Suppose $F$ is a field and there is a ring homomorphism from $\Bbb Z$ onto $F$. Then show that $F$ is isomorphic to $\Bbb Z_p$ for some prime number $p$.  I am facing difficulty to do the proof. I think that we have to make a map so that the kernel of the homomorphism goes to class $0$ of $\Bbb Z_p$.
122001
Here we have $F$ a field and $f: \mathbb{Z} \to F$ a ring epimorphism. We are to prove that $F$ is a finite field with non-zero characteristic.   I know that since $f$ is an epimorphism, we have that $\forall x\in F$, $\exists y \in \mathbb{Z}$ such that $f(x) = y$. (As an epimorphism is a surjective ring homomorphism).   I am unsure about two things: Am I to prove that F is finite, or just prove that it has a non-zero characteristic. Either way, I could use some guidance. Thank you :)   PS This is for HW, so I only require a good hint about how to approach the problem.
48784
Any homomorphism $φ$ between the rings $\mathbb{Z}_{18}$ and $\mathbb{Z}_{15}$ is completely defined by $φ(1)$. So from  $$0 = φ(0) = φ(18) = φ(18 \cdot 1) = 18 \cdot φ(1) = 15 \cdot φ(1) + 3 \cdot φ(1) = 3 \cdot φ(1)$$  we get that $φ(1)$ is either $5$ or $10$. But how can I prove or disprove that these two are valid homomorphisms?
65976
I'm trying to prove that every finite Boolean ring, $R$, is isomorphic to a finite number of copies of $\mathbb{Z} / 2 \mathbb{Z}$:  \begin{align} R \cong \mathbb{Z} / 2 \mathbb{Z} \; \times \cdots \; \times \; \mathbb{Z} / 2 \mathbb{Z} \end{align}  I'm aware that there are answers to this question on this website. However, almost all of the answers seem to use facts about modules and vector spaces defined over fields. This question is posed in section 7.6 in Dummit and Foote's Abstract Algebra. This section is part of the chapters which introduces the basics of ring theory (rings, homomorphisms, ideals, Chinese Remainder Theorem etc.).  I'm not even sure how to being answering this question. I know that if $R$ were an integral domain, then we would have that $R \cong \mathbb{Z} / 2 \mathbb{Z} $.   Please only give hints, considering that that the statement has to be proved using the material only included up till chapters 9 in the textbook.
65737
If $A$ is Noetherian, the surjectivity of a homomorphism from $A$ to itself is necessarily injective.   Thus, it's clear that if $\phi: A \to A$ is surjective, but not injective, the ring in question would have to be non-Noetherian.  Reformulating the question using the first isomorphism theorem. Given $\phi:A \to A$, we know that $A/ \ker \phi \cong \textrm{im} \phi \cong A$. When is it the case that the kernel is nontrivial?   Related Question: If every surjective homomorphism $A \to A$ is an isomorphism, does it follow that $A$ is Noetherian?
96660
I have a problem with something in abstract algebra. The problem is the following:  Prove or disprove that the mapping $\theta: \mathbb{C} \to \mathbb{C}$ defined by $\theta(a+bi)=a-bi$ is a ring isomorphism.  I know that it should be an isomorphism. I know how to show that the mapping is one-to-one, and I know how to show that the mapping respects both addition and multiplication as defined in $\mathbb{C}$. But how would you clearly show that $\theta$ is onto?  Thanks in advance for any help!
21732
How many isomorphisms are there from $\Bbb Z_{12}$ to $\Bbb Z_{4} \oplus \Bbb Z_{3}$?  I can see that if $f$ is the isomorphism, the isomorphism is determined by the value of $f(1)$ because $f(1) = (1,1)$ and $f(k) = f(k) + f(1) = f(k) + (1,1)$.  But since $\Bbb Z_{12}$ has $4$ generators, there are $4$ different ways to set $f(1)$ equal to a generator.  Therefore there are $4$ isomorphisms.  Is this correct?
30081
I am studying from Patrick Morandi's Field and Galois Theory, and I am stuck on Problem 6 from Section 1 (on page 13).     Verify the following universal mapping property for polynomial rings:         Let $A$ be a ring containing a field $F$. If $a_1,\dotsc,a_n \in A$, show that there is a unique ring homomorphism $\varphi \colon F[x_1,\dotsc,x_n] \to A$ with $\varphi(x_i) = a_i$ for each $i$.       Moreover, suppose that $B$ is a ring containing $F$, together with a function $f \colon \{ x_1,\dotsc,x_n \} \to B$, satisfying the following property: For any ring $A$ containing $F$ and elements $a_1,\dotsc,a_n \in A$, there is a unique ring homomorphism $\varphi \colon B \to A$ with $\varphi(f(x_i)) = a_i$. Show that $B$ is isomorphic to $F[x_1,\dotsc,x_n]$.      The  for the book says that we must assume that all the rings are commutative and the ring homomorphisms are $F$-homomorphisms (that is, they are ring homomorphisms that fix $F$).    I have proved the first part as follows: let $\mathbf{a} = (a_1,\dotsc,a_n)$. The evaluation map $\operatorname{ev}_\mathbf{a} \colon F[x_1,\dotsc,x_n] \to A$ given by $\operatorname{ev}_\mathbf{a}(f) = f(a_1,\dots,a_n)$ is an $F$-homomorphism with $\operatorname{ev}_\mathbf{a}(x_i) = a_i$ for each $i$. Moreover, if $\varphi$ is any $F$-homomorphism from $F[x_1,\dotsc,x_n]$ to $A$ with $\varphi(x_i) = a_i$, then we must necessarily have that $$ \begin{align} \varphi\left( \sum_{k_1,\dotsc,k_n} c_{k_1,\dotsc,k_n} x_1^{k_1} \dotsm x_n^{k_n} \right) &amp;= \sum_{k_1,\dotsc,k_n} c_{k_1,\dotsc,k_n} \varphi(x_1)^{k_1} \dotsm \varphi(x_n)^{k_n}\\ &amp;= \sum_{k_1,\dotsc,k_n} c_{k_1,\dotsc,k_n} a_1^{k_1} \dotsm a_n^{k_n}\\ &amp;= \operatorname{ev}_\mathbf{a}\left( \sum_{k_1,\dotsc,k_n} c_{k_1,\dotsc,k_n} x_1^{k_1} \dotsm x_n^{k_n} \right). \end{align} $$ Thus, the existence and uniqueness of $\varphi$ has been established.    I am stuck in proving the second part. From the previous result, we have a unique $F$-homomorphism $\Phi \colon F[x_1,\dotsc,x_n] \to B$ given by evaluation at $(f(x_1),\dotsc,f(x_n))$, and my intuition is that this is the required isomorphism. But I am not able to make much progress.  To show that $\Phi$ is injective, the idea I have in mind is as follows: given $0 \neq g \in \ker(\Phi)$, choose the ring $A$ and elements $a_1,\dotsc,a_n$ in such a way that $g(a_1,\dotsc,a_n) \neq 0$, thereby arriving at a contradiction. To show that $\Phi$ is surjective, my intuition is that the uniqueness of the map $\varphi$ should play a role, since we will have used only the existence in showing injectivity.  Am I on the right track? I haven't been able to convert these thoughts into a proof yet, and I would really appreciate some hints. Thanks in advance!
89299
Let $f \colon A \to B$ be a ring homomorphism, and let $N$ be a $B$-module. Regarding $N$ as an $A$-module by restriction of scalars, form the $B$-module $N_B = B \otimes_A N$. Show that the homomorphism $g \colon N \to N_B$ which maps $y$ to $1 \otimes y$ is injective and that $g (N)$ is a direct summand of $N_B$.   [Define $p \colon N_B \to N$ by $p (b \otimes y) = by$, and show that $N_B = \operatorname{Im} (g) \oplus \operatorname{Ker} (p)$.]   My idea was to define an isomorphism \begin{align*} \phi \colon N_B &amp; \to \operatorname{Im} (g) \oplus \operatorname{Ker} (p) \\ b \otimes y &amp; \mapsto (p (b \otimes y), b \otimes y - 1 \otimes (g \circ p) (b \otimes y)) \end{align*}  It's easy to check that $\phi$ is injective and surjective. If I take the $B$-module structure of $N_B$ to be $$ b' . (b \otimes y) = (b'b) \otimes y $$ I only manage to show that $\phi$ is a homomorphism of $A$-modules, because $g$ seems to be only a homomorphism of $A$-modules: $$ b.g(y) = b.(1 \otimes y) = b \otimes y \neq 1 \otimes (b.y) = g (b.y) $$ (where $\neq$ means "not equal in general").  However, the way the question is phrased seems to suggest that the isomorphism should be one of $B$-modules.  Should I take the $B$-module structure of $N_B$ to be something else, for example $$ b'.(b \otimes y) = b \otimes (b'.y) $$ or should I show the result only for $A$-modules?  I've been confusing myself about this issue of "as $A$-module" or "as $B$-module" and would appreciate if someone could clarify. (I also wonder if, since homomorphisms are only fancy functions of sets, it somehow doesn't matter.)
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.
45109
I'm currently trying to solve this problem.      Let $f: R \rightarrow S$ be a surjective ring homomorphism. Let $K = \ker(f)$. Assume $P$ is a prime ideal s.t. $K \subset P$. Show $f(P)$ is a prime ideal in $S$.   I solved the ideal part.   Let $y \in f(P)$, by definition then $y = f(x)$ for some $x \in P$. Now let $s \in S$. Then since $f$ is surjective $\exists r \in R$ s.t. $f(r) = s$. So then since $f$ is a homomorphism we have $f(x)\cdot s = f(x) \cdot f(r) = f(x\cdot r)$. Then since $P$ is ideal and $x \in P$, $x\cdot r \in P$, so $f(x \cdot r) = f(x)\cdot s \in f(P)$ so $f(P)$ is closed under multiplication by an element in $S$.   Now let $y, y' \in f(P)$, again by definition we have $x, x' \in R$ s.t. $f(x) = y, f(x') = y'$. By $f$ homomorphism we have $y - y' = f(x) - f(x') = f(x-x')$. Then by $P$ ideal we have $x-x' \in P$, thus $y-y' \in f(P)$.   Therefore $f(P)$ is an ideal.   The part I'm having trouble with is showing that $P$ prime implies $f(P)$ prime.   I started off let $AB \subset f(P)$, by definition we know we have $C \subset P$ s.t. $f(C) = AB$. From here I need to arrive at showing that either $A \subset f(P)$ or $B \subset f(P)$. I'm not really sure how to move forward from here but I know I somehow need to involve the fact that $K \subset P$ since I haven't used that yet.  Any help would be appreciated.  Note: I'm not assuming that $R$ or $S$ is commutative, I already know how to solve it if they are using the commutative definition of prime.
38531
How does one prove $\Bbb{Z}_{100} \times \Bbb{Z}_{10}$ is not isomorphic to $\Bbb{Z}_{1000}$  Do you have to show one is cyclic and the one isn't? (why does this work?)
90225
The Question:  Find all primes $p$ such that $\phi(x)=x^{13}$ is a homomorphism $\Bbb Z/p\Bbb Z \rightarrow \Bbb Z/p\Bbb Z$    My Thoughts:  So from what I understand  \begin{align} \ &amp; \phi:\Bbb Z/p\Bbb Z \rightarrow \Bbb Z/p\Bbb Z \quad\text{is a homomorphism} \\ \ \iff &amp; \phi(xy) \equiv \phi(x)\phi(y) \pmod p \quad \forall \; x,y\\ \ \iff &amp; (xy)^{13}\equiv x^{13}y^{13} \pmod p\quad \forall \; x,y \end{align}  But isn't $(xy)^{13}\equiv x^{13}y^{13} \pmod p\quad \forall \; x,y$ always true, regardless of $p$?  Or is there something I am not understanding?
111203
1) Prove that $\operatorname{Hom}_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0$.       2) Show that $\Bbb{Q}$ is not a projective $\Bbb{Z}$-module.    1) We know that $\Bbb{Q}$ is an injective $\Bbb{Z}$-module.  This implies that every short exact sequence $$0 \rightarrow \Bbb{Q} \xrightarrow{f} A \xrightarrow{g} B \rightarrow 0$$ is split exact.   In particular,   $$0 \rightarrow \Bbb{Q} \xrightarrow{f}  \Bbb{Z} \xrightarrow{g} A \rightarrow 0$$  is split exact.   But this implies that we have a homomorphism $k: \Bbb{Z} \rightarrow \Bbb{Q}$  such that $kf = 1_{\Bbb{Q}}$.  So we need to look at what possible homomorphisms we could have from $\Bbb{Z}$ to $\Bbb{Q}$. Since $\Bbb{Z}$ is cyclic, the homomorphism $k$ is determined by where it sends $1$. Suppose that we send $1$ to $q \in \Bbb{Q}$ such that $q \not= 1$ and $q \not= 0$. Let $m \in \Bbb{Z}$. Then  $$k(m) = mk(1) = mq$$   But then   $$qmq = k(1)k(m) \not= k(m) = mq$$   So either $1 \in \Bbb{Z}$ must be sent to $1 \in \Bbb{Q}$ or $k$ can be the trivial homomorphism.  But if $1_{\Bbb{Z}}$ is sent to $1_{\Bbb{Q}}$, then $k$ must be the inclusion map. However in order to for $kf = 1_{\Bbb{Q}}$ to hold, $f$ must also sent $z$ to $z$ for all $z \in \Bbb{Q}$. But if that's the case, then for any $a/b \in \Bbb{Q}$ where $b \not= 0, 1$, we have (for $n \in \Bbb{Z}$) $$f(a/b) = n = f(n)$$  and this contradicts the fact that $f$ must be injective (since the sequence is exact).      So the only possible $\Bbb{Z}$-module map from $\Bbb{Z}$ to $\Bbb{Q}$ is $0$.   Do you think my answer is correct?   2) I was wondering if anybody could give a hint on this one, because I couldn't really get started.  Thanks in advance.
37080
Let $A$ be a ring. I'm trying to prove that there is only one ring homomorphism (different from the zero one) from $\Bbb Q$ into $A$ or there are no ring homomorphisms between $\Bbb Q$ and $A$.    I have proven that if $A=\Bbb Z$ or $A=\Bbb R$ then there are no ring homomorphisms between $A$ and $\Bbb R$ (not taking into account the zero one) but I don't know how to prove it for any $A$.
73025
So I want to find the isomorphism $\phi$ that takes $F = \mathbb{Z}_3/\langle x^3 - x - 1\rangle$ to $E = \mathbb{Z}_3/\langle x^3 - x + 1\rangle$.  I understand that these are both finite fields of size 3^3 since $x^3 - x - 1$ and $x^3 - x + 1$ are both irreducible polynomials of degree 3.  I feel like a good start would be to take the subfield $\mathbb{Z_3}$ in $F$ to the subfield $\mathbb{Z_3}$ in $E$.  Or maybe find the right generator of each multiplicative group of size $3^3-1$ of each field and send those to each other.
78325
Let $A$ be a finitely generated abelian group and $f:A\rightarrow A$ a surjective homomorphism. How do I prove that $f$ is an isomorphism?  And if $f$ were injective instead of surjective would the statement still hold?
21328
The isomorphism $\mathbb{Z}[i]/(a+bi) \cong \Bbb Z/(a^2+b^2)\Bbb Z$ is , when the integers $a$ and $b$ are coprime. But what happens when they are not coprime, say $(a,b)=d&gt;1$?      — For instance if $p$ is prime (which is not coprime with $0$) then $$\mathbb{Z}[i]/(p) \cong \mathbb{F}_p[X]/(X^2+1) \cong \begin{cases} \mathbb{F}_{p^2} &amp;\text{if } p \equiv 3 \pmod 4\\ \mathbb{F}_{p} \times \mathbb{F}_{p} &amp;\text{if } p \equiv 1 \pmod 4 \end{cases}$$ (because $-1$ is a square mod $p$ iff $(-1)^{(p-1)/2}=1$).  — More generally, if $n=p_1^{r_1} \cdots p_m^{r_m} \in \Bbb N$, then  each pair of integers $p_j^{r_j}$ are coprime, so that by CRT we get $$\mathbb{Z}[i]/(n) \cong \mathbb{Z}[i]/(p_1^{r_1}) \times \cdots \times \mathbb{Z}[i]/(p_m^{r_m})$$  I was not sure how to find the structure of $\mathbb{Z}[i]/(p^{r}) \cong (\Bbb Z/p^r \Bbb Z)[X] \,/\, (X^2+1)$ when $p$ is prime and $r&gt;1$.  — Even more generally, in order to determine the structure of $\mathbb{Z}[i]/(a+bi)$ with $a+bi=d(x+iy)$ and $(x,y)=1$, we could try to use the CRT, provided that $d$ is coprime with $x+iy$ in $\Bbb Z[i]$. But this is not always true: for $d=13$ and $x+iy=2+3i$, we can't find Gauss integers $u$ and $v$ such that $du + (x+iy)v=1$, because this would mean that $(2+3i)[(2-3i)u+v]=1$, i.e. $2+3i$ is a unit in $\Bbb Z[i]$ which is not because its norm is $13 \neq ±1$.  — I was not able to go further. I recall that my general question is to known what $\mathbb{Z}[i]/(a+bi)$ is isomorphic to, when $a$ and $b$ are integers which are not coprime (for instance $a=p^r,b=0$ or $d=(a,b) = a^2+b^2&gt;1$).  Thank you for your help!