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If $\mathfrak{a}$ is a modular right ideal, then every right ideal $\mathfrak{b}$ containing $\mathfrak{a}$ is modular as well, because $r-er\in\mathfrak{a}$ implies $r-er\in\mathfrak{b}$. Note also that $e\notin\mathfrak{a}$, if $\mathfrak{a}$ is a proper ideal. Applying Zorn's lemma is indeed the way to go. Consider the family of right ideals containing ...

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The two ideals are in fact equal. Let $ra + na \in (a)$ and $sb + mb \in (b)$ where $r, s \in R$ and $n, m \in \mathbb{Z}$, then $$(ra + na)(sb + mb) = rsab + mrab + nsab + nmab = (\underbrace{rs + mr + ns}_{\in\ R})ab + (\underbrace{nm}_{\in\ \mathbb{Z}})ab \in (ab).$$ As $(a)(b)$ consists of finite sums of elements of the form $(ra + na)(sb + mb)$ and ...

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After inverting primes in $\Sigma$, we can assume $A=A_\Sigma$. Hensel's lemma is being used as the following assertion: if $C$ is a finite étale $A$-algebra, then any section $C \to B$ lifts uniquely to a section $C \to A$. Apply this with $C = A[x]/(x^n - a)$, for any $n$ prime to $\Sigma$ and the section $C\to B$ corresponding to the identity element.

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I think what you looking for is the following proof which use the axiom that any non-truvual subset of the natural numbers contain aminimal element. Let $a,b\neq 0 \in \mathbb{N}$. Then there exist $q,0\leq r<|b|$ (and unique), such that $$a=bq+r.$$ You define the following set $$W=\{a-bq\geq 0:\quad q\in \mathbb{Z}\}.$$ It is easy to show that $W$ is ...

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I will restrict to commutative rings. If $R$ has dimension $d$, then $R[x]$ has dimension $d+1 \leq \dim(R[x]) \leq 2d+1$. So one property is "being of finite dimension". Further properties listed here are "integral domain", "noetherian ring", "normal ring", "UFD". Other properties are "reduced", "connected", "smooth" (over some base ring), "finitely ...

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I can only give an inefficient answer; others will do much better. I hope you know about and are thinking of Newton Polygon. Remember that it’s described by using the additive valuation on $\Bbb Q_p$, $v(p)=1$, $v(ab)=v(a)+v(b)$, $v(a+b)\ge\min(v(a),v(b))$, $v(0)=\infty$. For $f=\sum_ia_ix^i$, you plot all points $(i,v(a_i))$ in the plane, erect a vertical ...

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I assume you mean $(-1)^{2} = 1$ for this answer. Furthermore, I assume you mean rings, not groups as in the tag, because I assume the square is the multiplicative operation, yet the minus sign refers to additive inverse. These things should be clarified. The answer is yes. In general, we can show that $(-1)(x) = -x$ in the following way: $x + (-1)(x) = ... 2 In the semigroup theory S is called a left ideal in the semigroup R under multiplication. See any semigroup theory book(J.M.Howe: Semigroup theory for example) 4 The general criterion is that no number can be found with more than one valid, distinct factorization. This might sound like I'm merely rephrasing the question, but it's actually a reframing of the question. Plenty of numbers (infinitely many, to be precise) in$\mathbb{Z}[\sqrt{-5}]$have more than one factorization.$6$is just the easiest to find. To ... 3 Yes, there is. But first there is the matter of integral closure (I prefer to think of it as "completeness") to attend to. If$d \equiv 1 \pmod 4$, then you need$\frac{1 + \sqrt{d}}{2}$instead of$d$. For example, with$d = 13$,$1 + \sqrt{13}$is an algebraic integer, but so is$\frac{1 + \sqrt{13}}{2}$. For fun, try this in your calculator: ... 2 You can relate the commutator for a Lie group$G$to the (ring-type) commutator of its Lie algebra$\mathfrak{g}$. Say$g$lies infinitesimally close to the group identity,$g = \exp(\epsilon\,X)$for some$X \in \mathfrak{g}$. Likewise say$h = \exp(\epsilon\,Y)$for some$Y \in \mathfrak{g}$. Then,$ghg^{-1}h^{-1}$also lies infinitesimally near the ... 3 If zero is considered to be a correct counterexample for this problem, then the problem is not a very good one, having a trivial answer for a pedantic reason. We can surmise this was not the intended effect, and that the author meant for something meaningful to happen. In all likelihood, the author probably meant for zero to count as a zero divisor. This is ... 0 This is well-studied under the heading of "cancellability," and Lam's crash course on the topic is very nice. Are there any simple conditions on$R$-modules$M,A$and$B$... The readiest one is that if$R$has stable range 1 and$M$is finitely generated and projective, then it cancels from$M\oplus A\cong M\oplus B$. You can find this, for example, in ... 1 (answer from a discussion with @peter a g) Our objective is to form a homomorphism from$M''$to$M$. Call$\{e_1,e_2,...,e_n\}$a basis of the free module$M''$. As it is a basis, every element of$M''$can be written as a linear combination of the basis elements in a unique way. At this point, we construct our homomorphism from$M''$to$M$by first ... 2 A beautiful combinatorial answer was found by Gjergji Zajmi. For an expanded version, see the solution to Exercise 2 in PRIMES 2015 reading project problem set #1. (Ignore Section 0.1; it has nothing to do with this.) Unlike Martin's answer, this one directly gives a formula for$\dbinom{xy}{n}$as a nonnegative linear combination of products ... 3 An example of a difference:$\mathbb{Z}/3\mathbb{Z}$is projective (and free) as a$\mathbb{Z}/3\mathbb{Z}$-module but is not projective (nor free) as a$\mathbb{Z}$-module. However things like simplicity and indecomposability will remain the same, which is easy to prove since the action of$A/\mathfrak a$is given by the action of$A.$The only thing ... 2 Take$\mathbb{Z}\times R$for$R$any nontrivial ring with only finitely many invertible elements. 15$\mathbb{Z}\times\mathbb{Z}$. There are only$(-1,1),(-1,-1),(1,1),(1,-1)$as invertible elements but infinitely many zero divisors$(a,0)$for every$a$. 3 The power set of$\: \{\hspace{-0.02 in}0,\hspace{-0.05 in}1\hspace{-0.02 in},\hspace{-0.04 in}2,\hspace{-0.04 in}3,...\hspace{-0.04 in}\} \:$is an example of such a ring. 2 In both cases, commutators measure how far the object is from being commutative. A group is commutative when its operation is commutative. A ring is commutative when its multiplication is commutative. (Addition in a ring is always commutative.) Hence, the definition of commutator in each case reflects what it means to commute, as it should. 6 Another perspective on this comes from considering the relationship between a Lie group and its corresponding Lie algebra. Details are at On the relationship between the commutators of a Lie group and its Lie algebra, but the upshot is that the ring commutator$[A,B]=AB-BA$can be regarded as an infinitesimal version of the group commutator ... 1$\operatorname {Frac}(\mathbb Z[[x]])\subsetneq \mathbb Q((x))$because$\sum\frac {1}{n!}x^n\in \mathbb Q((x))\setminus \operatorname{Frac}(\mathbb Z[[x]])$29 Here is a more abstract and unified approach to commutators. I could phrase this whole answer in terms of category theory (and this is probably the best way to think of it), but I won't in order to make the answer more accessible. Let$G$be a group. We want to make it commutative. What does that mean? We want to find a commutative group$A$and a ... 38 The commutator of a group and a commutator of a ring, though similar, are fundamentally different, as you say. In each case, however, the commutator measures the "extent" to which two elements fail to commute. That is, given elements$a,b$, we wish to "compare"$ab$and$ba$. In the context of a group, we only have one operation: "multiplication". One ... 0 I'm going to say no (even though the answer is really yes) and here's why: The process you are describing above is an algorithm -- given a finite set of numbers define recursively the next number and then solve. For algebraic objects, of course there are constructions that you could imagine mimic these operations but they are very different. For example, a ... 7 Although I am not much of an Algebraic Geometry person, but it is a very important application of commutative ring theory. Just to give you a flavor, here is a brief idea. Consider the ring$\Bbb{C}$. Now we do like to study surfaces/curves (that is an application, right!) , and we define them by equations in rings$R=\Bbb{C}[x_1,x_2,\dots , x_n]$. Now ... 1 EDIT: see the comments below. Whether this answer is correct depends on exactly how "homomorphism" is defined; if homomorphisms must preserve multiplicative unity, then this is correct, but otherwise this is incorrect. Basically, this hinges on whether we have a "primitive symbol" for the multiplicative identity or not. One way to do it is to look at the ... 1 Hint: if$f$is a homomorphism from a ring of characteristic$p$into one of characteristic$q$, then$qf(1)=f(p1)=pf(1)=0$. Use this to show that$f(1)=0$. 1 You know that$r(x)\in\mathbb Q[x]$and$Lr(x)\in P$for some$L\in\mathbb Z$,$L\ne0$. Now multiply$Lr(x)$by a non-zero integer$a$such that$ar(x)\in\mathbb Z[x]$, and find$Lar(X)\in P$. Then$ar(x)\in P$and$\deg ar(x)=\deg r(x)<\deg m(x)$. You have to be careful when start with$m\in P$,$\deg m\ge1$of minimal degree to suppose that$m$is ... 2 We use that$k\subset k'$is separable, a result of Grothendieck which says: If$k'$and$K$are extension fields of$k$such that either$k'$or$K$is finitely generated over$k$and if$k'$is separable over$k$, then$k'\otimes_kK$is regular. and the Theorem 33.2(i) from Matsumura, CRT, which asserts: If$A\to B$is a faithfully flat ... 1 Clément has already provided a nice answer. Here is an alternative solution, based on Mariano's comment here. Comments and criticism are welcome! Let$I$be the principal ideal in$R$generated by the element$(xy)^2$. It suffices to show that$R/I$is not a finitely generated$k$-algebra. It is not hard to see that a general element of$R/I$is of the ... 1 Let k be a field, let A be the enveloping algebra of the three-dimensional Heisenberg Lie algebra g, and let I be the ideal of A generated by any nonzero element of the center of g. This satisfies all your conditions. 5 The subring$R$is a finitely generated$k[x,y]$-module since, as an ideal it is$(x)$. However, as a$k$-algebra it cannot be finitely generated. Take$(f_1,...,f_r)$to be a finite set in$R$and set$S$the algebra generated by$(f_1,...,f_r)$. Set$d_i$the degree of$f_i$in$x$. Now we are interested in the degree$1$in$x$part of elements of$S$. ... 2 By Gauss'lemma,$\underline f(x)$and$\underline g(x)$are irreducible in$K(R)[x]$, where$K(R)$is the field of fraction of$R$. Since$K(R)[x]$is a pid$(\underline f(x), \underline g(x))=(d(x))$for an opportune polynomial, we may assume that$d$is a polynomial with coefficients in$R$. But both$\underline f(x)$and$\underline g(x)$are irreducible ... 2 Let P be the prime ideal$P \subseteq \Bbb{Z}[x]$and let it be not a principal ideal, so$P$contains two non zero elements$f,g$with no common factor in$\Bbb{Z}[x]$. We can prove this easy lemma- Lemma-$f,g$has no common factor in$\Bbb{Q}[x]$. Sketch- If they have a common factor, say$h\in \Bbb{Q}[x]$with degree$\ge1$, then writing$h=ah_0$... 2 Yes. Consider homomorphism$\phi: (R_1 \oplus R_2 ) \to (R_1 / \langle x_1\rangle ) \oplus (R_2 /\langle x_2\rangle) $which sends$(a,b)\to (a+\langle x_1\rangle, b+ \langle x_2\rangle)$. Clearly$\phi$is onto. Now the ker$(\phi)=\{(a,b)\ |\ a=r_1x_1, b=r_2x_2\ \text{for some}\ r_1,r_2\in R \}=\langle(x_1,x_2)\rangle$. And apply First isomorphism ... 1 If$S=K[X,Y]$with the usual grading, that is,$S_n=\sum_{i+j=n} KX^iY^j$, then the ideal$I=(X+Y^2)$is not graded (why?). Even simpler,$S=K[X]$with$S_n=KX^n$and$I=(X+1)$. 2 Let$\mathscr{I}$be the set of ideals of$R$containing$I\cup\{a\}$as a subset. First show that if$J\in\mathscr{I}$, then$i+ra\in J$for each$i\in I$and$r\in R$; this will show that $$\{i+ra:i\in I\text{ and }r\in R\}\subseteq\bigcap\mathscr{I}=(I,a)\;.$$ Then show that$\{i+ra:i\in I\text{ and }r\in R\}$is an ideal; this will show that ... 1 Yes, exactly, you are supposed to prove that$(I,a):=\{i+ra\mid i\in I,\,r\in R,\,a\in A\}$is the smallest ideal of$R$that contains$I$and$a$. Prove that$(I,a)$is an ideal. Note that it contains$I\cup\{a\}$. Prove that any ideal$J$that contains$I$and$a$, also contains$(I,a)$. 2 Other answers have been given, but I'll throw my 2 cents anyway. For proving uniqueness you just need that$a$is left invertible, that is, there exists$c\in R$such that$ca=1$. Indeed, if$ax_1=b=ax_2$, you get$a(x_1-x_2)=0$. Thus$ca(x_1-x_2)=c0=0$and therefore$x_1-x_2=0$, because$ca=1$. Right invertibility of$a$provides existence: if$ad=1$, ... 1 If$ax_1 = ax_2 = b, \tag{1}$with$a \in R$a unit, then since we have$c \in R$with$ac = ca = 1_R$,$x_1 = 1_R x_1 = (ca)x_1 = c(ax_1)= c(ax_2) = (ca)x_2 = 1_R x_2 = x_2, \tag{2}$so the solution is unique. We further note that$ax = b \tag{3}$yields$x = 1_R x = (ca)x = c(ax) = cb; \tag{4}$the unique solution to (3) is thus$cb$. 4 Your solution is close to being great. First off, they told you that$R$is ring with unity because only those can have units. Second, your proof doesn't require knowledge that it is an integral domain. Towards the end, you have $$a(x_1-x_2)=0$$ Then, because$a$is a unit,$a^{-1}$exists, and $$a^{-1}a(x_1-x_2)=a^{-1}0=0$$ $$x_1-x_2=0\implies x_1=x_2$$ ... 2 Let$R$be a unitial ring with$a,b\in R$where$a$is a unit. Then$ax=b$if and only if$x=a^{-1}b$. Indeed, if$ax=b$, then $$x=1_Rx=a^{-1}ax=a^{-1}b$$ Conversely $$a(a^{-1}b)=(aa^{-1})b=1_Rb=b$$ 5 The identity$\binom{x+y}{n} = \sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k}$is well-known, it is called the Vandermonde identity. The answer for$\binom{xy}{n}$can be explained using the notion of a$\lambda$-ring, where here we consider the binomial ring$\mathbb{Z}$with$\lambda^n(x)=\binom{x}{n}$. The main theorem on symmetric polynomials enables us to ... 0 Yes. If$R$has infinite left global dimension, then for every$n$there is a left module$M_n$with projective dimension greater than$n$. But then$\bigoplus_{n\in\mathbb{N}}M_n$has infinite projective dimension. The same argument works for weak dimension. 1 We can present a module$M$over a ring as follows - take a free module$F$. Its generators are called "the generators". Now form the quotient$M=F/H$by a submodule$H$of$F$. The generators of$H$are called "the relations". We have the exact sequence $$F\rightarrow H \rightarrow M \rightarrow 0.$$ 1 As has been pointed out, this exercise is as easy as you thought. In particular, your argument is complete. The only mistake is when you said "zero divides everything"; I believe you meant everything (other than the zero) divides zero. 3 Let$R\subset\mathbb Z_2^{\mathbb N}$be the set of sequences with elements in$\mathbb Z_2$which are constant from a rank on. Then$R$is a boolean ring. If$R\simeq\prod_{i\in I}R_i$with$R_i$indecomposable, then$R_i$is isomorphic to a subring of$R$, hence$R_i$is boolean. Moreover, since$R_i$is indecomposable we must have$R_i\simeq\mathbb Z_2$. ... 0 HINT: Try$R$the ring of continuous functions from$C= \{0,1\}^{\mathbb{N}}$to$\mathbb{R}$. 2 As$R$is a ring, it contains a multiplicative identity$1_R$. As such, there is a unique ring homomorphism$\phi : \mathbb{Z} \to R$given by$\varphi(n) = n\cdot 1_R$; note,$n\cdot 1_R$does not denote the multiplication of two elements of$R$, rather it is shorthand for the sum of$n$copies of$1_R$if$n$is positive, or$|n|$copies of$-1_R\$ (the ...

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