By "pseudo ring homomorphism", I mean a map $f: R \to S$ satisfying all ring homomorphism axioms except for $f(1_R)=f(1_S)$.  Even if we let this last condition drop, there are only two ring pseudo-homomorphisms from $\Bbb Z$ to $\Bbb Z[\omega]$ where $\omega$ is any root of unity, for example. They are the identity homomorphism and the "$0$ pseudo-homomorphism".  I couldn't find a non-zero example. It is easy to conclude that $0=f(a)(1_S-f(1_R))$, so I know that the counterexample will have to be some $S$ with zero-divisors.
18785
Let $R$ and $S$ be rings with unity $1_R$ and $1_S$ respectively. Let $\phi\colon R\to S$ be a ring homomorphism. Give an example of a non-zero $\phi$ such that $\phi(1_R)\neq 1_S$  In trying to find a non-zero $\phi$ I've done the following observation: Since for $\forall r\in R$ $\phi(r) = \phi(r\times1_R) = \phi(r)\times\phi(1_R)$ we must have that $\phi(1_R)$ is an identity of $\phi(R)$ but not an identity of $S$. We must therefor construct a $\phi$ that is not onto and which have this property. I can't come up with any explicit example though, please help me.
116
Though I've seen several cool axiomizations of $\mathbb{R}$, I've never seen any at all for $\mathbb{Z}$.  My initial guess was that $\mathbb{Z}$ would be a ordered ring which is "weakly" well-ordered in the sense that any subset with a lower bound has a minimal element.  However, after seeing  of a discrete ordered ring, I'm less sure. I made that guess under the impression that the fundamental characteristic of $\mathbb{Z}$ is that every nonzero element has exactly one representation of the form $\pm (1+1+\dots+1)$, but that seems to be shared by every DOR.   Presumably, this definition wouldn't exist if every instance of it was isomorphic to $\mathbb{Z}$, so can someone give me an example of another discrete ordered ring? More to the point, what is a sufficient characterization of $\mathbb{Z}$? (and a proof sketch of uniqueness would be nice)  I'm aware that $\mathbb{Z}$ is pretty easily constructible from $\mathbb{N}$, but I want to use this for a seminar and given the audience I am expecting, I would rather not deal with Peano. (And I guess it feels like cheating to say "$\mathbb{N}$ is a well-ordered rig")
72862
Refer to localization of rings in Lang's Algebra p. 108. In particular, let $A$ be a commutative ring, $S$ a subset of $A$ that is a submonoid of the multiplicative monoid structure of $A$, containing the identity. Then  $S^{-1}A$ is the set of elements $\frac{a}{s}$ with $a \in A$ and $s \in S$.  We have $\frac{a_1}{s_1}=\frac{a_2}{s_2}$ if there exists $\bar{s} \in S$ such that $\bar{s}(s_2a_1 - s_1 a_2)=0$. Now consider the injection of $A$ into $S^{-1}A$ given by $x \mapsto \frac{x}{1}$. This is a ring homomorphism. Let $J$ be an ideal of $A$. Ring homomorphisms preserve ideals and so the injection of $J$ into $S^{-1}A$, call it $J^{*}$, will be an ideal. My problem is that i don't see why for any $\frac{a}{s} \in S^{-1}A$ and any $\frac{j}{1} \in J^{*}$ we will have that $\frac{aj}{s}$ will be in $J^{*}$, i.e. why there must exist an element $\kappa \in J$ such that $\frac{aj}{s}=\frac{\kappa}{1}$.
28511
TRUE/FALSE TEST:  There is a non-trivial group homomorphism from $S_3$ to $\mathbb Z/3\mathbb Z.$   My Attempt:  True: Choose $a,b\in S_3$ such that $|a|=3,|b|=2.$ Then $S_3=\{1,a,a^2,b,ba,ba^2\}.$ Define $f:S_3\to\mathbb Z/3\mathbb Z:b^ia^j\mapsto j+3\mathbb Z$ Then $f$ is a nontrivial homomorphism.  Is the attempt correct?
96113
My questions are:     $1.$ Is there a field morphism $\Bbb R(X) \hookrightarrow \Bbb R$ ?      $2.$ If the answer to $1.$ is "yes", are $\Bbb R$ and $\Bbb R(X)$ isomorphic as fields?   $ $    $ $  For $1.$, here are some comments:   This is true for $\Bbb C$, see . The key point here is the fact that $\Bbb C$ is algebraically closed, of cardinality $2^{\aleph_0}$, and of characteristic $0$. The same argument doesn't work for $\Bbb R$. We can actually find a subfield $F$ of $\Bbb R$ such that $F \cong F(X) \cong F(X,Y) \cong \cdots$, see .  that $\Bbb Q(X)$ does not embed in $\Bbb Q$ (as a field).  This is equivalent to: Is there an injective ring morphism $\Bbb R[X] \hookrightarrow \Bbb R$ ? (By taking the fraction fields).   As for 2. :   They are isomorphic as abelian groups, and actually as $\Bbb Q$-vector spaces. They are not isomorphic as $\Bbb R$-algebras, because they haven't the same transcendence degree over $\Bbb R$. Let's try to see how to construct such a field isomorphism. We could try$^{[1]}$ to find a subset $B$ of $\Bbb R$ such that $\Bbb R = \Bbb Q(B)$ and $C \subsetneq B \implies \Bbb R \neq \Bbb Q(C)$.  The set $B$ must contain transcendental elements, pick such an $x_0 \in B$.  Then $\Bbb R = \Bbb Q(B) = \Bbb Q(B \setminus \{x_0\})(x_0)$ and we could try$^{[2]}$ to find a field isomorphism $\Bbb Q(B \setminus \{x_0\}) \cong \Bbb Q(B)$, so that $\Bbb Q(B \setminus \{x_0\})(x_0) \cong \Bbb Q(B)(x_0) \cong \Bbb R(X)$.  For ${[1]}$, I tried to use Zorn's lemma on $\mathscr B = \{B \subset \Bbb R \mid \Bbb R = \Bbb Q(B)\}$ and the partial order $B_1 ≤ B_2 \iff B_1 \supset B_2$. But this is not clear to me why this should be an inductive set.  For ${[2]}$, my idea was to define $f : \Bbb Q(B \setminus \{x_0\}) \cong \Bbb Q(B)$ by choosing a bijection $b : B \setminus \{x_0\} \to B$ (these sets should be uncountable), and then trying to extend $f\vert_{B \setminus \{x_0\}} = b$ to a field morphism.     Edit for $^{[1]}$: in  article (MINIMAL GENERATING SETS OF GROUPS, RINGS, AND FIELDS, by LORENZ HALBEISEN, MARTIN HAMILTON, AND PAVEL RUZICKA) the theorem 2.4. proves that $\Bbb R$ has no minimal generating set as field (or $\Bbb Q$-algebra, i.e. as ring). The argument uses Artin-Schreier theorem. Minimal generating sets of modules are   .  I think that $B$ is a minimal generating set of a subfield $K \subset \Bbb R$ over $\Bbb Q$ if and only if any element of $B$ cannot be written as a rational fraction of other elements of $B$. Under these conditions and using (transfinite) induction, we can probably prove (and this was my first idea) that any function $f: B \to \Bbb C$ such that $b$ and $f(b)$ are conjugate over $\Bbb Q$ (i.e. have the same minimal polynomial) extends to a (unique) field morphism $\sigma : K = \Bbb Q(B) \to \Bbb C$.  Actually, the collection $\mathscr B$ defined above doesn't have to be an inductive set, at least if we replace $\Bbb R$ by $K =\Bbb Q(\sqrt 2, \sqrt[4]{2},\sqrt[8]{2},\dots)$, because the descending chain $\left(E_n = \{\sqrt[2^m]{2} \mid m \geq n\}\right)_{n \geq 1}$ satisfy $K = \Bbb Q(E_n)$, but $K \neq \Bbb Q\left(\bigcap_{n≥1} E_n\right)=\Bbb Q$, so the chain has no upper bound w.r.t. $≤$ in $\mathscr B$.
113148
I am trying to understand the proof of Theorem 3.4.3 from Weibel's book Introduction to homological algebra.  The statement is the following.     Let $R$ be a ring. Given $R$-modules $A$ and $B$, an extension $\xi$ of $A$ by $B$ is an exact sequence $0\to B\to X\to A\to0$. Then the connecting homomorphism $$\delta:\text{Hom}(A,A)\to\text{Ext}^1(A,B) \tag{1}$$ induces the 1-1 correspondence $\theta$ between equivalence classes of extensions of $A$ by $B$ and $\text{Ext}^1(A,B)$, which is given by $\theta:\xi\mapsto\delta(\text{id}_A)$.   In the beginning of the proof it is written the following     Fix an exact sequence $0\stackrel{j}\to M\to P\to A\to0$ with $P$ projective. Applying $\text{Hom}(-,B)$ yields an exact sequence $$\text{Hom}(P,B)\to\text{Hom}(M,B)\stackrel{\delta}\to\text{Ext}^1(A, B)\to0.\tag{2}$$   I don't understand why the connecting homomorphism in the sequence (2) is denoted by the same letter as in (1). Are they really the same?
40143
Let $D$ be a domain which is not a field. If there exists $b \in D - \{0\}$ such that for all $a \in D - \{0\}$, $a|b$, then is it true that $b = 0$?   This is certainly true if one has a UFD with infinitely many irreducibles (that are not associates) or a Jacobson domain which is not a field (or more generally a domain where the intersection of all non-zero prime ideals is the zero ideal), but I can't seem to be able to prove this for a domain in general or find a counterexample.  If this statement is false for an arbitrary domain, then is there some additional weak hypothesis on the domain which makes the above statement true?   I was actually trying to answer a question asked yesterday on MSE, which was the following:  If $D$ is a domain which is not a field and $Q = Frac(D)$, then $Hom_D(Q,D) = \{0\}$. Note that if $\varphi$ is any such $D$-linear map, then for all $a \in D- \{0\}$, $a\varphi(1/a) = \varphi(a/a) = \varphi(1)$. Thus, I get for all $a \in D - \{0\}$, $a|\varphi(1)$, and I want to conclude that $\varphi(1) = 0$, but cannot.
69063
I know ring homomorphisms map units to units, which made me curious about the following. Suppose $f:R\to R'$ is a surjective ring homomorphism, mapping $1$ to $1'$. Is it necessarily surjective from $U(R)\to U(R')$?  I know if $f(u)$ is a unit in $R'$ with $f(w)$ its inverse, then $f(uw)=f(wu)=1'$, but I see no reason to conclude $uw=wu=1$ in $R$ without assuming $f^{-1}(1')=\{1\}$. But I can't find a counterexample, so I'm not sure whether it's true or not.
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!
106259
Dummit and Foote, 7.4.33(a): Let $R$ be the ring of all continuous functions $[0,1] \to \mathbb{R}$ and let $M_c$ be the kernel of evaluation at $c \in [0,1]$, i.e. all $f$ such that $f(c) = 0$. Show that if $M$ is a maximal ideal in $R$ then $M = M_c$ for some $c$.    I am aware of proofs that use the compactness of $[0,1]$; however I have been told that a simpler proof exists using only the isomorphism theorems for rings along with basic facts about ideals. I have tried on my own to prove it this way but still have no success; is it possible?  (My attempts go something like: since $M$ is maximal, the quotient ring $R/M$ is a field. Then it would be nice to use the fact that $R/M$ has no zero divisors, but I can't quite make it work. Alternatively, consider $M \cap M_c$.)
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 $\nu$).
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.
15640
Is an automorphism of the field of real numbers $\mathbb{R}$ the identity map? If yes, how can we prove it?  Remark An automorphism of $\mathbb{R}$ may not be continuous.
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.
25982
I face the problem of finding how many non-trivial ring or group homomorphisms there are from $\mathbb{Z}_m$ to $\mathbb{Z}_n$, where $m&lt;n$. Is there any general formula?   At the moment, I want to know how many ring homomorphisms there are when $m=12,n=28$.  Please help.
60139
I had an exam this morning, one of the questions asked about the truth of the statement      There is a space $X$ such that $S^1$ is homeomorphic to $X\times X$.   I said that this was false and this was my reasoning... was I correct? (Note, an earlier part of the question asked me to prove $\pi_1(Y\times Z)=\pi_1(Y)\times\pi_1(Z)$, I took this as a hint.)  Suppose for the sake of contradiction that such an $X$ exists. Then $\Bbb{Z}\cong G\times G$ where $G\cong\pi_1(X)$. Let $\Phi$ be the isomorphism.  Let $\Phi(1)=(g_1,g_2)\in G\times G$. Then at least one of $g_1,g_2$ is nonzero, since $\Phi(0)=(0,0)$ (the identity on $G\times G$, and $\Phi$ is injective). Without loss of generality $g_1\ne 0$. Then which element $n\in\Bbb{Z}$ has $\Phi(n)=(kg_1,0)$? Since $$kg_1=\underbrace{g_1+\dots +g_1}_k\in G,$$ and $\Phi$ is surjective, such an $n$ should exist. But this is not possible since $\Phi(n)=(kg_1,0)$ implies (since $\Phi$ is a homomorphism) that $kg_1=ng_1$ and $ng_2=0$, and since $k$ was arbitrary and this need not hold if $k\ne 0$.
71157
I have the following problem:  Is it possible to construct ring homomorphism from $M_2(\mathbb Z)\to \mathbb Z$, or in other words, a homomorphism from ring of all $2\times2$ matrices over the integers into integers?  I tried determinant, trace and mapping that maps matrix to it's element in the position $(1,1)$ but none of that obviously works, which led me to believe there might not be such a homomorphism. Determinant doesn't work because it is obviously not a linear map.  Trace doesn't work because it doesn't respect multiplication.  And the mapping to the position $(1,1)$ also doesn't work because $$\begin{pmatrix}1&amp;1\\1&amp;1\end{pmatrix}\begin{pmatrix}1&amp;1\\1&amp;1\end{pmatrix}=\begin{pmatrix}2&amp;2\\2&amp;2\end{pmatrix}.$$