I'm totally confused by this question. I appreciate any help or answer. thanks in advance why is the nontrivial left ideal of M_2(D) minimal? that D is a division ring
122471
please first read these then answer my question  Let $D$ be a division ring and r, n be two positive integer numbers such that $0 ≤ r ≤ n$. Then $H_r(D)$ denotes the left ideal of $M_n(D)$ containing all matrices whose $j$th column is zero, for every $r &lt; j ≤ n$.     Theorem A: Let $D$ be a division ring and $n$ be a positive integer number.   Then for every left ideal $I$ of $M_n(D)$, there exist an invertible matrix $P$ and integer $r$,   $0 ≤ r ≤ n$, such that $I = PH_r(D)P^{−1}$.   Now from theorem A how I can conclude that each left ideal of $M_2(D)$ is minimal? I'm totally confused with this
119458
Let  $R$  be  a  ring  such  that  the  only  right  ideals  of  $R$  are  $(0)$  and  $R$.  Prove  that  either  $R$  is  a  division  ring  or  that  $R$  is  a  ring  with  a  prime  number  of  elements  in  which  $ab=  0$  for  all  $a,b\in R$.    I  don't  want  the  proof.  I  am  stuck  at  one  point.  Why  does $R$  have  to  be  finite  here?
42647
I'm struggling with some concepts in my algebra class. I've chosen these two completely unrelated examples that demonstrate my struggling topics (characteristics and function rings) the two are the following:  $1$) Let $X$ be a set, and let $S$ be a ring. Give a proof that the left distributive law holds for $(S^X, +, *)$.  $2$) Suppose $Y^3 = Y$ for each element $Y$ of a unital ring $R$. Show that $R$ has a finite characteristic d that is a divisor of $6$.  Any example solutions for these proofs would be helpful. Thank you.
55777
Let $R$ be a commutative ring with finitely many maximal ideals $\mathfrak m_1,\ldots,\mathfrak m_n$. Let $M$ be a finitely generated module. Then there exists an element $x\in M$ such that $\frac{x}{1}\not\in\mathfrak m_iM_{{\mathfrak m}_i}$ for every $i=1,\dots,n$.    I cannot prove that such an element there exists. I was trying to prove it by induction on $n$. If $n=1$ it is true by Nakayama, so suppose it is true for $n-1$. Then for every $i$ I can find an $x_i\in M$ such that $\frac{x_i}{1}\not\in \mathfrak m_jM_{\mathfrak m_j}$ for every $j\neq i$. If $\frac{x_i}{1}\not\in \mathfrak m_iM_{\mathfrak m_i}$ for some $i$ we are done, so suppose $\frac{x_i}{1}\in\ m_iM_{m_i}$ for every $i$. I don't know how to go on, could you help me?
26946
Let $R$ be a commutative Noetherian ring with unit and let $I$ be a fixed ideal. I am sorry if the following turns out to be a very silly question.  1) Suppose $\operatorname{Ass}(R/I)$ are all minimal, then what can we say about $R/I$?  One conclusion is that $I$ has a unique primary decomposition. If $R/I$ is CM, then $\operatorname{Ass}(R/I)$ are all minimal, but I don't think the converse is true. Can someone provide a counterexample?  2) Suppose $\operatorname{Ass}(R/I)$ are all minimal and have the same height. I still dont think we can conclude that $R/I$ is CM. Can someone provide a counterexample? Theorem 2.1.6 of Bruns and Herzog, CM Rings, gives a result in this direction but it needs additional hypothesis. It says that if every ideal $I$ is generated by height $I$ elements and $\operatorname{Ass}(R/I)$ are all minimal, then $R$ is CM.  It would be nice if someone can say something nice about the ring in the special situation described in 1) and 2).  Thanks.
5528
This result appears to be ubiquitous as an algebra exercise. How do you prove this result?     Let $A$ be an integral domain with field of fractions $K$, and let $A_{\mathfrak{m}}$ denote the localisation of $A$ at a maximal ideal $\mathfrak{m}$ considered as a subring of $K$. Prove that $$A = \bigcap_{\mathfrak{m}} A_{\mathfrak{m}}\,,$$ where the intersection is taken over all maximal ideals $\mathfrak{m}$ of $A$.
73097
If $F$ is a field, then $F[x]$ is a principal ideal domain   By a previous theorem, we know that $F[x]$ is an integral domain. Now, let $I$ be an ideal in $F[x]$. If $I = \{0\}$, then $I = \langle 0 \rangle$. If $I \neq \{0\}$, then among all the elements of $I$, let $g(x)$ be one of minimum degree. We will show that $I = \langle g(x) \rangle$. Since $g(x) \in I$, we have $\langle g(x) \rangle \subseteq I$. Now let $f(x) \in I$. Then, by the division algorithm, we may write $f(x) = g(x)q(x) + r(x)$, where $r(x) =0$ or deg $r(x) \lt$ deg $g(x)$. Since $r(x) = f(x) - g(x)q(x) \in I$, the minimality of deg $g(x)$ implies that the latter condition cannot hold. So, $r(x) = 0$ and, therefore, $f(x) \in \langle g(x) \rangle$. This shows that $ I \subseteq g(x)$.  My question is: What is a polynomial of minimum degree in this context?  And how does this minimality imply that deg $r(x) \lt$ deg $g(x)$ cannot hold?
88343
I'm trying to find a ring which has an element that has many(more than two) distinct left inverses.  I first tried the (square) matrices but then realized that if a matrix is a left unit, then it has determinant greater than 0, thus it is also a right unit and has unique inverse.  And then next I tried to do it with functions with usual addition and composition, and I found this example(in the first answer, by user23211):   But I realized that the space of functions does not form a ring since the distribution law fails.  To make distribution law hold, we can only have homomorphisms; while the elements in the above example are not.  I also have seen that we can construct a similar examples with infinite vector spaces, but they are not rings.  Is there any ring like this?
63629
I have to study the prime elements of the ring $ \mathbb{Z} \! \left[ \dfrac{i \sqrt{3} - 1}{2} \right] $. For the moment, I cannot find the general form of such elements. Can you help me?  Thanks! :)
42366
In $\mathbb{Z}\left [ x \right ]$, let $I = \left \{ f\left ( x \right ) \in \mathbb{Z}\left [ x \right ]: f\left ( 0 \right ) \text{ is an even integer} \right \}$. Is $I$ a Prime ideal of $\mathbb{Z}\left [ x \right ]$?  Let $f_{1}\left ( x \right ),f_{2}\left ( x \right ) \in \mathbb{Z}\left [ x \right ].$ Let $f_{1}\left ( x \right )f_{2}\left ( x \right ) \in I$. Clearly, I is a proper ideal of $\mathbb{Z}\left [ x \right ]$ since the elements in $\mathbb{Z}\left [ x \right ]$ evaluated at 0 can produce an odd integer but this cannot be true in I. Now, $f_{1}\left ( 0 \right )f_{2}\left ( 0 \right )$ is an even integer in I. Hence, either $f_{1}\left ( 0 \right )$ has an even constant integer or $f_{2}\left ( 0 \right )$ has an even constant integer. Thus, I is a prime ideal by definition. Is this correct? I would now like to show if I is a maximal ideal. Again, clearly, I is a proper ideal by the above argument. Let B be an ideal of $\mathbb{Z}\left [ x \right ]$ and I $\subseteq B\subseteq \mathbb{Z}\left [ x \right ]$. How should B be defined so that I would conclude either $B=I$ or $B=\mathbb{Z}[x]$?
26230
Suppose that $k$ is an algebraically closed field. Then what do the prime ideals in the polynomial ring $k[x,y]$ look like?   As far as I know, the maximal ideals of $k[x,y]$ are of the form $(x-a,y-b)$ where $a,b\in k$. What can we say about the prime ideals? Are there similar results? And what about $k[x,y,z], k[x,y,z,w]$ and so on. Would someone be kind enough to give me some hints or referrence on this topic? Thank you very much!
43419
Let integer $n\geq 1$. I have obtained that for any field $k$, the matrix ring $M_n(k)$ is simple, i.e., $M_n(k)$ contains no nonzero proper two sided ideals. Now I want to prove that: for any ring $A$ (not necessarily commutative), the matrix ring $M_n(A)$ is simple if and only if $A$ is simple.   How to prove?
12858
Say I have a commutative ring $R$ with a maximal ideal $m$. Then $m/m^k$ is a maximal ideal in $R/m^k$ for any $k$. Is it the only maximal ideal, i.e. is $R/m^k$ a local ring?    This is a well known result for $k = 1$, as $R/m$ is a field. It seems to be true in other cases, e.g. $p\mathbb{Z}\subset \mathbb{Z}$ for prime $p$ and for $(x,y) \subset \mathbb{F}[x,y]$, the maximal ideal generated by $x$ and $y$ in a polynomial ring over the field $\mathbb{F}$.   Equivalently, if an element $x\notin m$, is $x + m^k$ an invertible element in $R/m^k$?
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!
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
58724
Let $R$ be a unital commutative ring, $P$ $\subseteq$ $R$ a prime ideal, $X\subseteq P$ a subset. Show there exists a minimal (inclusion minimal) prime ideal contained in $P$ which contains $X$.   My approach: Let $\langle X\rangle$ be the ideal generated by $X$. $P^c$ is a multiplicative set, as $P$ is prime. $\langle X\rangle\subseteq P\subseteq R$ and $P^c\cap\langle X\rangle=\emptyset $ so by Zorn's lemma there exist prime ideal $I_1$ satisfying $\langle X\rangle\subseteq I_1\subseteq P$. As $I_1$ is prime, it's complement is a multiplicative subset which does not intersect $\langle X\rangle$, hence I can find prime ideal $I_2$ satisfying $\langle X\rangle\subseteq I_2 \subseteq I_1$ and so on... I've got an infinite descending chain of prime ideals and I can't be sure that the process will stop.  Hints will be appreciated, thank you.   Edit: Can I conclude that every prime ideal contains a minimal prime ideal?
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.