Let $R$ be a commutative ring with $1$. Let $a\in R^\times$ and $b\in\text{Nil}(R)$. Show that $a+b\in R^\times$.   Attempts:  Suppose $b^n=0$.  I've tried to figure out a "simple" case as $n=2$;  I've tried to develop the expressions $(a+b)^2=a(a+2b)$ and $a^{-1}(a+b)=1+a^{-1}b$ but nothing came up.  Also $\exists x,y\in R$ s.t. $1=xa+yb$ but it doesn't help because we get $x=a^{-1},y=0$.  Another direction i've failed is to show $\langle 0\rangle=R/\langle a+b\rangle$.
5478
If $ua = au$, where $u$ is a unit and $a$ is a nilpotent, show that $u+a$ is a unit.   I've been working on this problem for an hour that I tried to construct an element $x \in R$ such that $x(u+a) = 1 = (u+a)x$. After tried several elements and manipulated $ua = au$, I still couldn't find any clue. Can anybody give me a hint?
79747
In a ring $R$ with identity 1, show that $$a(-1) = (-1)a = -a \qquad\forall\, a \in R$$ I have started with $a + (-a) = 0$ but cant proceed from here.
50644
Let $(A, +, \cdot)$ be a ring with $1$.  An element $a\in A$ is nilpotent if there exists $n\in \mathbb{N}$ so that $a^n=0$.      Show that if $a$ is nilpotent then $1+a$ and $1-a$ are invertible.
2133
So as seen on this question , we know that if $(n,m) \neq 1$, then $\mathbb{Z} /mn \mathbb{Z} \ncong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$, because the right hand side does not have an element of order $nm$.  But take a more general setting. Let $R$ be a commutative ring, and let $A,B$ be ideals in $R$. $A$ and $B$ are said to be comaximal if $A+B=R$. If $A, B$ are comaximal, then we have: $R/AB \cong R/A \times R/B$. In this setup, is the converse true?      If we have $A, B$ ideals in $R$ such that $R/AB \cong R/A \times R/B$, do we always have that $A$ and $B$ are comaximal?
118234
I am stuck in a question for Algebraic Structures:  I need to prove: If given any ring $R$, for any $a$ in $R$ $a\cdot a=0$ then $a+a=0$.  (of course $+$, $*$ and $0$ regarding $R$)  I didn't find any question regarding this question.
51212
Let $(A,+,\cdot)$ be a unity ring with the property that if $x \in A$ and $x^2=0$ then $x=0$. Consider the set $M=\{a\in A | a^3=a\}$. Prove that:      a) $2a\in Z(A)$, $\forall a\in M$, where $Z(A)$ denotes the centre of the ring $A$;   b) $ab=ba$, $\forall a,b\in M$.   My attempts revolved around the fact that an idempotent element in a reduced ring is central. So, since for $a\in M$ we have that $(a^2)^2=a^2$, it follows that $a^2\in Z(A)$, $\forall a\in M$. The next thing I wanted to use in order to solve a) was that $Z(A)$ is a subring of $A$, so if I had proved that $(a+1)^2 \in Z(A)$, $\forall a\in M$, then we would have reached the desired conclusion. However, I couldn't prove this and I honestly doubt that it is true. Another idea that I had was to prove that $M$ is a subring of $A$. Of course, this didn't work out because I cannot even prove that $M$ is closed under addition. Again, I don't know if this is true and it most likely isn't. As for b), I think that a) should be of use, but I don't know how. It is a well-known problem that a ring with $x^3=x$ for any $x$ in that ring is commutative, but since $(M,+,\cdot)$ is almost definitely not a ring, this doesn't help. EDIT: Is there any chance that this question is simply wrong? I tended to believe this before asking it here too, but since nobody has made any progress on it until now I am even more inclined to think so.
28637
Let $R$ be a commutative ring with 1, we define  $$N(R):=\{ a\in R \mid  \exists k\in \mathbb{N}:a^k=0\}$$  and  $$U(R):=\{ a\in R \mid a\mbox{ is invertible} \}.$$  Could anyone help me prove that if $a\in N(R) \Rightarrow 1+a\in U(R)$?  I've been trying to construst a $b$ such that $ab=1$ rather than doing it by contradiction, as I don't see how you could go about doing that.
73961
Let $R$ be a commutative ring, $a_1, ..., a_n$ its elements and  $\phi: R[x_1, ..., x_n] \to R$  defined by  $ \phi(f(x_1, ..., x_n)) = f(a_1, ... ,a_n)$  a ring homomorphism.  Prove: $\ker \phi = (x_1-a_1, ..., x_n-a_n)$  It is obvious that $(x_1 - a_1, ..., x_n -a_n) \subseteq \ker \phi$. I'm not sure how to prove the converse.  At this point I don't know any division algorithms for multivariable polynomials, only for the ones in $R[x]$(and the book from where I taken the exercise doesn't assume the reader to know something beyond basics of rings and ideals, and the division algorithm for $R[x]$). Though I know this could be solved for $i = 1$ by dividing by $x-a$:   Let $f(x) \in \ker \phi$, divide by $x-a: f(x) = q(x)(x-a) + r, f(a) = r = 0$, so $f(x) = q(x)(x-a) \in (x-a)$.
32213
Let $R$ be a commutative ring with identity, $P$ be a prime ideal in $R$ and define $$X := \lbrace t \in P \mid t^2=t \rbrace. $$ Also let $J$ denote the smallest ideal of $R$ that contains $X$.  Prove that if $$a^2+J=a+J$$ in $R/J$, then $$a+J=J$$ or $$a+J=1+J$$  I am trying to solve this but I am new to algebra (aprox. 3 months), so I generally don't know how to start such questions. So any help is appreciated.
32240
How can we prove that (the set of all polynomials in infinitely non commutative indeterminants over $k$) is a ring, where $k$ is a ring with unity?   Adding two polynomials is a polynomial, but for commutativity  How can we show the following: $ax_1+bx_2=bx_2+ax_1$ (commutativity of addition) If we have finite number of variable it is easy to prove it, (we take the $n-1$ variables...), but what about the infinite number of variables?  Can someone give me a hint?
35475
$\DeclareMathOperator{\ass}{Ass}$     Let $R$ be a Noetherian ring and $S$ be a multiplicative set in $R$. Then show that   $\ass_R(S^{-1}M)=\ass_R(M)\cap \{P\in \text{Spec}(R):P\cap S=\emptyset\}$   Attempt: Let $P\in \ass_R(S^{-1}M)$. We argue that $S\cap P=\emptyset$. Say $P=\text{Ann}(m/s)$ for some $m/s\neq 0$ in $S^{-1}M$. If $r\in P\cap S$, then from $r\cdot (m/s)=0$ we get $(r/1)\cdot (m/s)=0$. But $r/1$ is invertible in $S^{-1}R$ and thus we have $m/s=0$, a contradiction. So we must have $S\cap P=\emptyset$.  The main difficulty I am having is in showing that $P\in \ass_R(M)$. I suspect that $P=\text{Ann}(m)$. Clearly, if $r\cdot m=0$, then $r\cdot (m/1)=0$ and therefore $r\in P$. So $\text{Ann}(m)\subseteq P$. On the other hand, if $r\in P$, then $(r\cdot m)/s=0$, giving an element $u\in S$ such that $(ur)\cdot m=0$. But I am getting nowhere with this.
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.
23270
A ring $R$ is a Boolean ring provided that $a^2=a$ for every $a \in R$. How can we show that every Boolean ring is commutative?
6346
Suppose $R$ is a ring with no zero divisors and with identity $1_R$ not equal to $0_R$.  Suppose that $a,b$ are in $R$ and that $ab$ is a unit.  Prove that $b$ is a unit.   My thoughts: I know a unit is basically a unit that (for this example) would mean $abu = 1_R$ for some nonzero $u$ in $R$.  I am really stuck after that.  Not seeing a clear path to manipulate the variables to prove b is a unit by itself.
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 $ba_0=ba_1=\cdots=ba_n=0$?
42711
Let $R$ be a commutative ring with $1 \neq 0$.  It is known that $R^{n} \hookrightarrow R^{m}$ implies $n \leq m$ and $R^{n} \twoheadrightarrow R^{m}$ implies $n \geq m$, and I might use this without mentioning (though I don't think I did). Isomorphisms are $R$-module maps if not mentioned otherwise.  Let $m \geq n$ and consider $R^{n} = R^{n} \times 0 \times \cdots \times 0 \subseteq R^{m}$. We have $\text{Hom}_{R}(R^{n}, R^{m}) \simeq M_{m \times n}(R)$ via $\phi \mapsto M(\phi) = (c_{ij})$ where $\phi(e_{j}) = \sum_{i=1}^{m}c_{ij}e_{i}$. Take any $R$-linearly independent $y_{1}, \cdots, y_{n} \subseteq R^{m}$ and define an $m \times n$ matrix $A = (a_{ij})$ via $y_{j} = b_{1j}e_{1} + \cdots + b_{mj}e_{m}$.  If we understand $e_{j}$ with $1 \leq j \leq n$ as $n \times 1$ column matrix in $R^{n}$ when we multiply an $m \times n$ matrix to it on its left, then $Ae_{j} = y_{j}$ for $1 \leq j \leq n$. Thus, we can understand this matrix as an injection $A : R^{n} \hookrightarrow R^{m}$.  By definition $A$ is an $m \times n$ matrix, but if what I wrote is correct, then $A|_{R^{n}}$ must be an isomorphism $R^{n} \simeq \bigoplus_{j=1}^{n} Ry_{j}$. I am not trying to find an invertible matrix to $A$, but is it okay to think of it as an embedding like how I described?
13698
I am given that $R$ is a commutative ring, $A$ is an ideal of $R$, and $N(A)=\{x\in R\,|\,x^n\in A$ for some $n\}$.  I am studying with a group for our comprehensive exam and this problem has us stuck for two reasons.  FIRST - We decided to assume $n\in\mathbb{Z}^+$ even though this restriction was not given. We decided $n\ne 0$ because then $x^0=1$ and we are not guaranteed unity. We also decided $n\notin\mathbb{Z}^-$ because $x^{-1}$ has no meaning if there are no multiplicative inverses. Is this a valid argument?  SECOND - We want to assume $x,\,y\in N(A)$ which means $x^m,\,y^n\in A$ and use the binomial theorem to expand $(x-y)^n$ which we have already proved is valid in a commutative ring and show that each term is in A so $x-y$ is in $N(A)$. The biggest problem is how to approach the $-y$ if we are not guaranteed unity. Does anyone have any suggestions?  Thank you in advance for any insight.