Let $A$ be a local unital algebra, that is, it contains a unique maximal left ideal $J$ ($J \neq A$). I want to show that $J$ is also a right ideal. For that, I want to use the fact that an element $y \in A$ is left invertible if and only if $y \not \in J$. So, let $x \in J$ and $a \in A$, and we want to show that $xa \in J$. If this is not the case, then $xa$ is left invertible: there exists $b \in A$ such that $b(xa) = 1$. How do I conclude?
120944
I have been trying to prove the statement in the title, however I seem to get stuck at a certain point. Let $M$ be the maximal left-ideal of $R$. Then consider $M.r$ for $r \in R$. If $M.r \neq R$, we have that $M.r \subseteq M$. Now if this assumption would be true, we would have that $M$ is a right ideal as well. However when trying to prove that $M.r \neq R$ for all $r \in R$ I don't get very far. I know that if we assume equality, one would get $m.r=1$ for one $m \in M$. However I don't know how to go from here. Thank you in advance!
60832
It well known that if $R$ is a commutative ring then every ideal of $M_n(R)$ is $M_n (I)$ where $I$ is an ideal of $R$. My question is: Suppose that $I$ and $J$ are ideals of $M_n(R)$, i.e., $I=$M_n(A)$ and $J=$M_n(B)$. When is the statement $M_n(AB)=M_n(A)M_n(B)$ correct?
10576
A homework question from my algebra class asks:     Show that in a local ring $R$ with maximal ideal $M$, every element outside $M$ is a unit.   My argument is that since $M$ is maximal $R /M $ is a field and so for any $ x \in R \backslash M $, $ x + M $ has a multiplicative inverse, which implies $ x $ is a unit.  I don't see where we need the fact that $R$ is a local ring.
40621
Let $K$ be a field and let $R$ be a $K$-algebra with unity which is finite dimensional as a $K$-vector space. Prove that $R$ has only finitely many prime ideals all of which are maximal. (Hint: Chinese remainder theorem)   My attempt: We have $R \simeq K^n$, for some $n$, as vector spaces. I claim that there are finitely many ideals of $K^n$. Let $I$ be an ideal of $K^n$ and $(a_1, a_2,...,a_n) \in I$. For every entry $a_i$, we have two choices, either $a_i=0$ or not.  So we get $2^n$ many cases, and any case give a particular ideal of $K^n$ which are distinct from others.  Is my proof correct or am I missing some point? If it is true, how can we give a proof by using the Chinese remainder theorem? Thanks!
52534
Suppose $u$ is an element of a ring with a right inverse. I'm trying to understand why the following are equivalent.   $u$ has at least two right inverses $u$ is a left zero divisor $u$ is not a unit   If $v$ and $w$ are distinct right inverse of $u$, then $u(v-w)=0$, but $v-w\neq 0$, so $u$ is a left zero divisor. It's also clear that if $u$ is a left zero divisor, it cannot be a unit (else I could cancel $u$ from $ub=0$ to see $b=0$).  I'm having a heck of a time seeing why $u$ is not a unit implies $u$ has at least two right inverses. I tried the contrapositive, but saw no good approach. What am I missing?
19973
Let $R$ be a commutative ring and $I_1, \dots, I_n$ pairwise comaximal ideals in $R$, i.e., $I_i + I_j = R$ for $i \neq j$. Why are the ideals $I_1^{n_1}, ... , I_r^{n_r}$ (for any $n_1,...,n_r \in\mathbb N$) also comaximal?
8996
Let $A$ be a commutative ring with identity and $B$ a finitely generated $A$-algebra that is integral over $A$.  If $\mathfrak{p}$ is a prime ideal of $A$, there are only finitely many prime ideals $P$ of $B$ such that $P\cap A=\mathfrak{p}$.   Let me say that I am aware of  answer, but I can't follow through the hint.  Also, I don't know how to extend to work for algebras rather than extension rings.
13787
$R$ and $S$ are two rings. Let $J$ be an ideal in $R\times S$. Then there are $I_{1}$, ideal of $R$, and $I_{2}$, ideal of $S$ such that $J=I_{1}\times I_{2}$.  For me it's obvious why $\left\{ r\in R\mid \left(r,s\right)\in J\text{ for some } s\in S\right\}$ is an ideal of $R$ (and the same for $S$) so I can prove $J$ is a subset of the product of two ideals and also that the product of these two ideals is an ideal. But I can't see how to prove equality without assuming existence of unity or commutativity. Trying basically to show that if $\left(r,s\right)\in J$ then also $\left(r,0\right),\left(0,s\right)\in J$. Thank you for your help!
9858
Well, the problem I'm trying to solve is this:      Let $A$ be the ring of all continuous functions from $\mathbb{R} \to \mathbb{R}$. Show that $M = \{f \in A: f(0)=0\}$ is a maximal ideal of $A$.   I tried to show that if $J$ is an ideal of A that properly contains M, so $J = A$: we know that there is a function $g \in J$ with $g(0) \neq 0$. So, we have that $$J \supseteq \langle g, M\rangle=\{ag+f;a \in A, f\in M\}$$   I tried to show that 1 $\in$ $\langle g,M\rangle$. However I'd have to find some function $h(x)\in A$ such that $h(x)(ag+f) = 1$, but this function $h$ has to be different from 0 for all x $\in \mathbb{R}$, so I don't know how to find this function.  My second idea to solve this problem was prove that the quocient $A/M$ is a field. We have that $$A/M = \{h(x) + M; h(x) \in A\},$$ but again I have the problem to show that all elements of $A/M$ are units.  If someone could help, I would be really grateful (:
64815
Let $F$ be a field and let $(K,*)$ be an associative $F$-algebra which, as a vector space, is finitely generated over $F$. Given an element $a\in K$, do there necessarily exist elements $a_1,a_2 \in K$ satisfying $a_1*a_2=a$?  I'm not sure how to go about this. Since we have an associative $F$-algebra we know that $v*(w*y)=(v*w)*y$ for all $v,w,y \in K$.
27785
So I basically have to prove what is on the title.      Given $R$ a commutative rng (a ring that might not contain a $1$), with the property that $I+J=R$, (where $I$ and $J$ are ideals) we have to prove that $IJ=I\cap J$.   One inclusion is easy. If $x\in IJ$, then $x=\sum a_ib_i$ where $a_i\in I$ and $b_i\in J$. Thus for any fixed $i$, we have that since $a_i\in I$, we have that $a_ib_i\in I$, and the same argument shows that $a_ib_i\in J$, thus $\sum a_ib_i\in I$ and $\sum a_ib_i\in J$, this means that $x=\sum a_ib_i\in I\cap J$, and thus $IJ\subset I\cap J$.   I am having troubles proving the other inclusion. Any comments?  Thanks
65827
In the midst of attempting to prove that for a commutative ring $A$ with identity, and an ideal $I$ of $A$, $I = \operatorname{rad}(I)$, where $\operatorname{rad}(I) = \{x: x^m \in I, m &gt;0\}$, implies that $I$ is an intersection of prime ideals, I've run into a little bit of confusion.   An answer to another question on this site, , indicates a strategy that I am attempting to use. I know that if $x$ is an element of $A\setminus I$, then so too is $x^m$ for any positive integer $m$. I use this fact to consider a subset $X = \{x^m : m &gt; 0\}$, which is closed under multiplication, and use Zorn's lemma to find an ideal of $A$ containing $I$ maximal with respect to the constraint that said ideal, $J$, is disjoint from X.  I want to prove $J$ is prime by contradiction. I suppose that $J$ were not prime; this also means that $J$ is not maximal in $A$. Then there would exist two elements, $f$ and $g$, of $A$ so that $fg$ is in $J$, but neither $f$ nor $g$ is in $J$. Following the prompting of the aforementioned answer, I know that $J + \langle a\rangle$ and $J + \langle b\rangle$ would be ideals strictly containing $J$, and since J is the maximal ideal disjoint from $X$, both $J +  \langle a\rangle$ and $J + \langle b\rangle$ contain at least some elements of $X$ (I am tempted to say all of them, but I'm not entirely sure that's true, in case for example $\langle a\rangle$ started containing powers of $x$ at $x^2$, in which case it isn't necessarily true that $\langle a\rangle$ contains $x$ itself), and so both $\langle a\rangle$ and $\langle b\rangle$ must themselves contain elements of $X$.  And this is where I am stuck. I feel like there is something I should know about multiplicative subsets of rings that would help me arrive at the desired contradiction, but I can't remember what it might be.
115283
Let $R$ be an associative ring with $1$. Prove that $R$ is a division ring if and only if $R$ has no proper left ideals.    Clearly, if $R$ is a division ring and $I\neq\{0\}$ is a left ideal, then every nonzero $z\in I$ satisfies $1=z^{-1}z\in I$, hence $I=R$.  Now if $z\in R$ is nonzero and $R$ has no proper left ideals, then $I:=\{rz:r\in R\}$ is equal to $R$ and there exists $r\in R$ such that $rz=1$. However, we still need to prove $zr=1$. How can we do this?  Thank you.
46016
Ok, I'm reading some thesis of some former students, and come up with this proof, but it doesn't really look good to me. So I guess it should be wrong somewhere. So, here it goes:     Let $R$ be a unitary commutative ring, and $I$ be an ideal, and $N$ is an $R$-module. I'll now show that $$I \otimes_R N \cong IN\,.$$   First, since $1 \in R$, and $I$ is an ideal of $R$, we must have that $I = RI$. So then $$I \otimes_R N = RI \otimes_R N = R \otimes_R IN \cong IN\,.$$  But this must be so wrong, since if this proof is true, I could prove that any $R$-module is flat. So how come this proof is wrong?
97142
Please give me a hint for this problem.  Let $R$ be a ring, and $I$ a right ideal of $R$. Show if $M$ is a left $R$-module, then $$ f  \   \colon (R/I) \otimes_{R} M \to  M/IM$$  defined by $ f((r+I) \otimes m) = rm + IM$ is an isomorphism of commutative groups.   And show that if $L$ is a left ideal, then  $$ (R/I) \otimes _{R} (R/L) \cong R/(I+L)$$ as commutative groups.
99873
Let $A$ be a ring (commutative unital), and $\mathcal I$ be a nonempty family of proper ideals of $A$.  I will say that $\mathcal I$ has property $\dagger$ if for any $\mathfrak a\in\mathcal I$ and any $xy\in \mathfrak a$, one of $\mathfrak a+(x),\mathfrak a+(y)$ is in $\mathcal I$.  In particular, any maximal (w.r.t. inclusion) element of $\mathcal I$ is prime.  Does $\dagger$ (or maybe $\dagger$ + hypothesis of Zorn's lemma) have a name (a prime family, perhaps, as I suggest in the title)? As a side question, are there some interesting criteria for checking that a given family of ideals has property $\dagger$?  It seems to me that posets with property $\dagger$ are rather abundant in commutative algebra (e.g. in proof of Cohen's characterization of Noetherian rings), but I've yet to see $\dagger$ discussed on its own, or any name for it.
20649
I want to show: If $R$ is local and $I\neq R$ an ideal, then $R/I$ is also local.  We already know: A Ring $R$ is local if and only if $R-R^{\times} = \{r\in R \, | r \notin R^\times \}$ is an ideal. We also know that if $I$ is an maximal Ideal, $R/I$ is a field.  I have no idea how to show that other than using the above Lemma.