The task is to give the description of all (with the precision of isomorphism) commutative rings with unity of $8$ elements.   As for addition definition, I understood that the are two options for the sum of the units: $1+1=0$ and $8 = 0$, for if $n = 0$ we will get a vector space over $\mathbb{F}_n$, and its amount of elements in should be the power of $n$, which is possible only for $2$ and $8$: $2^3$ and $8^1$. In both cases addition is completely defined.  In the case of $8 = 0$ the ring is isomorphic to $\mathbb{Z}/(8)$.   In the second case we can consider the ring as vector space over $\mathbb{F}_2$ of dimension $3$, with elements $\{1, 0, a, a+1,b,b+1,a+b,a+b+1\}$. Now the structure of the ring depends on the definition of multiplication in it. The only fact that I understood about it is that if we define $a^2=a$ and $b^2=b$ then for any element $r$ of the ring $r^2=r$, hence it is boolean ring, which in turn is isomorphic to $\mathbb{F}_2 \times \mathbb{F}_2 \times \mathbb{F}_2$.  Also I know that if we define $a^2=a+1$, then $\{1, 0, a, a+1\}$ is a field of $4$ elements, but how is it connected with the structure of the complete ring?  What are the other options for multiplication and what do they mean in terms of the ring structure?
121466
I have read that there are $10$ commutative rings of order $8$ (in this question, rings are unital and associative by definition), but I haven't found a concrete list of them.  But I think that they are the following:   $\mathbb{F}_2 \times \mathbb{Z}/4$ $\mathbb{F}_2 \times \mathbb{F}_2 \times \mathbb{F}_2$ $\mathbb{F}_2 \times \mathbb{F}_4$ $\mathbb{F}_2 \times \mathbb{F}_2[x]/(x^2)$ $\mathbb{Z}/8$ $\mathbb{F}_8$ $\mathbb{F}_2[x]/(x^3)$ $\mathbb{Z}/4[x]/(2x,x^2)$ $\mathbb{Z}/4[x]/(2x,x^2-2)$ $\mathbb{F}_2[x,y]/(x^2,xy,y^2)$   Can someone confirm this? Also, is there a direct proof that these are all?  Since rings of order $2$ and $4$ are easy to classify and every finite commutative ring is a direct product of local commutative rings, we may restrict our attention to local commutative rings of order $8$. These will be the last $6$ in the list.  The paper "Associative rings of order p^3" by Gilmer and Mott is not very specific about the proofs, and the paper "Finite associative rings" by Raghavendran is rather confusing for me. The papers also consider non-unital and non-commutative rings.  Edit. Here is a proof that these rings $R_1,\dotsc,R_{10}$ are pairwise not isomorphic. Probably it is not optimal, but for the moment it just works.   $R_1,\dotsc,R_4$ are not local, but $R_5,\dotsc,R_{10}$ are. $R_2,R_3$ are reduced, but $R_1,R_4$ are not. $R_1$ has characteristic $4$, but $R_4$ has characteristic $2$. $R_2$ has a trivial group of units, but $R_4$ does not. $R_5$ is the only ring of characteristic $8$. $R_6$ is the only field. $R_7,R_{10}$ have characteristic $2$, but $R_8,R_9$ have characteristic $4$. $R_{10}$ has a unique maximal ideal $\mathfrak{m}$ satisfying $\mathfrak{m}^2=0$, but the one of $R_7$ only satisfies $\mathfrak{m}^3=0$. $R_9$ contains a root of $2$, but $R_8$ does not, because $R_8/(x) = \mathbb{Z}/4$ does not.
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 $R$?
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?
83586
This is an exercise in Artin,     Describe the ring obtained from $\mathbb{Z}$ by adjoining an element $\alpha$ satisfying the two relations $2\alpha - 6 = 0$ and $\alpha - 10 = 0$.    I have seen examples in the text where one relation is imposed on a ring, but not two. I would very much like to see how this is done. Thanks!
24821
This is an exercise from Chapter 3 of Golan's linear algebra book.  Problem: Show $\mathbb{Z}$ is not a vector space over a field.  Solution attempt:  Suppose there is a such a field and proceed by contradiction. I will write multiplication $FV$, where $F$ is in the field and $V$ is an element of $\mathbb{Z}$.   First we rule out the case where the field has characteristic 2. We would have   $$0=(1_F+1_F)1=1_F1+1_F1=2$$  a contradiction.  Now, consider the case where the field does not have characteristic 2. Then there is an element $2^{-1}_F$ in the field, and   $1=2_F(2^{-1}_F1)=2^{-1}_F1+2^{-1}_F1$  Now $2^{-1}_F1\in\mathbb{Z}$ as it is an element of the vector space, but there is no element $a\in\mathbb{Z}$ with $2a=1$, so we have a contradiction.  Is this correct?
86854
We have the ring $\displaystyle R=\mathbb{Z}_3[x]/I, \ I=(x^2-1)\mathbb{Z}_3[x]$ is a principal ideal. The question is to find the elements $a\in R$ such that $a^n=0_R$ for some $n\in\mathbb{N}$.    I'm trying to understand the form of the elements in $R$.  Firstly, the elements of $R$ have the form $f(x)+I, \ f(x)\in\mathbb{Z}_3[x]$. By the euclidean division there exist $\pi(x)\in\mathbb{Z}_3[x]$ and $a,b\in\mathbb{Z}_3$ such that $f(x)=\pi(x)(x^2-1)+(ax+b)$. The elements of the form $ax+b$ where $a,b\in\mathbb{Z}_3$ are the following:  $$A=\{0,1,2,x,x+1,x+2,2x,2x+1,2x+2\}$$  (Is that a correct approach to examine the form of the elements in $R$?)  We have that $x^2-1=(x+1)(x-1)=(x+1)(x+2)$ in $\mathbb{Z}_3[x]$. If $f(x)^n+I=0+I$ then $f(x)^n\in I\Rightarrow (x+1)(x+2)\mid f(x)^n$. By looking at the polynomials in $A$, this is true if $f(x)=0$. But how can I check what happens to the $n-$th power of the rest of the polynomials?  Secondly, I 'm trying to find the invertible elements of $R$. Now we are looking for elements $f(x)\in A$ such that $\gcd(f(x),x^2-1)=1$. Are these $\{1,2,x,2x\}$? How can I find the inverse of $2x$?   Thank you in advance.
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
104181
(a) Determine the set of units $(\Bbb Z/n\Bbb Z)^\times$ for $n = 4, 9, 15, 16$.      (b) For which pairs of natural numbers m and n does there exist a ring homomorphism $\Bbb Z/n\Bbb Z \to \Bbb Z/m\Bbb Z$?    Where $\Bbb Z$ = set of integers.   My attempts:  a) I know that $a$ in $\Bbb Z$ is a unit if there exists $b$ in Z such that $a.b = b.a = 1$ where $b$ is the inverse of $a$.   So for $n=4$, $(\Bbb Z/4\Bbb Z)$ ={$0,1,2,3$} and $(\Bbb Z/4\Bbb Z)^\times$ = {$(1.1),(3.3)$} where "." represents multiplication.  And for $n=9$, $(\Bbb Z/9\Bbb Z)$ = {$0,1,2,3,4,5,6,7,8$} and $(\Bbb Z/9\Bbb Z)^\times$  = {$(1.1), (2.5), (4.7), (7.9), (8.8)$}. I'll do the same procedure for $n= 15$ and $n=16$.   But is the above correct?   And could you please show me how part b) is done? We didn't talk about it in the lecture
120863
Are the units of the quotient ring $\mathbb{F}_2[x]/\langle x^k+1 \rangle$ known in general, where $\mathbb{F}_2$ is the finite field with two elements? I'm specifically interested in the case where $k$ is divisible by two, such as say $k=8$, or $k=12$.  If so, can they be constructed easily? I'd be grateful if you could provide a few examples.
80535
Show that $4x^2+6x+3$ is a unit in $\mathbb{Z}_8[x]$.  Once you have found the inverse like , the verification is trivial. But how do you come up with such an inverse. Do I just try with general polynomials of all degrees and see what restrictions RHS = $1$ imposes on the coefficients until I get lucky? Also is there a general method to show an element in a ring is a unit?
71948
$R$ is finite commutative ring without zero divisors which has at least two elements. How to show that $R$ is field?   I'm just starting with abstract algebra and I'd really appreciate if someone could explain it to me.
79441
There is an easy argument which shows that a finite integral domain (commutative unital ring with no zero divisors) is a field. Here I wonder whether this result still stands if the term "unital" is dropped.   In other words, can a finite commutative ring with no zero divisors always contain a multiplicative identity? More generally, if this is true, can we even generalize Wedderburn's little theorem: every finite ring with no zero divisors is a field?
114093
Show that a division ring is simple.   With a division ring I mean a ring such that all onzero elements are invertible. And with a simple ring I mean a ring which has exactly two two-sided ideals.  I'm doing some exercises to prepare for a new course that start next week. I'm trying to get my ring theory neurons back in the shape they used to be 6 months ago
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?)
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 $a = b = 0$.
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!
26257
How do I exactly find the zero divisors and units of a ring in the form of $\mathbb{Z}_n \times \mathbb{Z}_m$? For instance, how does one calculate zero divisors and units of $\mathbb{Z}_6 \times \mathbb{Z}_2$? I understand that a zero divisor is a non-zero element of $R$ such that $rs = 0$ for some other non-zero element $s$ in $R$. But I don't really quite understand how to apply that to a ring of the form $\mathbb{Z}_n \times \mathbb{Z}_m$. Also, the units of $\mathbb{Z}_6 \times \mathbb{Z}_2$ would be elements that send some element $s$ in $\mathbb{Z}_n \times \mathbb{Z}_m$ to the multiplicative identity, right? (which would be 1)? How do I apply these concepts to "product" rings? Thank you.