpha,\beta\in End_R(M)$.      $\alpha\beta$ is surjective if and only if both $\alpha$ and $\beta$ are surjective   This is what i wrote:  The implication  if both $\alpha$ and $\beta$ are surjective then $\alpha\beta$ is surjective is evident since the composition of two surjective functions is surjective:   Let $\alpha:M \to M$ and $\beta:M \to M$ be surjective functions.   We wish to show that $\alpha\circ \beta:M \to M$ is also surjective.  Now, we know that for every $c \in M$, there is a $b \in M$ such that $c = \alpha(b)$, by surjectivity of $\alpha$. And we know that for that same $b$, there exists an $a \in M$ such that $b=\beta(a)$, by surjectivity of $\beta$.  So what we have shown is that for every $c \in M$, there is an $a \in M$ such that $c=\alpha(\beta(a))=\alpha \circ \beta(a)$, i.e., $\alpha \circ \beta$ is surjective.  For the other inmplication let us suppose that $\alpha\beta$ is surjective. Let $a\in M$ by sujectivity of $\alpha\beta$ there exist $b\in M$ such that $\alpha\beta(b)=a$ hence there exist $c=\beta(b)\in M$ such that $\alpha(c)=a$ then $\alpha$ is surjective   But i am not able to prove that $\beta$ is also surjective. Please help me to do so. Thanks in advance.
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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.
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If $F,G: C \rightarrow D$ are functors between regular categories and $F \Rightarrow G$ is a natural isomorphism, is it true that if $F$ is faithful {resp. full) then $G$ is full?  I believe this is true but would like some confirmation.  What about if $F$ is essentially surjective?  Is it necessarily true that $G$ is essentially surjective?  Thanks.
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I am wondering how to prove that a non-zero degree map from $A \to A$ is surjective. For example, identifying $S^1 \subset \mathbb{C}$, we can take $f:S^1 \to S^1$ via $f(z) = z^k$ with $k\neq 0$. This is a map of degree $k$. How can I show that it is surjective?  Note that I am interested in general topological spaces, not the circle example in specific.
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Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function s.t $|f(x)-f(y)|\ge \frac{1}{2}|x-y|$ for all $x,y$. Then $f$ is injective can be seen easily as $x\neq y$ implies $f(x)\neq f(y)$.  But how to check that $f$ is surjective or not ?
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If $\varphi:R^{m}\to R^{n}$ is an epimorphism of free modules over a nontrivial commutative ring, does it follow that $m \geq n$?  This is obviously true for vector spaces over a field, but how would one show this over just a commutative ring? -----Edit Is there any way to use the following? If $\varphi : M \to M'$ is an epimorphism of left $S$-modules and $N$ is any right $S$-module then $id_N \otimes \varphi $ is an epimorphism.
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Let $f :R \to S$ be a surjective ring homomorphism , $M$ be a maximal ideal of $S$ , I am writing a proof showing $f^{-1}(M)$ is a maximal ideal of $R$ , Please verify whether it is correct or not .   Proof :- Let $J$ be an ideal such that $f^{-1}(M) \subset J \subseteq R$ , we want to show $J=R$ i.e. $R \subseteq J$. Now as  $f$ is surjective , $M=f(f^{-1}(M)) \subseteq f(J) \subseteq S$ . As $f$ is surjective , $f(J)$ is an ideal of $S$ . Now if it were possible that $M=f(J)$ , then $x \in J \implies f(x) \in f(J)=M \implies x \in f^{-1}(M)$ , so $J \subseteq f^{-1} (M)$  ,contarry to our assumption $f^{-1}(M) \subset J$ . Thus $M \ne f(J)$ , $f(J)$ is an ideal of $S$ containing $M$ ; since $M$ is maximal ideal in $S$ , so $f(J)=S$ . Then $x \in R \implies f(x) \in S=f(J) \implies \exists j \in J : f(x)=f(j) \implies f(x-j)=0_S \in M $  $\implies x-j \in f^{-1} (M) \subset J \implies x=x-j+j \in J$ , so $R \subseteq J$   Is the proof correct . Please comment .
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Assume R is noetherian and $f:R \to S$ is a ring homomorphism. Is $f(R)$ noetherian?     Reading Dylan Moreland commentary: "If $\mathfrak{b}$ is an ideal of $f(R)$, then $f^{-1}(\mathfrak{b})$ is an ideal of R and $f(f^{-1}(\mathfrak{b}))=\mathfrak{b}$ ." Then he writes: "if you have an  increasing chain of ideals in $f(R)$" . So because there are only finitely many ideals in $f^{-1}(\mathfrak{b})$, which is noetherian, there can also only be finitely many ideals in $f(R)$. So we can say from this that $f(R)$ is noetherian?  Reading Arturo Magidins commentary: I don't know the lattice isomorphism theorem. I can't find it on google either. Do you maybe have a link for me so I can read it? For the second definition of a noetherian ring: every ideal is finitely generated because: $f(a_{0},a_{1},....,a_{n}) = f(a_{0}),f(a_{1}),f(a_{2}),...,f(a_{n})= f(R)$ follows directly because f is a ring homomorphism. Is this a correct proof already?    Thank you.
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.
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I need help solving the next question, I think that it's actually a refutation:     let $f,g: \mathbb{N} \rightarrow \mathbb{N}$  so that $\forall n ∈ \mathbb{N} : f(n) = g(2n)$   Prove that if $f$ is surjective $g$ is not injective
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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 )
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This is part of the exercise, I'm stuck with it.  $A$ is a commutative ring with unit.  1) Suppose we have an homomorphism $\phi : A^{m} \to A^{n}$ surjective. Is true that $m \geq n $ ?  2) Suppose we have an homomorphism $\phi : A^{m} \to A^{n}$ injective. Is true that $m \leq n $ ?