Let $R=\mathbb{Z}[i]/\langle2-2i\rangle$.  I want to know the number of elements in $R$.  The function $N$ given by $N(a+bi)= a^2+b^2$ is the norm, so $N(a+bi) &lt; N(2-2i)$.  This leaves us with representatives $$ \{0,1,-1,i,-i,1+i,1-i,-1+i,-1-i,2,-2,2i,-2i\} \, . $$ So in this particular factor ring, $2 = 2i$. Then squaring both sides we see $1 = i$, so $R = \{I,1+I,-1-I, 2+I,-2+I\}$, where $I=\langle2-2i\rangle$.  So I find that the number of elements of $R$ is five. Is this solution correct?
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!
92340
Given the ring $ \mathbb{Z}/n\mathbb{Z} $ is always true that $ \mathbb{Z}/n\mathbb{Z}=[\text{zero divisors}]\cup[\text{units}] $      How can evaluate the zero divisors and units ?   I believe that $ a x=0 \pmod n $ for zero divisor  and $ ax=1 \pmod n $ for units    I know how to solve a congruence but what is $a$ ?? thanks.  What are the generators of the group $ \mathbb{Z}/n\mathbb{Z} $ under the addition '+' and product '$\times$'?
118281
[Update: Based on the hints provided by @zcn and @whacka, I believe I have found a solution. See my answer below.]  Below, $R$ is a commutative ring with $1$.  In John J. Watkins' Topics in Commutative Ring Theory, the author observes that if $a$ and $b$ are associates in $R$, then they generate the same ideal: $(a) = (b)$. This is clearly true, because $a = ub$ and $b = va$ for some units $u$ and $v$, and so $a \in (b)$ and $b \in (a)$, whence $(a) \subset (b)$ and $(b) \subset (a)$. Indeed, we didn't even require $u$ and $v$ to be units to reach this conclusion.  The author then claims that if $R$ is not a domain, then the converse may not be true: we may have $(a) = (b)$ even if $a$ and $b$ are not associates. As a "counterexample" he gives $R = \mathbb{Z}/(6)$, with $a = 2 + (6)$ and $b = 4 + (6)$. Clearly these elements generate the same ideal: $(a) = (b) = \{0, 2 + (6), 4 + (6)\}$. The author asserts that $a$ and $b$ are not associates, but this is false: $5 + (6)$ is a unit, and $(2 + (6))(5 + (6)) = 4 + (6)$, so $a$ and $b$ actually are associates.  I figured that the author botched the example but that surely the basic fact must be true: if $R = \mathbb{Z}/(n)$ then there may be nonassociates $a,b \in R$ with $(a) = (b)$. However, I was not able to find a counterexample and eventually resorted to writing a computer program and found that there are no counterexamples in $\mathbb{Z}/(n)$ for $n \leq 100$.  So now I'm starting to think that there is no counterexample in $\mathbb{Z}/(n)$ at all, but I have not yet been able to prove this.  My questions:   If I am correct that $(a) = (b)$ in $\mathbb{Z}/(n)$ implies that $a$ and $b$ are associates, I would appreciate a nudge in the right direction toward a proof. (But if the proof is elementary, please give only a hint, not a solution :-) I'll post my solution once I have one. In either case, could you please point me to a genuine counterexample? If there is one in $\mathbb{Z}/(n)$ for some $n$, that would be ideal (no pun intended), otherwise a counterexample in any commutative ring with $1$ would be great.
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!
68822
Question: If $R$ is a finite boolean ring, then show $R \cong \mathbb Z_2 \times \mathbb Z_2 \times \cdots\times \mathbb Z_2$.   I know that   $\mathrm{char}(R) =2$  $R$ has $2^k$ elements for some $k$  (this is because, if $p\neq 2$ is a prime with $p\mid|R|$, then there exists a non-zero $x \in R$ s.t. $px=0$ and $p $ is odd $\implies x=0$, contradiction).   Does this suffice to answer my question?
17288
Does there exist a square root of $-1$ in the ring $\mathbb{R}[x]/\langle(x^2+1)^2\rangle$?  Now, any element in the ring $\mathbb{R}[x]/\langle(x^2+1)^2\rangle$ is of the form $(a+ib)+e(c+id)$ where $i^2=-1$ and $e^2=0$. Now, setting $a=c=d=0$ and $b=1$, I think we would get a square of $-1$. Is this right? Any hints? Thanks beforehand.
28000
Somebody can to help me in such exercices:     (1) A ring R such that $a^2 = a$ for all $a\in R$ is called a Boolean ring. Prove that every    Boolean ring R is commutative and $a + a = 0$ for all $a \in R$.       (2) Let R be a ring with more than one element such that for each nonzero $a\in R$ there is a unique $b \in R$ such that $aba = a$. Prove:            (a) R has no zero divisors.           (b) $bab = b$.           (c) R has an identity.           (d) R is a division ring.       My greatest difficulty in the question 2 is to prove that R has no zero divisors.   Thanks!
37266
I'm considering the ring $\mathbb{Z}[\sqrt{-n}]$, where $n\ge 3$ and square free. I want to see why it's not a UFD.   I defined a norm for the ring by $|a+b\sqrt{-n}|=a^2+nb^2$. Using this I was able to show that $2$, $\sqrt{-n}$ and $1+\sqrt{-n}$ are all irreducible. Is there someway to conclude that $\mathbb{Z}[\sqrt{-n}]$ is not a UFD based on this? Thanks.
31358
Prove that $N(\gamma) = 1$ if, and only if, $\gamma$ is a unit in the ring $\mathbb{Z}[\sqrt{n}]$  Where $N$ is the norm function that maps $\gamma = a+b\sqrt{n} \mapsto \left | a^2-nb^2 \right |$  I have managed to prove $N(\gamma) = 1 \Rightarrow \gamma$ is a unit (i think), but cannot prove $\gamma$ is a unit $\Rightarrow N( \gamma ) = 1$  Any help would be appreciated, cheers
72407
I need to prove that the center of $A_n$ is trivial for $n \geq 4$.  $Z(A_3) = A_3$, since $A_3 = \mathbb{Z}/3\mathbb{Z}$ is commutative.  One idea is two use "counting" technique. First of, all we count cojugacy classes of even permutations in $S_n$. It's the classes correspodning to the types $[\lambda_1, ..., \lambda_r]$ where $n \equiv r \mod 2$  Now, the conjugacy class in $S_n$ correspoding to the type $[\lambda_1, ..., \lambda_r]$ splits into two equal-sized classes in $A_n$ if $\lambda_1, ..., \lambda_r$ are distinct odd numbers. If they aren't then it is preserved in $A_n$.  Then we can use "the counting formula"(  ) of elements in each "even" class of $S_n$ and divide one by $2$ if needed.  Still, I'm not sure if it can be done. Seems to be a lot of work there. Maybe there are other, he easier ways? Or, maybe, it is the way, in this case, I wouild appreciate any advices.
70385
The reasoning my book uses is as follows:   Since $(3+i)(3-i) = 10$, $10 + \langle 3+i\rangle = 0 + \langle 3+i\rangle $. Also, $i + \langle 3+i\rangle = -3 + \langle 3+i\rangle = 7 + \langle 3+i\rangle$. So, $\mathbb{Z}[i]/\langle 3+i\rangle = \{k + \langle 3+i\rangle \mid k = 0,1,\ldots ,9\}$, since $1 + \langle3+i\rangle$ has additive order $10$.   I do not understand why the additive order of $1 + \langle 3+i\rangle$ is $10$, or why the fact that $(3+i)(3-i) = 10$, shows the additive order or $1 + \langle 3+i\rangle$  is $10$. If someone could please explain this I would greatly appreciate it.   Note: I apologize for not using MathJax to write this post. I intend to learn it over the coming weekend.
44639
Find the cardinality of the quotient ring, $$\mathbb Z_5[i]/\langle 1+i\rangle$$  My Attempt:  Since, $$\mathbb Z_5[i]\cong Z[x]/\langle x^2+1,5\rangle$$  Hence, $$\mathbb Z_5[x]/\langle 1+i\rangle\cong \mathbb Z[x]/\langle x^2+1,x+1,5\rangle$$  In the quotient ring we have the following,  $x^2+1=0,x+1=0$ and $5=0$, i.e., $x=-1\;\; \implies 2=(-1)^2+1=0$  And, $5=0 \implies 1=0$. Thus, $\mathbb Z_5[i]/\langle 1+i\rangle\;\;=\{0+\langle 1+i\rangle\}$.   Is the above reasoning correct?  Initially, I had come across the following argument,$$1=(3+2i)(1+i)\implies 1\in \langle 1+i\rangle$$  And hence the quotient ring has just one element namely the zero element. This argument is quite good but it is not always easy to make such guesses. So I wanted to use the traditional method to find the cardinality.  Moreover, is there any different approach to the problem?
7354
I am having trouble with another homework problem (Hungerford, Algebra, Problem 12, Chapter III, Section 2).  Let $R$ be a ring without identity and with no zero divisors. Let $S$ be the ring whose additive group is $R \times \mathbb{Z}$ with multiplication given by $$(r_1,n_1)(r_2,n_2)=(r_1r_2+n_2r_1+n_1r_2,n_1n_2).$$ Let $A=\{(r,n) \in S \mid rx+nx=0 \text{ for all } x \in R\}$. (a) $A$ is an ideal in $S$. (b) $S/A$ has an identity and contains a subring isomorphic to $R$. (c) $S/A$ has no zero divisors.  I've done (a) and (b), but haven't been able to crack (c). Here's what I have so far: (c) Suppose $((r, n) + A)((s, m) + A) = (0, 0) + A$. Then $(r, n)(s, m) \in A$, so $(rs + mr + ns)x + (nm)x = 0$ for all $x \in R$. But $(rs + mr + ns)x + (nm)x = rsx + mrx + nsx + nmx = r(sx + mx) + n(sx + mx)$. Now here we have an expression involving $r(something) + n(something)$ and $s(something) + m(something)$, and I want to say that if it's zero for all $x$ then $rx + nx = 0$ for all $x$ or $sx + mx = 0$ for all $x$, but all I seem to be able to say is that $rx + nx = 0$ for all $x = sx' + mx'$, which is not (or not obviously) what I want. I appreciate any help!
72795
I have the following problem:  Let $R:=\mathbb{R}[X,Y]/(Y-X^2,Y+X)$ i) Show that $R\cong_\text{Ring} \mathbb{R}\times \mathbb{R}$ ii) Conclude that $|\operatorname{Spec}R|=2$. How do these two ideals look like?  So for the first one I got the hint to eliminate one variable at a time and see what happens. Doing that I got: $1.) \mathbb{R}[X]/(-X^2,X)\cong \mathbb{R}[X]/(-X^2)\times\mathbb{R}[X]/(X)\cong\mathbb{R}[X](-X^2)\times \mathbb{R}$ $2.) \mathbb{R}[Y]/(Y,Y)=\mathbb{R}[Y]/(Y)\cong\mathbb{R}$ using the chinese remainder theorem in $1.)$. But since $\mathbb{R}[X]/(-X^2,X)\times\mathbb{R}[Y]/(Y)\ncong\mathbb{R}[X,Y]/(Y-X^2,Y+X)$ and the fact that there is still $\mathbb{R}[X]/(-X^2)$ left in the first equation, I don't see how this leads me to the desired isomorphism. For the second one I don't see how an isomorphism can lead me to the number of ideals, because I'm pretty sure $A\cong B$ doesn't imply $\operatorname{Spec}A \cong \operatorname{Spec}B$ for rings $A,B$ (I've seen it in a book). Can someone help me with this problem? Thanks in advance.
104085
I'm trying to find the ring of integers of $\mathbb{Q}[\sqrt{17}]$, and it comes down to determining the set $\{(a,b)\in\mathbb{Q}^2\mid 2a\in \mathbb{Z}, a^2-17b^2\in\mathbb{Z}\}$. How can I determine this set?
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.
109765
For some of you, this question is going to seem extremely basic.  I think I understand what an ideal is. Such as $\langle 2, 1 + \sqrt{-5} \rangle$, it consists of all numbers in this ring of the form $2a + (1 + \sqrt{-5})b$.  But then what is $\langle 2, 1 + \sqrt{-5} \rangle^2$? My first thought was $\langle 4, -4 + 2 \sqrt{-5} \rangle$, but that seems wrong somehow. I also had something in the back of my mind saying $\langle 2 \rangle$, but I'm not sure about that one either.  Then I thought about trying to figure out $\langle 2, 1 + \sqrt{-5} \rangle \langle 2, 1 + \sqrt{-5} \rangle$ when I realized I don't actually understand how to multiply ideals to begin with.  I'm only using $\langle 2, 1 + \sqrt{-5} \rangle$ as an example (though that does draw in one question identified as similar that looks much more relevant than all the questions identified as "Questions that may already have your answer"). In a principal ideal domain, would it be correct to think that $\langle a \rangle \langle b \rangle = \langle ab \rangle$?  Any help would be much appreciated.