If a nonzero element $a$ of a ring $\mathbb{D}$ does not have a multiplicative inverse, then $a$ must be a zero divisor.   If $\mathbb{D}$ has finitely many elements, then the statement is true. However, I don't see why, as for the infinite case, 7 isn't a zero divisor only because every element $a &gt; 0$ when multiplied by a $b &gt; 0$ gives you $ab &gt; 0$ which has nothing to do with infinity or finite amounts of elements.
9155
Let $R$ be a finite ring with unity. Prove that every nonzero element of $R$ is either a unit or a zero-divisor.
61499
Let $R$ be a commutative ring with identity, and $S$ a multiplicative subset of $R$ not containing $0$. Which step in my reasoning is wrong?   Every element of $S$ is a unit in $S^{-1}R$. Every element of $S$ that's a zero divisor in $R$ is a zero divisor in $S^{-1}R$. But zero divisors can't be units. Therefore it's not possible to localize $R$ at $S$.
16851
I'm asked to show that there are infinitely many units in the ring $\mathbb Z[\sqrt 3]$.   But I don't really see a good approach to this one, so far.   Some thoughts: The inverse of $a+\sqrt3 b$ should be given by $\pm(2 - \sqrt 3 b)$, since the norm   $$N: \mathbb Z[\sqrt{3}] \to \mathbb Z, \qquad N(x+\sqrt{3} y) = x^2 - 3y^2$$  is multiplicative. So if $(a+\sqrt{3}b)^{-1}=x+\sqrt 3 y$, then   $$1 = N(1) = N((a+\sqrt{3}b)(x + \sqrt{3}y)) = (a^2 - 3b^2)(x^2 - 3y^2)$$  Hence we must have $\pm 1 = (a^2 - 3b^2) = (a+\sqrt{3}b)(a-\sqrt{3}b)$.  Therefore I need to show that there are infinitely many $a,b$ such that $a^2 - 3b^2 = \pm 1$.   Here I don't know how to proceed.   Maybe rewriting as $a = \sqrt{3b^2 \pm 1}$, and now trying to prove that there are infinitely many $b$ for which $3b^2 \pm 1$ is a square?  A hint would be appreciated! =) Thanks.
95915
I'm self studying abstract algebra. I see that in rings there's no requirement for a multiplicative inverse.  Is there something similar except with no requirement for an additive inverse. For example, all the non-negative rational numbers. Every number other than 0 has a multiplicative inverse, but no additive inverses. We have both the multiplicative and additive identities. Multiplication is still associative over addition.  Is there a name for such an algebraic structure, and has it been studied the way rings have?
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.
65578
Prove that in general the set of the right zero divisors is different from the set of left zero divisors in ring $R$.   The ring is arbitrary. Does exist general proof for it ?  Attention that the problem is comparing two sets. Thank you.
115409
I understand that for a number $a$ to have a multiplicative inverse in mod $n$, it must be coprime to $n$; therefore, all numbers that do not have a multiplicative inverse mod $n$ must share a factor with $n$.  Now, my prediction was that if $n$ is even, then the sum of all numbers that do not have a multiplicative inverse mod $n$ is $\frac{n}{2}\bmod{n}$, and if $n$ is odd, then the sum is $0\bmod{n}$.  Originally my proof was constructed something like this: if $n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, then the sum of all the elements of this array:  $$\begin{align} \begin{matrix} p_1 &amp; p_2 &amp; \cdots &amp; p_k \\ 2p_1 &amp; 2p_2 &amp; \cdots &amp; 2p_k \\ \vdots &amp; \vdots &amp; \cdots &amp; \vdots \\ n-2p_1 &amp; n-2p_2 &amp; \cdots &amp; n-2p_k \\ n-p_1 &amp; n-p_2 &amp; \cdots &amp; n-p_k \end{matrix} \end{align}$$   would cancel out in the correct way. However, while trying to prove this, I hit a roadblock  because I wasn't sure how to account for the multiplicity of the factors of $n$.  If I understand  correctly, this is the same as finding the sum of all numbers that are not units in the ring of integers modulo $n$, $\mathbb{Z}/n\mathbb{Z}$. Is there any way to do this? If the result is so nice, I feel like there is also a nice way to prove this.
117791
Let the ring of fractions be denoted as $S^{-1}R$.       Proposition (from Robert Ash's textbook "Basic Abstract Algebra")      Define $f: R \rightarrow S^{-1}R$ by $f(a) = a/1$. Then f is a ring homomorphism. If S has no zero divisors then f is a monomorphism, and we say that R can be embedded in $S^{-1}R$. In particular:      (i) A commutative ring R cna be embedded in its complete (or *full) ring of fractions ($S^{-1}R$, where S consists of all non-divisors of zero in R).       (ii) An integral domain can be embedded in its quotient field.    Why does the proposition say "If S has no zero divisors..."? If S really had zero divisors, then this wouldn't even be properly defined, right? Because we would have a/0 for some a in R.
113834
Does there exist a totally ordered ring with zero divisors? I can't think of an example right now.
68941
I think that this statement is false, but only in the trivial case. $\{ 0 \}$ is obviously a subgroup of $(\mathbb{Z},+)$ with infinite index.  In the case of a nontrivial subgroup of $(\mathbb{Z},+)$, intuition tells me that these subgroups will have finite index. How should I begin to prove this?  What I know:  Nontrivial subgroups of $\mathbb{Z}$ are of the for $n\mathbb{Z}$ for some $n\in\mathbb{N}$. These subgroups are infinite cyclic groups.  I read in another  posted here that I should try use the smallest positive element of $n\mathbb{Z}$ and Euclidean division to explicity describe all of $\mathbb{Z}$, but I'm not sure what that means, precisely.
48195
It is known that   Now consider a ring $R$ with exactly three ideals. Take an element $x$ such that $(x)\neq (1)$. For $y\in R$ and $y\notin (x)$, there are two possible cases:   $(y)=(1)$ $(y)\neq(1)$. Then $(y)=(x)$, that is $x=ys=xts$ for some $s,t\in R$. It follows that $$x(1-ts)=0$$ which implies that $x$ is a zero divisor.   However, if there is no such a $y$ of the second case, how to prove $x$ is a zero divisor?
53896
I need help proving  this stunning result.  Let me state it this way.  Let  $\mathbb L$ be a field and $\mathbb K$ be a field with $\mathbb K \subset \mathbb L$  Let $A$ be the set of algebraic numbers ie $A=\{a \in \mathbb L, \exists P \in \mathbb K [X], P(a)=0 \}$  I managed to prove that $A$ is a ring with the following reasoning:  If $a,b \in A$ then $ \mathbb K[a]$ is a ring.  And $ \mathbb K \subset \mathbb K[a] \subset \mathbb K[a][b] $  Some finite-dimension argument(ask details if needed) proves that $\mathbb K[a][b]$ has finite dimension as a vector space over the field $\mathbb K$  Reminding that $\mathbb K[a+b] \subset \mathbb K[a][b]$ and $\mathbb K[ab] \subset \mathbb K[a][b]$ proves that they both have finite dimension, hence $a$ and $b$ algebraic.  How should I proceed with inverses ?
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 $a = b = 0$.
99245
Everyone knows the rules of zero divisors like $$\forall \alpha,\beta\in\mathbb{R}\;:\;\alpha\cdot\beta = 0\Rightarrow\alpha=0\vee \beta=0.$$ But how can I prove it for $\mathbb{Z}$? My first try was this one: For $\alpha\cdot \beta=0$ and $\alpha\neq 0$ let $$0=\alpha^{-1}\cdot 0=\alpha^{-1}\cdot (\alpha\cdot\beta)=(\alpha^{-1}\cdot \alpha)\cdot\beta = \beta = 0\Rightarrow \beta = 0;$$ and the same for $\beta\neq 0\Rightarrow \alpha=0$, however i realized that the multiplicative inverse of a number $\alpha\in\mathbb{Z}$ is not defined in $\mathbb{Z}$ (because $(\mathbb{Z},\cdot)$ is not a multiplicative group). What now?  Furthermore the information: it's about basic multiplication and I should prove this via the basic "rules" neutrality of 0 and 1, comparability of 0 and 1, commutativity, associativity, distributivity, irreflexivity or transitivity. Group-theory should not be mentioned in the solution as out instructors don't want us to use these "advanced techniques"!  The rules (to use some from) are the following: $\forall a,b,c\in\mathbb{Z}:$   $a+0=a,\;\;\;\; a\cdot 1=a$ $0&lt;1$ $1+(-1)=0,\;\;\;\; 0-a=(-1)\cdot a$ $a+b=a+c \Rightarrow b=c$ $a\cdot b=a\cdot c,a\neq 0\Rightarrow b=c$ $0 &lt; a \Rightarrow a\neq 0$ $a&lt;b\wedge b&lt;c\Rightarrow a&lt;c$ $a&lt;b\rightarrow a+c&lt;b+c$ $a&lt;b\wedge 0&lt;c\Rightarrow a\cdot c&lt;b\cdot c$ $a&lt;b\wedge c &lt; 0\Rightarrow b\cdot c &lt; a\cdot c$
77699
Let $A$ be a commutative ring. I am trying to show that if $f(x_1,x_2,\ldots, x_r) \in A[x_1,x_2,\ldots, x_r]$ is a zero divisor then there exists $a$ in $A-\{0\}$ such that $af=0$ in $A[x_1,x_2,\ldots, x_r]$.    What I have tried so far as following. I am using induction on $r$ (not fixing the ring). The base case is for $r=1$ which I did assuming the minimal degree of $g$ for which $fg=0$ holds. So my induction assumption is for $r \geq 2$ whenever $f$ is in $A[x_1,x_2,\ldots x_r]$, $n&lt;r$ is a zero divisor there is $a$ in $A-\{0\}$ such that $af=0$. So for the final step I take $f(x_1,x_2,\ldots, x_r) \in A[x_1,x_2,\ldots, x_r]$ is a zero divisor. Now since $A[x_1,x_2,\ldots, x_r]=A[x_1][x_2,x_3,\ldots, x_r]$ by induction hypothesis there is $g$ in $A[x_1]-\{0\}$ such that $fg=0$.      I cannot proceed further. Am I correct so far? Please help me. Thank you.
120710
The problem I am attempting to prove is the following: In any finite group $G$, the number of elements not equal to their own inverse is an even number.  Caveat: I have had very limited experience with proofs. Any comments would be appreciated.  Let $a_{1}, a_{2}, ..., a_{n}$ be elements of $G$. Since $G$ is a group, every element in $G$ must have an inverse in $G$. Let the inverse of each element be represented by $a_{1}^{-1}, a_{2}^{-1}, ..., a_{n}^{-1}$. An element $a_{k}$ in $G$ that is not equal to its own inverse can be written as $a_{k} \neq a_{k}^{-1}$. For every element $a_{k}$, there is another element $a_{k}^{-1}$ that is not equal to its own inverse denoted by $a_{k}^{-1} \neq a_{k}$. So for every element $a_{k}$ in $G$ whose inverse is not equal to their own inverse, there will be 2 elements in $G$ namely $a_{k}$ and $a_{k}^{-1}$. Let the integer $q$ denote the number of elements ranging from $a_{1}, a_{2}, ..., a_{n}$ who are not equal to their own inverse. In addition for every value $q$, there must be another element in $a_{1}^{-1}, a_{2}^{-1}, ..., a_{n}^{-1}$ not equal to their own inverse. So the total number of elements not equal to their own inverse can be denoted by 2$q$, which is an even number.
22707
Let $R$ be commutative ring with no (nonzero) nilpotents. If $f(x) = a_0+a_1x+\cdots+a_nx^n$ is a zero divisor in $R[x]$, how do I show there's an element $b \ne 0$ in $R$ such that $ba_0=ba_1=\cdots=ba_n=0$?