I am trying to understand inverse limits and so I decided to try and compute one of such limits. We are going to work in the category of rings. Let $R$ be a ring. First, I set up an inverse system. Let $A_i:=R[x_1,\dots,x_i]$ denote a ring of polynomials over $R$ in $i$ indeterminates and let $f_{ij}:A_j \rightarrow A_i$   be morphism sending $x_j,\dots,x_{i+1}$ to $0$ and $x_1,\dots,x_i$ to themselves. Then $((A_i)_{i \in \mathbb{N}},(f_{ij})_{i \leq j \in \mathbb{N}}$) is my inverse system. Then, according to , the inverse limit $A$ of our inverse system is   \begin{equation*}A=\underleftarrow{\text{lim}}A_i=\Bigg\{\overrightarrow{a}\in\prod_\limits{i \in \mathbb{N}}A_i|a_i=f_{ij}(a_j) \; \forall \; i\leq j \in \mathbb{N} \Bigg\}. \end{equation*}   So if we look at some elements of $A$ (take $R=\mathbb{Z})$ we have, for example:  \begin{equation*} (\dots,3x_{3}^4-5x_3x_2x_{1}^2+5,5,5) \; \text{or} \; (\dots,x_{3}+x_2+x_1+4,x_2+x_1+4,x_1+4) \end{equation*}.   My suspicion is that $A \cong R[x_1,x_2,\dots]$ and so the inverse limit would be just that with nutaral projections. Is that correct? Is this the right way of looking at these sort of constructions?
122295
Define an inverse system of polynomial rings over a commutative ring $k$ by the canonical projection $k[x_1,...,x_n] \to k[x_1,...,x_m]\;(m&lt; n)$.   Question: What is the projective limit $\varprojlim_n k[x_1,...,x_n]$ ?
66341
I was reviewing my homework and it seems I overlooked something crucial while proving some ring has no Invariant Basis Number property. This is exercise VI.1.12 in Aluffi's Algebra: Chapter 0  The setup: $V$ is a $k$-vector space and let $R = \mathrm{End}_{k}(V)$.   Prove that $\mathrm{End}_{k}(V\oplus V) \cong R^4$ as an $R$-module Prove that $R$ doesn't satisfy the IBN property if $V = k^{\oplus \mathbb N}$.   For the first, I used to the fact that $V \oplus V$ is both the product and coproduct (in $k$-Vect) of $V$ with itself to get the isomorphism. What I just realized is I only showed that the two are isomorphic as groups not $R$-modules. So what would be the $R$-module structure on $\mathrm{End}_{k}(V \oplus V)$?   For the second, I used the fact that $V = k^{\oplus \mathbb N}$ implies $V \cong V \oplus V$ which in turn implies $R = \mathrm{End}_{k}(V) \cong \mathrm{End}_{k}(V \oplus V)$. Again, I just realized that I only showed the latter two are isomorphic as groups.   It may be obvious (and maybe why my professor let it pass?) but I can't come up with a good $R$-module structure that makes the two group isomorphisms $R$-linear.  Edit:  Explicitly, these are the isomorphisms I'm dealing with. Let $\pi_j, i_j$ be the natural projection/inclusion maps of the $j$-th factor resp. and $\psi: k^{\oplus \mathbb N} \oplus k^{\oplus \mathbb N} \to k^{\oplus \mathbb N}$ the isomorphism given by $\psi(e_i, 0)=e_{2i-1}$ and $\psi(0, e_i)=e_{2i}$.  Then the first isomorphism $\mathrm{End}_k(V \oplus V)\to R^4$ is given by $\varphi \mapsto (\pi_1\varphi i_1,\pi_2\varphi i_1,\pi_1\varphi i_2,\pi_2\varphi i_2)$  The second isomorphism $R \to \mathrm{End}_k(V \oplus V)$ is given by $\alpha \mapsto \psi^{-1} \alpha \psi$  The composition doesn't seem to be $R$-linear if I use the obvious $R$-module structure on $R$ and $R^4$.
58821
Let $T_1$ and $T_2$ be two Linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^n$.  Let $\{x_1,x_2,\cdots,x_n\}$ be a basis of $\mathbb{R}^n$. Suppose that $T_1(x_i)\neq 0$ for every $1\leq i\leq n$ and that $x_i\perp Ker (T_2)$ for every $1\leq i\leq n$.   Which of the following is true?   $T_1$ is invertible $T_2$ is invertible Both $T_1$ and $T_2$ are invertible Neither $T_1$ nor $T_2$ is invertible.   As $T_1(x_i)\neq 0$ for each $1\leq i\leq n$ we do not have $T_1(a_1x_1+a_2x_2+\dots+a_nx_n)=0 $ unless each $a_i=0$ i.e.,$a_1x_1+a_2x_2+\dots+a_nx_n=0$ i.e., $T_1$ is one one thus invertible.  I am not sure if $T_2$ is invertible or not.  we have $x_i\perp Ker (T_2)$ for all $x_i$. Would that be a good idea say something like  $\langle x_i :1\leq i\leq n\rangle \perp Ker(T_2)$ and as they span whole space we would have   $\mathbb{R}^n\perp Ker(T_2)$ and thus $Ker(T_2)=0$ so it is injective so it is invertible...  Thus both $T_1$ and $T_2$ are invertible?  I would be so thankful if someone can assure what it has been done here is sufficient/clear.  THank you
19185
Let $R$ be a commutative ring (with identity).   Let $f,g\in R\left[x\right]$ both be monic polynomials of degree $d$.   Then the underlying abelian groups of the rings $R\left[x\right]/\left(f\left(x\right)\right)$ and $R\left(x\right)/\left(g\left(x\right)\right)$ are isomorphic. Actually both are isomorphic to the direct sum $R^{\oplus d}$.  Questions:     What sort of relation between $f$ and $g$ will ensure that $R\left[x\right]/\left(f\left(x\right)\right)$   and $R\left(x\right)/\left(g\left(x\right)\right)$ are also isomorphic   as rings (with identities)?      What sort of relation between $f$ and $g$ will ensure that $R\left[x\right]/\left(f\left(x\right)\right)$   and $R\left(x\right)/\left(g\left(x\right)\right)$ are also isomorphic as   $R$-algebras?   Maybe my questions are too broad. Partial answers (e.g. with extra conditions on $R$) or references are also very welcome.
112879
If $R$ is a ring and $I$ an ideal of this ring, is that true that $(R/I)[x]=R[x]/I$?   Note that can $I$ be identified as a subset of $R[x]$ and we need that in order to $R[x]/I$ makes sense.  I think it's true because every polynomial $(a_0+I)+(a_1+I)x,\ldots,(a_n+I)x^n$ of $(R/I)[x]$ can be written as a polynomial of $R[x]/I$ and vice versa since we have this equality: $$(a_0+I)+(a_1+I)x,\ldots,(a_n+I)x^n=(a_0+a_1x+\ldots a_nx^n) +I$$  Is my reasoning right?  Thanks
59127
Given some category ${\mathcal C}$, then the opposite category will consist of the same objects with the morphisms "turned around."  Given $f:A\rightarrow B$, for $A,B$ objects of ${\mathcal C}$, then in general do we have a canonical $f^{op}:B\rightarrow A$ which is induced by $f$?    Slight Motivation:  This is something that's just been bothering me recently, and I feel like I am thinking about this concept incorrectly.  To make this question slightly less vague, if we had, say, the category of rings and we had the mapping $f:{\mathbb Z}\rightarrow {\mathbb Z}/2{\mathbb Z}$ by $f$ sends even numbers to 0 and odd numbers to 1, then what is the corresponding map in the opposite category, $f^{op}:{\mathbb Z}/2{\mathbb Z}\rightarrow {\mathbb Z}$ induced by $f$?  Also, I always see this $f^{op}$ that I've mentioned just written as $f$; is this shorthand, or is there something deeper there?
59340
If $f(x) = 8x^3+3x$ then, $$\lim_{x \to \infty} \frac{f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}$$ is?  My attempt: It is clear that the function cannot be easily inverted. So, there must be something in the limit given itself that may simplify the problem. Honestly, I have no clue what to do here. There are a few things which I could see is that the function has only $1$ root(i.e $0$) and is bijective on $x \in \mathbb R$. But that gave no benefit except showing that the inverse of the function exists. Any help would be appreciated.
15525
Let $a,b,c,x$ be elements of a unital non-commutative ring. Assume $c$ is an inverse of $1-ab$: $$ c(1-ab) = 1$$  How can I find an inverse for $1-ba$? What I tried:   Denote the unknown by $x$. Then $x (1-ba) = 1 = x - xba$. I tried to replace $1$ with $c(1-ab)$ but it didn't help because I cannot solve for $x$. I also tried to subtract $x (1-ba) $ from $ c(1-ab)$ but couldn't solve for $x$ either. Any suggestions?
15237
I've just learned what the integral closure is.     I would like to find what is the intergral closure of $\mathbb{Z}$ in $\mathbb{Q}[i]$.   Let $\mathcal{R}$ the integral closure of $\mathbb{Z}$ in $\mathbb{Q}[i]$. To determine $\mathcal{R}$ I started to note that $$\mathbb{Z}[i]\subset \mathcal{R}.$$ Indeed if $\alpha = x+iy\in \mathbb{Z}[i]$, $\alpha$ is a root of the monic polynomial $$f(X)=X^2-2xX+x^2+y^2\in\mathbb{Z}[X].$$  And we know that since $\mathbb{Z}$ and $\mathbb{Q}[i]$ are rings so is $\mathcal{R}$.  From there I'm not sure, is it true that there exist no ring $A$ such that $$\mathbb{Z}[i] \subsetneq A \subsetneq \mathbb{Q}[i] \text{ ?}$$ I've assumed that and show that $1/2\notin \mathcal{R}$ for example, otherwise it exists a monic polynomial $f\in \mathbb{Z}[X]$ such that  $$f\left(\frac{1}{2}\right) =\frac{1}{2^n}+a_{n-1}\frac{1}{2^{n-1}}+\cdots+a_1\frac{1}{2}+a_0 =0$$  So  $$2^n f\left(\frac{1}{2}\right) =0 \iff -1=2\underbrace{(a_{n-1}+\cdots +2^{n-2}a_1+2^{n-1}a_0)}_{\in \mathbb{Z}}$$  which is impossible.  In conclusion $\mathbb{Q}[i]\not\subset\mathcal{R}$ so $\mathcal{R}=\mathbb{Z}[i]$.     Is my proof correct ? Is there an other (easier) way to prove that ? And the most important, is there exist a general method to find the integral closure please ?   Thank you for your help.
30261
Find the multiplicative inverse of $1+ 3\sqrt{2}$ in the ring $\mathbb{Q}(\sqrt{2})$ and use it to solve the equation $(1+3\sqrt{2})x=1-5\sqrt{2}$.  I think that the inverse is the conjugate, so it would be $1-3\sqrt{2}$, but then I don't know where to use in the equation that needs to be solved.
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!
34331
Say $R$ is a commutative ring and $I\in R$ is an ideal. Let us consider the quotient $R/I$. It is created by taking every element $a\in R$, and adding all the elements of $I$ to it. The elements of $R/I$ are of the form $a+I$ and $b+I$, $\forall a,b\in R$.   Now $$(a+I).(b+I)=(a+i_{1}).(b+i_{2}),\forall (i_{1},i_{2})\in I \times I$$  $$(a+i_{1}).(b+i_{2})=ab+a.i_{2}+b.i_{1}+i_{1}i_{2}=ab+I$$  We know that $a.i_{2}+b.i_{1}+i_{1}i_{2}\in I$.   If $(a+I).(b+I)=ab+I$, then taking suitable $(i_{1},i_{2})\in I\times I$, we should be able to prove $(a+i_{1}).(b+i_{2})=ab+i_{3},\forall i_{3}\in I$. However, can every element in $I$ be generated by taking suitable $i_{1},i_{2}\in I$? And if not, is that the reason why $(a+I).(b+I)$ has to be defined as equal to $ab+I$ in violation of the distributive property of ring elements? This has confused me for a long time.   EDIT: I figured addition does not have to be defined as such because $(a+i_{1})+(b+i_{2})=a+b+i_{1}+i_{2}$, where $i_{1}+i_{2}\in I$. In fact, every element $i\in I$ can be constructed by taking $i_{1}=0$ and $i_{2}=i$. Hence, $(a+I) + (b+I)=a+b+I$ naturally. This is the logic I followed to determine that $(a+I).(b+I)$ doesn't quite work as nicey. I'm not sure if this logic is flawed or not as I haven't had the opportunity to ask anybody.   Thanks for your help in advance!
6294
$$\lim _{x\rightarrow 5}\frac{x^2-25}{x-5}  = \lim_{x\rightarrow 5} (x+5)$$  I understand that to evaluate a limit that has a zero ("hole") in the denominator we have to factor and cancel terms, and that the original limit is equal to the new and simplified limit. I understand how to do this procedurally, but I'd like to know why this works. I've only been told the methodology of expanding the $x^2-25$ into $(x-5)(x+5)$, but I don't just want to understand the methodology which my teacher tells me to "just memorize", I really want to know what's going on. I've read about factoring in abstract algebra, and about irreducible polynomials (just an example...), and I'd like to get a bigger picture of the abstract algebra in order to see why we factor the limit and why the simplified is equal to the original if it's missing the $(x-5)$, which has been cancelled. I don't want to just memorize things, I would really like to understand, but I've been told that this is "just how we do it" and that I should "practice to just memorize the procedure." I really want to understand this in abstract algebra terms, please elaborate. Thank you very much.
49881
Let $R$ be a ring. An element $x$ in $R$ is said to be idempotent if $x^2=x$. For a specific $n\in{\bf Z}_+$ which is not very large, say, $n=20$, one can calculate one by one to find that there are four idempotent elements: $x=0,1,5,16$. So here is my question:     Is there a general result which tells the number of the idempotent elements of ${\bf Z}_n$?
11622
Given a ring $R$ (with unity), prove that if $a \in R$ is nilpotent, then $a+1,a-1$ are both invertible.   Suppose $a^n=0$. Then $1=1-a^n=(1-a)(1+a+\ldots+a^{n-1})$, so $1-a$ is invertible.  If $n$ is odd we can use $1+a^n=(1+a)(1-a+\ldots+a^{n-1})$.  But what if $n$ is even?
15284
I know the inverse limit of $k[x]/(x^n)$ is $k[[x]]$ but i can't show the homomorphism between these is onto.   For example $\alpha:k[[X]]\to\varprojlim k[x]/(x^n)$ by F(x)=family of$ F(x)+(x^n)$. for example $(1+(x),0,5+7x^2+(x^3),3+11x+132x^2+(x^4),0,...)$ is image of things in k[[x]]?
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 $a = b = 0$.