How to describe an element of the Ring $\Bbb Z[x]/(x^2-3,2x+4)$?   Or is it isomorphic to any well known ring?    What is meant by quotienting $\Bbb Z[x]$ by a kernel like that?
54767
I need to describe a ring $R=\mathbb{Z}[X]/(X^2-3,2X+4)$    I know that its element would be of the form $\{f(x)+(X^2-3,2X+4)|f(x)\in \mathbb{Z}[X]\}$ After that will i divide $f(x)$ by $X^2-3$ first and then by $2X+4$ to reduce the elements  in $R$?
84383
Problem 4.   Let $\Bbb F_2$ be the field with 2 elements and let $R=\Bbb F_2[X]$. List, up to isomorphism, all $R$-modules with 8 elements.      Solution.   We use the classification theorem of modules over a PID. Since $R$ is a finite module, it is in particular an $\Bbb F_2$ vector space. We can write   $$M\cong R/n_1R\oplus R/n_2 R\oplus\cdots\oplus R/n_r R$$   for polynmials $n_1\mid n_2\mid\cdots\mid n_r$. In our case, we have $\sum_{i=1}^r\deg n_i=3$, so we have three options: $r=1,\deg n_1=3$, $r=2,\deg n_2=2$, and $r=3,\deg n_3=1$. The first case yields 8 options. For the second case, we need $n_2$ to be reducible, so we have $X(X+1),X^2,(X+1)^2$ as choices. The first choice yields 2 decompositions, and the latter choices yield 1 decomposition each, for a total of 4. For the linear case, we need the same linear term repeated thrice, which is 2 choices. Therefore there are 14 in all, listed by invariant factors below:   $$\begin{align} &amp;\{X^3+(0/1)X^2+(0/1)X+(0/1)\}\\ &amp;\{X^2,X\},\{X^2+X,X+(0/1)\},\{X^2+1,X+1\}\\ &amp;\{X,X,X\},\{X+1,X+1,X+1\}. \end{align}$$   My question is the problem above. I can understand the solution, but how can I see that the solutions given are from distinct isomorphism classes? For instance, is the below true? $$R/(X)\oplus R/(X)\oplus R/(X)\cong R/(X+1)\oplus R/(X+1)\oplus R/(X+1)\\ \cong F_2\oplus F_2\oplus F_2$$  Thanks for any help.
41740
That's it. I want to do this quotient,  $\mathbb Z [\sqrt{-11}] / (1+\sqrt{-11})$. My first idea was to see which elements are in the ideal:  $(a+b\sqrt{-11})(1+\sqrt{-11}) = a+ a\sqrt{-11} + b\sqrt{-11} -11b$  So I want to make and application who send this elements to $0$, and then see the Kernel of the application and apply the Isomorphism Theorem.  Any ideas?
13573
I'm currently considering the ideal $(2,X+1)\subset\Bbb{Z}[X]$. I'm trying to figure out if this is a prime ideal, a maximal ideal, or neither. So far I have attempted to look at the quotient ring $\Bbb{Z}[X]/(2,X+1)$. I know that if this quotient ring is a field, then our ideal is maximal, since $\Bbb{Z}[X]$ is commutative. However I'm stuck on how I should be viewing this quotient ring.
19439
I used this isomorphism today but now I'm having trouble justifying it. The norm function isn't additive so I can't come up with a ring isomorphism to prove the following:     For any $\,a+bi\in\Bbb Z[i],\,\gcd(a,b)=1$, we have a ring isomorphism   $$\Bbb Z[i]/\langle a+bi\rangle\,\cong\Bbb Z/(a^2+b^2)\Bbb Z=:\Bbb Z_{a^2+b^2}.$$   Could someone show me an isomorphism between these rings to prove this?
92095
I am confused about this quotient ring.  I know that $\mathbb{Z}_7 [x]/\langle x^2 + 1\rangle =  \{ f(x) + \langle x^2 + 1\rangle \mid f(x) \in \mathbb{Z}_7[x] \}$. Here $x^2 + 1$ is a zero element in $\mathbb{Z}_7 [x]/\langle x^2 + 1\rangle$, so if we replace $x$ by $i$ we get $\mathbb{Z}_7 [x]/\langle x^2 + 1\rangle =\mathbb{Z}_7 (i) $.  I know the distinct element of $ \mathbb{Z}_6 [x]/\langle 2x +4\rangle$ are $\langle 2x +4\rangle,\;x + \langle 2x +4\rangle, \;x^2 + \langle 2x +4\rangle, \ldots$ but I do not  know how to prove this.  Please tell what could we do similar here in case of $ \mathbb{Z}_6 [x]/\langle 2x +4\rangle$? Since $1$ is the zero element of the polynomial $2x +4 $. Why we can not replace here $x$ by $1$?  Any help would be appreciated. Thank you.
46430
Possible Duplicate:          An element $a$ of a ring $R$ is nilpotent if $a^n = 0$ for some positive integer $n$. Let $R$ be a commutative ring, and let $N$ be the set of all nilpotent elements of $R$.  (a) I'm trying to show that $N$ is an ideal, and that the only nilpotent element of $R/N$ is the zero element.  (c) What are the nilpotent elements of $R = \mathbb{Z}_{24}$? And what is the quotient ring $R/N$ in that case? (what known ring is it isomorphic to?)
34098
An exercise from Dummit &amp; Foote:      Determine the units of the ring $A = \mathbb{Z}[X]/(X^{3})$ and the structure of the unit group $A^{\times}$.   Help would be great.  Thanks!
99257
How many ring homomorphisms there is between $\mathbb{Z}[x,y]/(x^3+y^2-1)$ and $\mathbb{Z_7}$? Here $\mathbb{Z_7}$ denote ring of integers mod 7.  I don't know how to approach this problem,so far I've only worked with one variable ring so I can't tell the properties of $\mathbb{Z}[x,y]/(x^3+y^2-1)$. Thanks in advance
95400
I am confused by expressions like $\mathbb{R}/\mathbb{Z}$, especially if I see something of the sort $SL(2,\mathbb{R})/SO(2,\mathbb{R})$. I don't know what to make of it. I understand what a quotient of a group for an equivalence looks like, but I can't get my head around expressions like above.   Is a quotient like this always a group and what do elements in the set $SL(2,\mathbb{R})/SO(2,\mathbb{R})$ look like?
120588
In general, I struggle working with quotients of polynomial rings, so I was hoping someone might be able to help with the following exercise:  Show that $\mathbb{Q}[x]/(x^2-2x)$ is isomorphic to $\mathbb{Q}[x]/(x^2-1)$, but the latter is not isomorphic to $\mathbb{Q}[x]/(x^2)$.  Mostly I want to understand what these quotients all &quot;look like.&quot; We did few examples in class and few homework problems with such quotients, and the ones we do work with are much simpler (typically involving $\mathbb{Z}_2$ or $\mathbb{Z}_3$ or $\mathbb{R}$ instead of $\mathbb{Q}$). Any help is appreciated, thank you!
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 $\mathscr B$.
88261
Can anyone show me how to find invertible elements in $\mathbb{Z}/m\mathbb{Z}$, $m$ is say $28$?  Also I'm not very clear about what it mean by 'invertible element in $\mathbb{Z}/m\mathbb{Z}$'
107078
I need to prove that the ring $\mathbb Z[\sqrt{-2}]= \{a+b\sqrt{-2} ; a\in \mathbb Z,b\in \mathbb Z \}$ has a Euclidean algorithm, and to decide whether there are infinitely many primes in this ring.   How do I approach this kind of question?  Thank you.
25001
Let $R$ be a ring of order $n$ and suppose $n$ has no square in its prime decomposition. How do I see that $R$ is isomorphic to $\Bbb Z/n\Bbb Z$?   I bet that the map $\Bbb Z  \to R, \, 1\mapsto 1_R$ descends to an iso $\Bbb Z/n\Bbb Z \to R$ but I don't see how $n$ having no squares implies the desired descent.
25570
I want to prove that the kernel of the evaluation map $s_a : \mathbb{C}[x_1,\dots,x_n] \rightarrow \mathbb{C}, x_i \mapsto a_i$ where $a = (a_1,\dots, a_n) \in \mathbb{C}^n$ is the ideal generated by $\{x_1 - a_1, \dots, x_n - a_n\}$.    The proof in the book first shows it easily for the case $a = 0$, and says that by substitutions of variables $x_i' = x_i - a_i$, you can show it for the rest of the cases.  I know that $f \in \ker s_a$ iff $f(a) = 0$.  And $\phi : \mathbb{C}[x_1,\dots, x_n] \rightarrow \mathbb{C}[x_1 - a_1, \dots, x_n-a_n], x_i \mapsto x_i - a_i$ is a surjective ring homomorphism.  Please give a hint and not the full answer.  Thanks.
3951
A very basic ring theory question, which I am not able to solve. How does one show that    $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$. Extending the result: $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$, if $a,b$ are relatively prime.   My attempt was to define a map, $\varphi:\mathbb{Z}[i] \to \mathbb{Z}/10\mathbb{Z}$ and show that the kernel is the ideal generated by $\langle{3-i\rangle}$. But I couldn't think of such a map. Anyhow, any ideas would be helpful.