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## New answers tagged ring-theory

0

$\mathbb{Z_6}[X]/(2x+4)\simeq\mathbb{Z_2}[X]/(2x+4)\times\mathbb{Z_3}[X]/(2x+4)\simeq\mathbb{Z_2}[X]\times\mathbb{Z_3}$

0

What is $(a+I)(1+I)=(1+I)(a+I)$ ? where $a+I\in R/I$

0

Note $1\notin I$ as $I\neq R$. Let $\phi:R\rightarrow R/I$ be the canonical projection. Then if $\bar{y}\in R/I$ (where $\bar{y}=\phi(y)$), we have $\bar{1}\bar{y}=\phi(1)\phi(y)=\phi(y)=\bar{y}$.

1

Indeed, an alternative definition of Noetherian ring is a ring where every ideal is finitely generated. As you noticed, a ring $R$ admits an infinitely generated ideal if and only if there is an infinite ascending chain of ideals. Indeed, suppose there is an ideal of $R$ which can only be generated by an infinite set $\{e_i\}_{i \in I}$ and extract a ...

2

Your seem to have the main idea, but let me try to clarify. The map $\pi : R \to R/I$ is given by $\pi(r) = r + I$. In order to show $\pi$ is surjective, we need to show that given any element $y \in R/I$, there exists an element $x \in R$ such that $\pi(x) = y$ (this is just the definition of surjectivity of a map of sets). Now, as $R/I = \{r + I \mid r ... 1 Hint: look at$A=F[x]/(x^2)$with$A=M$, and$L=I=(x)/(x^2)$0 If the uniformizing parameter is a prime number multiple of the multiplicative identity, then there is no ring homomorphism$A\to k[[x]]$. In particular, if we let$A=\Bbb Z_p$(the$p$-adic integers, which is complete) or$A=\Bbb Z_{(p)}=\{\frac{x}{y}\in\Bbb Q:p\nmid y\}$(whose completion is$\Bbb Z_p$) then the uniformizing parameter will be$t=p$and the ... 6 Hint A ring isomorphism is in particular an isomorphism of the underlying abelian group under addition, but the abelian groups$n \Bbb Z$,$n > 0$, each have only two automorphisms; using this one can show that there are only two isomorphisms (of abelian groups)$n \Bbb Z \to m \Bbb Z$for any$n, m > 0$. 8 In$2\mathbb Z$, there exists a generator$a$of the additive group such that$a+a=x^2$has a solution. This is not the case for$3\mathbb Z$. 6 If$A$is any ring, then$A$and$M_n(A)$have equivalent categories of modules, and usually$A$and$M_n(A)$are not isomorphic. This is the simplest example of a Morita equivalence. 1 When you quotient by$x-1$you are saying that$x-1 \equiv 0$(or$x \equiv 1). With this in mind \begin{align*} a_nx^n+a_{n-1}x^{n-1}+ \dotsb +a_1x+a_0 & \equiv a_n(1)^n+a_{n-1}(1)^{n-1}+ \dotsb + a_1(1)+a_0 \pmod{x-1}\\ & \equiv a_n+a_{n-1}+ \dotsb + a_1+a_0 \pmod{x-1}\\ \end{align*} Thus every polynomial is equivalent to its value atx=1$. Thus ... 0 Your result:$\begin{pmatrix} x & x \\ y & y \end{pmatrix}\begin{pmatrix} a & a \\ b & b \end{pmatrix}=\begin{pmatrix} x(a+b) & x(a+b) \\ y(a+b) & y(a+b) \end{pmatrix}$shows that in$S$does not exists a left identity. 2 A variant on this way of seeing things is that when you do operations on representative of cosets, each time you meet some$x^2+1$, you can replace it by$0$($x^2+1$is ‘killed’ in the quotient. In other words, each time you have an$x^2$, you replace it with$-1$, an$x^3$is replaced with$-1$,$x^4$with$1$, &c. 3 You seem to be focusing on understanding the elements of a single coset. But really, that is not so important. The important thing to understand the quotient is to understand the cosets all together. A good way to do this is to realize (in this case) that each coset has a unique element of degree at most$1$. Now you can play around with multiplying and ... 0 If$P$vanishes on a nonempty open subset$U$of$Z(Q)$, then it vanishes on all of$Z(Q)$. This is simply because$Z(P)$is closed, and if$U\subseteq Z(P)$, then$Z(Q)=\overline{U}\subseteq Z(P)$. The closure of any nonempty$U$is equal to$Z(Q)$because$Z(Q)$is irreducible. Hence,$P\in \sqrt{\langle Q\rangle}=\langle Q\rangle$as required. We choose ... 0 Not sure, whether this is necessary (meaning, that there is probably different more general approach - as shown in the other answer. But if we want to use the identity of the degrees we need to split it), but let's divide it into two cases. First when the degree of$f, g$is$0$and then else: In the first case$\operatorname{deg}(f(x)) = 0$and ... 4 You can see here more generally: Theorem. If the coefficient ring$R$is an integral domain, then so is also its polynomial ring$R[X]$. Let$f(X)$and$g(X)$be two non-zero polynomials in$R[X]$and let$a_f$and$b_g$be their leading coefficients, respectively. Thus$a_f≠0,  b_g≠0$, and because$R$has no zero divisors,$a_fb_g≠0$. But the ... 1 One way to describe the algebraic closure is that it is in some sense a "maximal" algebraic extension: it's an algebraic extension into which every other extension embeds. So it seems to me like the following question is a more basic one that should be answered first: What's an algebraic extension of commutative rings? There are various ways to answer ... 1 I'll assume what Pavel Čoupek indicated in his comment to the question. The desired statement then essentially follows from the following lemma (this is Lemma 3.32 from the book Algebraic Geometry 1 by Görtz and Wedhorn where you will also find a more complete proof on p. 79): Let$B$be a ring and let$A$be a$B$-algebra. Assume that there are finitely ... 0 Hint$\,\ (a_1,\ldots,a_j) \subseteq (b_1,\ldots,b_k)\iff a_1,\ldots,a_j \in (b_1,\ldots,b_k)\iff $$$\iff \begin{bmatrix} a_1\\ \vdots\\ a_j\end{bmatrix}\,=\,\begin{bmatrix}c_{11} &\ldots &c_{1k}\\ \vdots &\ddots & \vdots\\ c_{j1}&\ldots &c_{jk} \end{bmatrix}\begin{bmatrix} b_1\\ \vdots\\ b_k\end{bmatrix}\ \ \text{for some }\ ... 3 Let m,n\in\Bbb Z. We wish to show that \langle m\rangle\subset\langle n\rangle if and only if n\mid m. First, suppose that \langle m\rangle\subset\langle n\rangle. Since m\in\langle m\rangle it follows that m\in\langle n\rangle. That is, there exists a k\in\Bbb Z such that nk=m. Hence n\mid m. Conversely, suppose that n\mid m. To show ... 1 The ring in question is \{\frac{a}{b}\in\mathbb{Q} \mid b\text{ is odd}\}, also known as \mathbb{Z}_{(2)}. The key property of this ring is that R^\times = \{\frac{a}{b}\in\mathbb{Q} \mid a\text{ is odd, }b\text{ is odd}\} = R\setminus 2R, from which we can conclude that 2R is the unique maximal ideal. Presumably, you have identified all the ... 2 Well, actually your proof relies on the fact that (-1)\cdot x=-x. This does not follow immediately the definition, but it's a lemma that is usually proven as a corollary of 0\cdot x=0. You are not giving an alternative proof, though. Moreover, (-1)\cdot x=-x doesn't hold for rngs (rings without 1), while 0\cdot x=0 does. So I see little point in ... 2 I would say this is not correct. You see, the proof looks fine because you assume that (-1)x=-x. However to show that this is true you most likely have to use the fact that 0x=0. If you found a way to show (-1)x=-x, without using that 0x=0 then you're in the clear. I doubt that you can find one though. But if you (or anyone) can, I would love to ... 4 No, there is no such example, because if the polynomial ring R[x] is a PID, then R is a field, so that R[x] is also Euclidean. For references see here. 0 My reference is this post. Note that it is not necessary for A to have an identity. Let \mathscr{C} and \mathscr{D} denote, respectively, the collection of ideals of A containing I, and the collection of ideals of A/I. Define: f:\mathscr{C}\to\mathscr{D} by f(J) = \{a + I \mid a\in J\}\subset A/I. Define g:\mathscr{D}\to\mathscr{C} by ... 0 Hint: In the setting of Grothendieck Universe, we can talk about the size the of the set of small groups (denoted by Grp) and that of small rings (denoted by Rng), though not very meaningful I think. Then the mapping which sends each small group to the corresponding group ring is a injection from Grp to Rng. I also came up with some injections from Rng to ... 3 By definition an element f of a ring R is irreducible if whenever f=uv for u,v\in R, either u or v is a unit. By definition if f is nonprimitive in \Bbb{Z}[x], f=ng(x) where n is the gcd of it's coefficients, which is not \pm 1, and hence not a unit, and g(x) is a primitive polynomial in \Bbb{Z}[x] of positive degree and hence not ... 15 There are proper-class-many groups and rings - in fact, there are proper class many Xs, for any reasonable kind of mathematical structure X. One way to prove this is the following: given a cardinal \kappa, the direct product of \kappa-many rings is still a ring, and ditto for groups. So is there a sense in which there are "more" groups than rings? ... 8 The collection of all groups and the collection of all rings are, like the collection of all sets, proper classes (colloquially, they are "too big to be a set"). For any set S whatsoever, you can form the free group F_S and the commutative ring \mathbb{Z}[S], and if you have distinct sets S\neq T then F_S\neq F_T and ... 5 A proper and nonzero ideal I in \mathbb{C}[x] is generated by a single non constant monic polynomial f. Decompose f as$$ f(x)=(x-a_1)^{m_1}(x-a_2)^{m_2}\dots(x-a_k)^{m_k} $$with a_1,\dots,a_k pairwise distinct and m_i>0 (for i=1,2,\dots,k). Then the Chinese remainder theorem applies and$$ \mathbb{C}[x]/I\cong ... 3 Yes, as a DVR is a local ring there is only one prime number that is not invertible. (Note that a nontrivial ideal cannot include two distinct prime numbers and each non-unit is contained in some maximal ideal.) In more detail: Every non-zero subring of$\mathbb{Q}$must contain$\mathbb{Z}$and since a field is not a DVR the ring must not equal ... 2 Let$x\in S$. Then there are two chances:$x\in A$, or$x\notin A$. If the latter happens then$x^{-1}\in A$and then$x^{-1}\in\mathfrak m$, so$x^{-1}\in\mathfrak m_S$. But$x\in S$, so$1=xx^{-1}\in\mathfrak m_S$, a contradiction. 0 For question 1, this link seems to provide an answer (that's rather extensive for planetmath). With regards to the second question, I'm not sure what you're aiming at. If you consider a lattice algebraically, with binary operations$\wedge$and$\vee$, your structure will be commutative. What would it, for instance, mean if$a\wedge b\neq b\wedge a$? I ... 0 The ideal in a lattice meant in the theorem is definitely the order theoretic notion of an ideal. It has nothing to do with ring theory apart from some similarity in the definition. Regarding (2): Left and right ideals are notions from (non-commutative) ring theory, so that they don't apply here. In plain English the theorem tells you roughly the ... 1 I think the famous theorem of Jacobson about such a ring being commutative holds, even if there is no identity. The hypothesis implies the ring is von Neumann regular. Therefore every ideal is idempotent, including the entire ring. Since the ring is finite, by Nakayama's lemma, we have that R has an identity. It would follow that$R$is a finite product ... 0$k[X_1, X_2, X_3]/(X_1^3-X_3,X_2^2-X_3)\simeq k[X_1, X_2]/(X_1^3-X_2^2)$and continue like here. 2 Yes,$f\notin m^2$for any maximal ideal$m$is sufficient for proving the regularity. For showing$f\notin m^2$you need to check that not all the partial derivatives of$f$at$(\alpha,\beta,\gamma)$are$0$. (If$f(\alpha,\beta,\gamma)\ne 0$there is no need to go further.) 6 Automatically if$I=IJ$then$I=IJ\subseteq J$i.e.$I\subseteq J$. But equality isn't necessarily true. Consider$R=F\times F\times F$and the ideals$I$and$J$generated by$(1,0,0)$and$(1,1,0)$. Moreover, equality can occur when$I=I^2$but can't occur if$I\ne I^2$. 0 See http://math.stackexchange.com/a/1286492/62967 for (ii), where also a correction is made to the statement ($J$must be a proper ideal). For every proper ideal$J$of$R_P$you have$J=i(I)R_P$for some ideal$I$of$R$, with$I\subseteq P$. It's clear that for two ideals$I_1$and$I_2$of$R$, both contained in$P$, we have that, if$I_1\subseteq I_2$, ... 0 For (iii), take any ideal$J$of$R_P$. Then$J = i(I)R_P$for some$I\subseteq P$; it follows that$i(I)\subseteq i(P)$, so that$J\subseteq PR_P$. The same argument also shows that it is the only maximal ideal. 2 Assume$\varphi(\frac rs)=0$. Then$0=\varphi(s\cdot \frac rs)=\varphi(r)=\varphi'(r)$. 1 I assume what they mean is that this homomorphism makes$M$into an$R$-module and the kernel is the annihilator of$M$over$R$. 3 Let$J$be an ideal of$R_P$. Consider $$I=\{x\in R: x/1\in J\}$$ It's easy to prove that$I$is an ideal in$R$: obviously$0\in I$; if$x,y\in I$, then $$\frac{x+y}{1}=\frac{x}{1}+\frac{y}{1}\in J$$ and therefore$x+y\in I$; if$x\in I$and$r\in R$, then $$\frac{xr}{1}=\frac{x}{1}\frac{r}{1}\in J$$ and thus$xr\in I$. Suppose$x\in I$and$x\notin ...

0

"Sharp" in his book "Steps in Commutative Algebra" has: Note that your statement, $I\subseteq P$, is correct when $J$ is proper; by He even adds :

0

$\ker f$ is a two-sided ideal of $M_n(K)$, so $\ker f=(0)$. (For the form of two-sided ideals in matrix rings you can look here.) This tells us that $f$ is injective. Then $B=f(M_n(K))$ is a simple subalgebra of the central simple algebra $A=M_m(K)$, and $[A:K]=[B:K][C_A(B):K]$ where $C_A(B)=\{a\in A:ab=ba,\forall b\in B\}$ is the commutant subalgebra of ...

0

Taking into account your comment explaining what is $\;D\;$ , you could try to do as follows: first, every element in $\;\Bbb R[x]\;$ can be divided by $\;x^2\;$ with residue: $$f(x)=x^2g(x)+r(x)\;\;,\;\;\text{with}\;\;\deg r<2\;\;\;or\;\;\;r\equiv 0$$ Thus, we can write $\;f(x)=x^2g(x)+(ax+b)\;,\;\;a,b\in\Bbb R\;$ , and we can thus define \phi:\Bbb ...

2

There is a famous example of Macaulay which shows that there are prime ideals in $\mathbb C[X_1,X_2,X_3]$ having at least $n$ generators for any $n\ge 1$. For more details and a proof see this paper. In Macaulay's words, the example is constructed as follows: "Consider $l(l-1)/2$ straight lines through the origin $O$ in $3$-dimensional space, not ...

3

No, $(x,y)^n = (x^n,x^{n-1} y,\dotsc,x y^{n-1},y^n)$ is an ideal of $K[x,y]$ which can not be generated by $n$ elements.

1

Given a prime ideal $P$ of $\Bbb Z$, the localization of $\Bbb Z$ at $P$ is the set of all fractions where in reduced form, the denominators are not in $P$. As such it's a subring of $\Bbb Q$. For instance, $\Bbb Z_{(5)}$ is the ring of all fractions $a/b$ where $a$ and $b$ are coprime integers, and $b \notin (5)$, so $192/17$ is in there, but $3/40$ is ...

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