I am looking at the first answer to this question:  Why does the existence of an automorphism between $x$ and $x+1$ imply that the quotient rings are isomorphic? I understand that when you substitute $x$ with $x+1$ in the original expression you get $$(x+1)^3 + x+1 +1 = x^3 + 3 x^2 + 4 x + 3 = x^3 + x^2 +1$$ in $\mathbb{Z}_2[x]$, but I don't understand the connection to the quotient ring. I am trying to apply the logic to a similar question, to show that $$\mathbb{Z}_2[x] / \langle x^2 \rangle \cong \mathbb{Z}_2[x]/\langle x^2-1\rangle$$ But I cannot convince myself why this automorphism leads to the conclusion that the quotient rings are isomorphic.
7288
I still struggle with elementary knowledge about quotient rings.  For example, why can I just write down the second following isomorphism:  $$\frac{\mathbb{C}[x, y, z]}{(x^2 + y^2 - 1, y \pm iz)} \simeq \frac{\frac{\mathbb{C}[x, y, z]}{(y \pm iz)}}{\frac{(x^2 + y^2 - 1,y \pm iz)}{(y \pm iz)}} \simeq \frac{\mathbb{C}[x, y]}{(x^2 + y^2 - 1)} $$  I know the first isomorphism is right by the third isomorphism theorem. But why can I write down the second one?
77272
I am trying to find the automorphisms of the polynomial ring $\mathbb{Z}[x]$.  So far, I have shown that if $\varphi$ is an automorphism, then $\varphi(n)=n$ for $n \in \mathbb{Z}$, and $\varphi(x)$ is a polynomial of degree one, say $mx+n$. If we consider what gets sent to $x$ by $\varphi$, we see that $m$ must be a unit of $\mathbb{Z}$, so $m=\pm 1$. So I think the automorphisms fix $\mathbb{Z}$ and send the indeterminate variable $x \mapsto x+n$ or $x \mapsto -x+n$ for $n \in \mathbb{Z}$. Is this correct?
41643
I have a ring $\mathbf R =(R, +, -, ., 0, 1)$ (note that there is no invers for multiplication, $R$ is not $\mathbb R$, it is any set for the given algebra).  How do you prove that the following does not apply generally (meaning for all $a \in R$)?  $$\mathbf R[x]/(x^2−a)\mathbf R[x] \simeq \mathbf R^2$$  I know that the following does apply:  $\mathbf R[x]/(x−a)\mathbf R[x] \simeq \mathbf R$ , you can prove that using the first isomorphism theorem by finiding a homomorphism by substitution   $\Omega_a : \mathbf R[x] \to \mathbf R$ $\Omega_a(f(x)) = f(a)$  then $Ker(\Omega_a) = (x - a)\mathbf R[x]$ and $Im(\Omega_a) = \mathbf R$ therefore the two are isomorphic.  Please help, Thanks
117005
Are the groups $\mathbb{C}$ and $\mathbb{R}$ isomorphic under addition?  And how could I prove this ?  What about $\mathbb{Q}$ and $\mathbb{Q}[i]$ ?
81501
In Dummit and Foote, Exercise 15.18 (p. 727 3rd edition) one is asked to prove that for a commutative ring with unity $R$, and $f$ a non-nilpotent element of $R$, that the localization $R_f$ is isomorphic to $R[x]/(xf - 1)$.  I have done this, using the strategy outlined in the problem, using the universal property of $R_f$ proved in Theorem 36. The problem I have is this:  If $R_f \cong R[x]/(xf - 1)$ and $\phi:R[x] \to R_f$ is a surjective homomorphism, it stands to reason that $\text{ker}(\phi)$ should be $(xf - 1)$. I cannot show this directly. The surjective homomorphism I have in mind is, for:  $p(x) = a_0 + a_1x + \cdots + a_mx^m$  $\phi(p) = \dfrac{a_0}{1} + \dfrac{a_1}{f} + \cdots + \dfrac{a_m}{f^m}$  It is clear to me that $\phi$ annihilates $(xf - 1)$, so that ideal is contained in the kernel. What I cannot show is the other containment, or equivalently:  If $\phi(g) = 0_{R_f}$ then $g(x) \in (xf - 1)$. The trouble I am running into is that $R$ might be a "bad" ring, containing a lot of zero-divisors, $f$ might even be a zero-divisor (being non-nilpotent is a considerably weaker property), so the usual polynomial tricks of the division algorithm or using the irreducibility of $xf - 1$ aren't much use, here.  I feel I am missing something obvious....what is it?
119144
Does there exists an Isomorphism between $\mathbb{C}^* $ under multiplication with $\mathbb{C}$ under addition.  Attempt In my opinion It does not exists.  If there exists such an isomorphism say $\phi :\mathbb{C}^* \to \mathbb{C}$ then, from the fact that Isomorphism takes identity to identity we have $$\phi \left( 1 \right)=0$$  From the fact      Given $\phi:G \to H$ be an isomorphism, then For a fixed integer $k$ and a fixed group element $b$ in $G$, the   equation $x^k=b$ has the same number of solutions in $G$ as does   the equation $x^k=\phi\left(b\right)$ in $H$.   We have $x^4=1$ and $x^4=0$ must have same number of solutions, but the solutions to these equations are 4 and 1 only.
86616
I am trying to prove that $\mathbb Z[i]/ \langle 1+2i \rangle$ is isomorphic to  $\mathbb  Z_5$.   The only thing that came to my mind was trying to apply the first isomorphism theorem using an appropiate function. If I consider the euclidean function $N: Z[i] \setminus \{0 \}  \to  \mathbb N$ defined as $N(a+bi)=a^2+b^2$, then I can express any element $z$ in $\mathbb Z[i]$ as $z=(1+2i)q+r$ with $N(r)=1,2,3,4$ or $r=0$.   If I define $f:\mathbb Z[i] \to \mathbb  Z_5$ as $f(x=(1+2i)q_x+r_x)=\overline{r_x}$, then it is clear that $f(x)=0$ if and only if $x \in  \langle 1+2i \rangle$. The problem with this function is that it doesn't satisfy $f(x+y)=f(x)+f(y)$ and it is not surjective for if $r_x=f(x)=3 (5)$, then if $r_x=a+bi$, we have $a^2+b^2=3$, which is absurd.  I don't know what else to do, any suggestions would be appreciated.
43828
Let $I$ and $J$ be two ideals of a commutative ring $R$ with $1.$ Give a necessary and sufficient condition so that    $$R/IJ\cong R/I\times R/J.$$   Prove your claim. Then decide whether the following ring isomorphisms are true or not:   $$\mathbb{Q}[x]/\langle x^2-1\rangle\cong \mathbb{Q}[x]/\langle x-1\rangle\times \mathbb{Q}[x]/\langle x+1\rangle,$$   $$\mathbb{Z}[x]/\langle x^2-1\rangle\cong \mathbb{Z}[x]/\langle x-1\rangle\times \mathbb{Z}[x]/\langle x+1\rangle.$$   It has been a while since I took Abstract Algebra and I am preparing for the prelims. I am not sure how to tackle this one. Any help/suggestion/hint will be much obliged. This is what I have so far:     This is a simplified version of the chinese remainder theorem. The map   $R\to R/I\times R/J$ defined by $r\mapsto (r+I,r+J)$ is a ring homomorphism with kernel $I\cap J.$ If the ideals $I$ and $J$ are comaximal, then this map is surjective and $I\cap J=I\cdot J$, so    $$R/(I\cdot J)=R/(I\cap J)\cong R/I\times R/J.$$   My question is about the second part of the question, do I have to check whether: $\langle x-1\rangle$ and $\langle x+1\rangle$ are comaximal and $\mathbb{Q}[x]$ and/or $\mathbb{Z}[x]$ are ring homomorphisms? I believe that these ideals are comaximal in both cases but forget how to prove it.
51084
To show: $(\mathbb{Q},+)$ is not isomorphic to $(\mathbb{R},+)$  Now, the equation $x^{2} =3$ has a solution in $\mathbb{R}$, but not in $\mathbb{Q}$.  Hence they are not isomorphic to each other.  Is that right, or do I need to prove something else?  Thanks.
91403
I have been having a lot of trouble teaching myself rings, so much so that even "simple" proofs are really difficult for me. I think I am finally starting to get it, but just to be sure could some one please check this proof that $\mathbb Z[i]/\langle 1 - i \rangle$ is a field. Thank you.  Proof: Notice that $$\langle 1 - i \rangle\\  \Rightarrow 1 = i\\  \Rightarrow 2 = 0.$$ Thus all elements of the form $a+ bi + \langle 1 - i \rangle$ can be rewritten as $a+ b + \langle 1 - i \rangle$. But since $2=0$ this implies that the elements that are left can be written as $1 + \langle 1 - i \rangle$ or $0 + \langle 1 - i \rangle$. Thus  $$ \mathbb Z[i]/ \langle 1 - i \rangle = \{ 0+ \langle 1 - i \rangle , 1 + \langle 1 - i \rangle\}. $$  This is obviously a commutative ring with unity and no zero-divisors, thus it is a finite integral domain, and hence is a field. $\square$
15433
Is there a nice way to perform calculations within a quotient ring? for instance, an exercise I had was to show $\mathbb{Z}[X]/(X^2+1) \cong \mathbb{Z}[i]$, but this got me thinking; If I didn't know this, how could I calculate in $\mathbb{Z}[X]/(X^2+1)$ easily?  (and of course any other general quotient ring)?
70945
I'm supposed to find all the ring homomorphisms $\mathbb Z[x]\rightarrow \mathbb Q$.  Here's my attmept: $\mathbb Z[x]$ is generated by $\left\{ x^n \right\} _{n\geq 0}$, so the images of the generators determines the homomorphism. Since ring homomorphisms also respect multiplication, the images of $1,x$ determine everything. Homomorphisms respect units, so the image of $x$ determines our homomorphism. Hence, the only homomorphisms are the evaluations: for each $q\in \mathbb Q$ we have $\mathrm{eval}_q:\sum_na_nx^n\mapsto \sum _n a_n q^n$.  Is this correct? If so, isn't the codomain completely irrelevant? It seems we can say the same for any ring $R$ in place of $\mathbb Q$.
117076
How to show that $\mathbb Q(\sqrt 2)$ is not field isomorphic to $\mathbb Q(\sqrt 3)?$   My text provides the hint as: Any isomorphism from $\mathbb Q(\sqrt 2)\to\mathbb Q(\sqrt 3)$ is identity when restricted to $\mathbb Q.$  Ya! I can see that: $\mathbb Q(\sqrt 2)\simeq\mathbb Q[x]/\langle x^2-2\rangle$ and $\mathbb Q(\sqrt 3)\simeq\mathbb Q[x]/\langle x^2-3\rangle.$ Now if $\phi$ be such an isomorphism then $\phi:1+\langle x^2-2\rangle\mapsto1+\langle x^2-3\rangle.$ Hence the restriction of $\phi$ over $\{q+\langle x^2-2\rangle:q\in\mathbb Q\}$ remains the identity.  Now what?
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!
96440
I want to show that on a ring $R$ where $x^2 = x$ for all elements $x$, we must have that $2x = 0$ and $R$ is commutative. Does my answer look sound?  Well, if $x = 0$ we have that $x^2 = x$ is trivially true.  If $x \neq 0$ then $x^2 = x$  $\Longleftrightarrow x \cdot x = x$  $\Longleftrightarrow x \cdot x \cdot x^{-1} = x \cdot x^{-1}$  $\Longleftrightarrow x = 1$  I.e. the only non-zero element that we can possibly find in $R$ is $1$.  So $R = \{0, 1\}$ and as $1 + 0 = 1$ the only way we can satisfy the additive inverse ring axiom is if $1 + 1 = 0$.  I.e. $2(1) = 0 \Longleftrightarrow 2x = 0$  $R$ must be commutative because if we take two elements from $R$, e.g. take $a = 0$ and $b = 0$ such that $ab \neq ba$ we will have that $ab$ is either $0$ or $1$.  $ab = 1$ is not possible as $a = b = 0$.   If $ab = 0 \implies ba = 1$ and this is not possible as again $a = b = 0$.
80487
Show that $\mathbb{Z}[i]/(2+3i)$ is a finite field.    By applying some isomorphisms and realizing $\mathbb{Z}[i]$ as the quotient ring of $\mathbb{Z}[x]/(x^{2}+1)$, I found that:  $\mathbb{Z}[i]/(2+3i) \cong \mathbb{Z}[x]/(x^2+1,2+3x)$  However, I run into trouble as I can't divide $x^{2}+1$ by $2+3x$ in $\mathbb{Z}[x]$. What would be another way to approach the problem?  Thanks for the help.
87898
Are the following modules $M=\mathbb{Z^3}/\langle(2,1,0),(0,-1,0)\rangle$ and $N=\mathbb{Z^3}/\langle(1,0,0),(0,1,0)\rangle$ isomorphic? Are they free? Which rank do they have?  My attempt: I would say that they are isomorphic since when i reduce the matrix made of (2,1,0), (0,-1,0) in RREF i get (1,0,0), (0,1,0), so they would span the same space? But what are elements of these two modules? And which can be a isomorphism between them? And what about their rank and fact that they're free? They are free if they have a base, but i am not sure how can I write the base in this case. I am thankful for any sort of help or reference to literature or anything