Surjective map between $\mathbb{Z^{+}}$ and $\mathbb{Z}$. I am confused as to how this is even possible to create.
120464
I've come to this: $$f: \mathbb{N} \to \{\ldots, -6,-4,-2,0,2,4,6, \ldots\},\qquad f(n) = \begin{cases} 2n &amp; \text{ if } n \text{ is odd} \\  -n &amp; \text{ if } n  \text{ is even} \end{cases}$$  I don't know what to do with this though. I never know how to format a proof correctly.
71626
Prove $\mathbb{Z}_3[x]/(x^2+1)$ and $\mathbb{Z}_3[x]/(x^2+x-1)$ are isomorphic by finding an explicit isomorphism. My question is how I can define the map. Here are what I tried:  $\mathbb{Z}_3[x]/(x^2+1)=\{ a+b\alpha \,|\, \alpha^2 + 1 = 0, \quad a,b\in\mathbb{Z} \}$  $\mathbb{Z}_3[x]/(x^2+x-1)=\{ a+b\beta \,|\, \beta^2 + \beta - 1 = 0, \quad a,b\in\mathbb{Z} \}$  Then I am looking for an isomorphism of the form $f(a+b\alpha) = a+bf(\alpha)$ for all $a,b\in\mathbb{Z}$ where $f(\alpha^2)+1=0$. Here I'm stuck.  My professor said $f(a+b\alpha) = a+b(1-\beta)$ so that the restriction of $f$ to $\mathbb{Z}_3$ is the identity morphism. I don't understand why we have $f(\alpha) = 1-\beta$.  He also gave another problem: $\mathbb{Z}_5[x]/(x^2+2)$ and $\mathbb{Z}_5[x]/(x^2-x+1)$, where $f(a+b\alpha) = a+b(2+\beta)$.  Could someone please explain to me why $f(\alpha) = 1-\beta\,$ in the first problem and $f(\alpha)=2+\beta\,$ in the second problem?  Thank you very much.
83064
So I've got a bijection. It clearly has an inverse, but how exactly do I prove that the inverse is a linear map as well?     Suppose that the linear map $T:U\to V$ is a bijection. So $T$ has an inverse map $T^{-1}:V\to U$. Prove that $T^{-1}$ is a linear map.
1896
This seems to be a common result. I've been trying to follow the bijective proof of it, which can be found easily online, but the explanations go over my head. It would be wonderful if you could give me an understandable explanation of the proof and let me know how I'd go about finding such a bijection.
74387
I'm trying an old problem here:   Suppose $n\mid m$, and I have a natural ring homomorphism $\varphi\colon \mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $\varphi(j)=j\mod{n}$. I can verify that this is a ring homomorphism, but why does it induce a surjective group homomorphism on the unit groups $(\mathbb{Z}/m\mathbb{Z})^\times\to(\mathbb{Z}/n\mathbb{Z})^\times$?  My question in particular is why is it surjective? I take $k\in(\mathbb{Z}/n\mathbb{Z})^\times$. Then there exist integers $s,t$ such that $sn+tk=1$. Since $n\mid m$, I can also write $m=na$. So multiplying through I get $sm+tka=a$. I'm lost here. How can I find a unit in $(\mathbb{Z}/m\mathbb{Z})^\times$ which maps to $k$ to show surjectivity?
65852
Prove or disprove : There exists a continuous surjection from $\mathbb{R}^3- S^2$ to $\mathbb{R}^2-\{(0,0)\}$ (here $S^2\subset \mathbb{R}^3$ denotes the unit sphere defined by the equation   $x^2+y^2+z^2=1$),.     This question had appeared in TIFR GS-2018 exam for PhD admissions. How should I think about such a map?
17830
I am getting bored waiting for the train so I'm thinking whether there can exist a $C^1$ injective map between $\mathbb{R}^2$ and $\mathbb{R}$. It seems to me that the answer is no but I can't find a proof or a counterexample... Can you help me?
113703
I am doing an exercise about algebraic geometry, where the exercise  tell us to provide example of the image of polynomial map $f:\mathbb{C}^{m}\rightarrow \mathbb{C}^{n}$ that is not an algebraic set. But I am thinking that the image of every polynomial is $\mathbb{C}$ thus every polynomial map is surjective. What have I misunderstand?
15293
In John Fraleigh's book, A First Course In Abstract Algebra Exercises 15.7 and 15.11, one shows that  $$ (\mathbb{Z} \times \mathbb{Z})/ \langle (1,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \  \mbox{ and  } \ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2 $$  One does this with the first isomorphism theorem. With the same idea I proved for example that  $$ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,3)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \ \mbox{ and }\ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,4)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2 $$ So I conjectured that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_{k}$ where $k=\mathrm{gdc}(m,n)$. For the previous four cases, the homomorphism $\phi: \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z} \times\mathbb{Z}_k$ given by $$ \phi(x,y)=\left(\frac{nx-my}{k}, \ x \ \ (\mathrm{mod} \ k) \right) $$ is surjective with kernel =$\langle (m,n)\rangle$. However, this is not the case when $(m,n)=(4,6)$. So, I cannot use what I did for the four cases to prove the general case.     What I want to know is if my conjecture is true. If so, how can I give a general homomorphism? If it is not true, how can I find $k$, such that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_k$?   Thanks in advance for any help/hint/comment!
59008
Possible Duplicate:             I wanna know if $\mathbb{Z^+}\times \mathbb{Z^+}$ is countably infinite and how should one construct a bijection and check if it is a bijective function
122369
Prove that $\mathbb{Z}$ is not isomorphic to additive group of any vector space over any field.  Proof.  Surpose that: $\phi : (A, +) \rightarrow  \mathbb{Z} $ is an isomorphism. Then there is some $r∈A $ such that $ϕ(r)=1_\mathbb{Z}$. We have $$ 1_\mathbb{Z}=ϕ(r)=ϕ(2(r/2))=2ϕ(r/2)$$  or equivalently that $ϕ(r/2)=\frac{1}{2}$. But this is not in $ \mathbb{Z}$, so there can be no such morphism.  Is my proof above correct? Thanks
123150
The second part of an exercise asks that I show:     There does not exist a surjective bounded linear map from $\ell^2(\mathbf{N})$ to $\ell^1(\mathbf{N})$.     The first part was to show that these two spaces are separable in their respective $L^p$ norms.    I've really hit a wall on this one.  I know, in fact, that there does exist a surjective bounded linear map from $\ell^1(\mathbf{N})$ to $\ell^2(\mathbf{N})$.  So my intuition says that if there were also one going the other way, then the two spaces would have to be isomorphic, which is false .  But I don't know of any such theorem that I could apply to "the category of Banach spaces".    My only other idea is to argue by contradiction using the open mapping theorem, but nothing is coming to me...  My guess is that separability is also going to be important, given the first part of the exercise.  -Thanks.
25878
Does there exist any surjective group homomorphism from $(\mathbb R^* , .)$ (the multiplicative group of non-zero real numbers) onto $(\mathbb Q^* , .)$ (the multiplicative group of non-zero rational numbers)?
112549
Does there exist an isomorphism between the semigroups $S(4)$ and $\mathbb{‎‎‎‎‎Z}_{‎256}$?‎   $S(4)$ is the set of all maps from the set $X$ to itself and $X =\{1, 2, 3, 4\}$,  $S(4)$ is a semigroup under the composition of mappings, and $‎‎‎‎‎‎\mathbb{Z}_{‎256}=\{0, 1, 2, \dots , 255\}$ is the semigroup under multiplication modulo 256.
107131
If we set $A$ and $B$. Assuming that there exist an injective map from $A$ to $B$ and an injective map from $B$ to $A$. How to prove that there is a bijection between $A$ and $B$.
41674
In self studying abstract algebra and I've come upon the following problem which I could not solve directly. For any $d\in \mathbb{Z}$ we are asked to show that $\mathbb{Z}[\sqrt d]=\{a+b\sqrt{d} ~\colon a,b\in \mathbb{Z} \}$ is an integral domain. This reduces to showing that the ring has no zero divisors which I tried to do through algebraic manipulation.  I have seen two solutions to this problem that use higher level techniques. One defines conjugate and norm and proves the claim using simple consequences of these definitions. The other says that since the given ring is a subring of the field $\mathbb{C}$ it can have no zero divisors.  I don't like either approach because the first uses ideas that haven't been introduced in the text and isn't a natural place to start (for me). The second, while both true and simple, is not satisfying to me. That $\mathbb{C}$ is a field has not been established and assuming it just makes the problem trivial.   My question is whether or not naive tactics like algebraic manipulation are capable of demonstrating the result. Am I lacking in cleverness or is such an approach impossible?
36141
Show that $(\mathbb{Z}[x]/(x^{n+1}))^{\times}\cong \mathbb{Z}/2\mathbb{Z}\times\Pi_{i=1}^n\mathbb{Z}$.   Anyway how and what method is used to prove this. I still have no idea now. really thanks for your help