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4

To complement Martin's answer in accordance with the wishes expressed in the comments of the OP, why is the kernel of the canonical map $A[x]/(q,p)\to (A/q)[x]/(\bar{p})$ zero? Suppose $f=a_nx^n+\cdots+a_1x+a_0\in A[x]$ is in the kernel. This means that if we reduce the coefficients modulo $(q)$, to get $\bar{f}=\bar{a}_nx^n+\cdots+\bar{a}_1x+\bar{a}_0$, ...


1

I think $\Bbb{Z}[x]/(x^2)$ with the basis $\{1+x,x\}$ is a counterexample.


2

If $\sqrt{0}^n=0$, then in particular $(2 e_n)^{n-1}=0$, i.e. $2^{n-1} = 0 \bmod 2^n$, a contradiction.


1

$\Sigma^{-1} A$ represents the functor of homomorphisms on $A$ which invert all elements in $\Sigma$. Since $\Sigma$ is covered by the $S_i$, this means that all the $S_i$ are made invertible. Hence, we get unique lifts to $S_i^{-1} A$, and these are clearly compatible. The construction is invertible, so that $\Sigma^{-1} A$ represents the same functor as ...


2

When studying factorization theory in domains, one is usually not interested in the trivial factorizations of $\,0,\,$ so one employs a definition of irreducible that proves most convenient for theorems about factorizations of other (nonzero) elements. So, to avoid exceptional cases, and to avoid having to frequently write "nonzero irreducible", it is ...


3

Tensor products commute with direct sums, so $S\otimes_R \bigoplus_i R \cong \bigoplus_i (S\otimes_R R) \cong \bigoplus_i S$. If you write down these isomorphisms explicitly, you will get the conclusion you want. EDIT: Adam Hughes has made the important observation that our ring morphism must send $1$ to $1$. This is usually an assumption in commutative ...


1

Hint: In a ring, the addition table is the table of a group. What are the groups of order $4$? Partial solution: Full solution:


1

In rings with zero-divisors, factorization theory is much more complicated than in domains, e.g. $\rm\:x = (3+2x)(2-3x)\in \Bbb Z_6[x].\:$ Basic notions such as associate and irreducible bifurcate into at least a few inequivalent notions, e.g. see the papers below, where three different notions of associateness are compared: $\ a\sim b\ $ are $ $ ...


1

The standard proof can be modified: Let $P = \sqrt{\mbox{ann }x}$ for $x \in M$. We have two cases. First suppose that for every $r$ such that $rx \in M'$ we have $r^n x = 0$, for some $n$ (i.e. $r \in P$). Then for the image $\overline x \in M''$, we have that $r \overline x = 0$ implies that $r \in \sqrt{\mbox{ann }x}$. $\ $ Hence $\mbox{ann } x ...


1

The key idea is an ideal-theoretic generalization of $\rm\color{#c00}{Euclid's}$ key idea in his proof that there are infinitely many primes $\rm p.\,$ Namely, if $\,f\,$ is divisible by every prime $\,p\,$ then $\,1+ nf\,$ has no prime divisors so is a unit. In a similar way, replacing "divides" by "contains", and $\,n\,$ by $\rm\,x\,$ we obtain Hint ...


1

In general, if $f$ is any element of a ring contained in the jacobson radical, then for every other element $r$ we have that $1+fr$ is a unit. Otherwise it would be contained in a maximal ideal, but $fr$ too, so that $1$ would be contained, a contradiction.


1

Pinter in "A book of abstract algebra" says it thus: Thus, we are led to the modern notion of algebraic structure. An algebraic structure is understood to be an arbitrary set, with one or more operations defined on it. And algebra, then, is defined to be the study of algebraic structures. It is important that we be awakened to the full generality of the ...


3

Here's a high-level proof - perhaps someone else will find a more elementary one. $\DeclareMathOperator{\nil}{nil}$ $\DeclareMathOperator{\Ass}{Ass}$ $\DeclareMathOperator{\Spec}{Spec}$ $\DeclareMathOperator{\Min}{Min}$ Set $B := A/\nil(A)$. The assumption on $A$ implies that every nonunit in $B$ is a zerodivisor (notice: this is stronger than saying that ...


0

The problem is that any ring that is a non-trivial product of rings will have non-trivial idempotent elements and in particular lots of zero-divisors. The notion of "atom" or "irreducible" is not so clear. There are several possible routes to take for what you define as an atom (all of which are equivalent in a domain, but none of which are equivalent in ...


1

Try to find a field $\;\Bbb F\;$ s.t. you can define a homomorphism $\;\phi:\Bbb Z\oplus\Bbb Z\to \Bbb F\;$ with $\;\ker\phi= A\;$, and looking at the form of the elements in $\;A\;$ a good idea is to choose $\;\Bbb F=\Bbb F_3\;$: $$\phi(n,m):=n\pmod 3$$ Do some checking: $$\begin{align*}\bullet&\;\;\phi((n,m)+(n',m')):=\phi(n+n',m+m')=(n+n')\pmod ...


3

The last line "$\implies (Z \oplus Z) / (3Z \oplus Z) \approx Z_3 \oplus (Z \oplus Z)$" does not make much sense. It certainly doesn't accomplish the goal you had of proving the quotient is a field, so it should definitely be reconsidered. Given an element of $\{(a,b) + (3Z \oplus Z) ~|~a,b \in Z\}$, the element is equivalent to one of the form ...


1

Let $x \in R-I$. Consider the ideal $J=\langle I, x \rangle$. Observe that $|J|>10$. Moreover this must be a subgroup of $R$ but based on Lagrange its order should divide $30$. But $I \leq J$ as well, therefore $10$ divides $|J|$. Thus $J=R$, hence $I$ is maximal.


2

Instead of commenting on this proof, which is too complicated in my opinion, here is a fast proof. Assume that $a$ is not nilpotent. Then the localization $A_a$ is non-zero (otherwise $1/1=0$ in $A_a$, which would mean $a^n \cdot 1 =0$ for some $n$). Hence, it has a prime ideal $\mathfrak{p}$. The preimage in $A$ is a prime which doesn't contain $a$ ...


1

I think freely generating $R^\times/\cong$ is what you're looking for. The requirement that the image of $\xi$ freely generates the monoid $R^\times/\cong$ implies that the elements of $\xi$ are irreducible. (If an element of $\xi$ were reducible, factorization of its factors would lead to a nontrivial relation among the generators.) Another observation is ...


1

One construction is simply to take all algebraic elements of $\mathbb{C}$ But a more general method is to enumerate all polynomials with integer coefficients, $p_n(x)$ and define $F_{n+1}$ to be $F_n(\alpha_1, \ldots ,\alpha_k)$ where $\alpha_1, \ldots ,\alpha_k$ are the roots of $p_{n+1}(x)$. Then $F=\bigcup F_n$ will be algebraically closed. For if $\beta$ ...


1

This question becomes even easier if you know the universal properties (which some people would call the "right definitions") of localization, of polynomial rings, and of quotients. $A^{-1}R$ is the universal example of a commutative ring with a homomorphism from $R$ that sends all elements of $A$ to invertible elements. That's equivalent to just sending ...


1

1) Yes, all the $A_i$ are rings. 2) Here the term "direct sum" means internal direct sum, which basically just means "sum", but with the additional property that the ideals intersect trivially. 3) Yes, the product is component-wise (just like the sum).


1

Yes and 3.: Addition and multiplication is defined component-wise. So $a_1 \oplus a_2 = (a_1, a_2, 0, ..., 0)$.


0

Homomorphisms preserves structure of group/ring; i.e: if you draw a table that show the "$.$" of group, then a homomorphism preserves structure on that table: At the row "$a$" and column "$b$" you have the element "$a.b$". Under homomorphism "$f$", at the row "$f(a)$" and column "$f(b)$" you have the element "$f(a.b)$"


0

Since $R$ is finite, suppose $|R|=p_1^{s_1}p_2^{s_2}\cdots p_n^{s_n}$, the distinct prime divisor factorization of $|R|$. For each $i=1,2,...,n$, let $R_i=\{r\in R\mid p_i^s r=0 \mbox{ for some }s\in \Bbb{N}^+\}$. It is easy to verify $R_i$ is an ideal in $R$ and $R_i\cap (R_1\cap \cdots \cap R_{i-1}\cap R_{i+1}\cap \cdots\cap R_n)=\emptyset$. Consider ...


1

Consider the ring $R$ of continuous functions $\mathbb R\to\mathbb R$ with compact support. There are no non-zero idempotents in this ring, yet your condition holds. Indeed, if $a\in R$, let $c\in R$ be any function which is equal to $1$ on the support of $a$. Later This ring does not have a unit, and you wwanted it to have one. But if $R$ does have a unit ...


8

You can easily construct an isomorphism using the universal properties of quotient and polynomial rings. Start as follows: We have $A \to A/q \to A/q [x] \to A/q[x]/(p)$ and the element $x$ in $A/q[x]/(p)$. Hence, we get $A[x] \to A/q[x]/(p)$. Clearly $q$ and $p$ lie in the kernel, so that we get $A[x]/(p,q) \to A/q[x]/(p)$. Now construct the inverse ...


3

$n\mathbb{Z}_k$ has a multiplicative identity if and only if there is an $a\in \{1,\dotsc,k-1\}$ (we exclude the case $k = 1$ here) such that for all $0 \leqslant b < k$ we have $$(na)(nb) \equiv nb \pmod{k}.\tag{$\ast$}$$ Write $(\ast)$ in the form $(na-1)(nb) \equiv 0 \pmod{k}$. Setting $b = 1$, we see that the necessary and sufficient condition is ...


0

If $n=md$ and $k=hd$ where $m$ and $k$ are coprime, then $n$ must be coprime to $h$.


2

I'm afraid there are shortcomings in your attempts. (i) The choice $i=0$ gives $a^i=1$, so you cannot conclude that $a$ is of finite order in thise way. The question has been covered many times recently. I recommend this answer by blue. (ii) I don't think the number $1+\sqrt2$ is in your set $H$, so $H$ is not closed under addition and hence not a ...


0

Yes; consider the injective ring homomorphism $$f:A=\mathbb{C}[x_1,x_2,x_3,\ldots]\to\mathbb{C}[x_1,y_1,x_2,y_2,\ldots]/(x_1y_1,x_2y_2,\ldots)=B.$$ It is easy to see that for any $b\in B$, $\sqrt{\operatorname{ann}_Bb}=\operatorname{ann}_Bb\subset B$ is not a prime ideal, so $\operatorname{ModerateAss}_BB=\varnothing$, hence ...


1

Of course, the finite ring already has its own multiplication, and you would not be at liberty to redefine it to some other product... But on the other hand, if you show your candidate is actually the only multiplication possible, you would have a solution. The primary components of the underlying groups are surely rngs, but to show that the big ring is a ...


2

The statement is in general false in the absence of special restrictions on $x$. As a counterexample, simply take $R = Z$, $0, 1 \ne x \in Z$, and $0 \ne s, t \in Z$; then if $(st)x = (sx)(tx) =(st)x^2, \tag{1}$ we have, since $st \ne 0$, that $x = x^2; \tag{2}$ but in $Z$, (2) implies that $x =0$ or $x = 1$, contradicting our assumption that $x \ne 0, ...


1

I get the form $$\mathcal{C}(x,y) = (a, b, c, d) = (1, 3, 3, -1) = x^3 + 3 x^2 y + 3 x y^2 - y^3$$ corresponding to the image of the Davenport-Heilbronn map but there are infinitely many acceptable binary cubic forms, in a class corresponding to the image, related to each other by a substitution $\begin{pmatrix}{x}\\{y}\end{pmatrix} = M ...


0

$Spec \mathbb{Z}$ which is the spectrum of the integers has a closed point for every prime number $p$. Furthermore every $p$ corresponds to the maximal ideal $(p) \subset \mathbb{Z}$ along with a non-closed generic point (whose closure is the whole space) corresponding to the zero ideal $(0)$. Thus the closed subsets of $Spec \mathbb{Z}$ are the finite ...


3

It's easy to see this by using the fact that $n:= \text{char}(R)$ is the natural number that generates the kernel of the ring morphism $f \colon \mathbb Z \to R$, $k \mapsto k \cdot 1_R$. The image of this morphism is central in $R$, and heuristically speaking any argument about the characteristic occurs in this image, so commutativity of $R$ is irrelevant. ...


6

Let's see how far we get without assuming commutativity. First, for every $x\in R\setminus\{0\}$, there is an ideal $A_x \subset \mathbb{Z}$ such that $n\cdot x = 0 \iff n\in A_x$. Let $a_x$ be the non-negative generator of $A_x$. Then, since $R$ has no zero divisors, for every $x\in R\setminus\{0\}$, $a_x$ is either $0$ or a prime. For suppose $a_x = ...


2

I wouldn't say that $\Bbb R^n$ is a division algebra under "the usual multiplication," but rather using the multiplications that $\Bbb C$ and $\Bbb H$ induce. The fact that the only finite dimensional associative $\Bbb R$ division algebras are the reals, the complexes, and the quaternions is precisely the Frobenius theorem. It's too long to prove here, but ...


3

Question 0: Every identity in the language of rings is equivalent to one of the form $p(\overline{x}) = 0$, where $p(\overline{x})\in \mathbb{Z}[x_1,\dots,x_n]$ for some $n$. So the question amounts to: if a polynomial in $\mathbb{Z}[x_1,\dots,x_n]$ is $0$ at all points of $\mathbb{Z}^n$, is the zero polynomial? The answer is yes. There are fancy ways to ...


3

The other posters did a good job explaining what abstract algebra is, so I'll try to help you understand groups. You can think of a group, ring, field etc. as being a set with a certain structure attached to it. The intuition usually comes from concrete examples so I'll include a few. (Disclaimer: This will not be rigorous in the slightest. I'm shooting for ...


3

Hint: suppose that $a, b \in R$, $a$ has additive order $n$ and $b$ has additive order $m > n$ or $\infty$. Consider the additive order of $ab$. Can you get a contradiction?


1

Hint: either the characteristic of $R$ is $0$ or it is a prime number. Let's not assume $R$ has a unit and is not the zero ring. Consider $$ I=\{n\in\mathbb{Z}:nr=0\text{ for some }r\in R, r\ne0\}. $$ If $n\in I$ and $m\in\mathbb{Z}$, then obviously $mn\in I$. If $m,n\in I$ and $mr=0$, $ns=0$, with $r\ne0$ and $s\ne0$, then $$ (m+n)(rs)=(mr)s+(ns)r=0 $$ ...


4

When you first came to learn maths, what did it show you? For instance, we wrote a lot of equations in terms of x,y,z and so on, but essentially, what they had common is that they were representing some unknown numbers. Abstract algebra is a bit more broad concept: here (in intuitive sense) letters represent pretty much anything, rather than numbers. I ...


23

We learn math with numbers early on. We learn how to apply operations to numbers to get new numbers. We learn rules, and consequences of those rules. All of that is pretty straightforward. But, the real numbers are not the only things we might want to examine in detail. The properties of how elements interact under operations is a more general, abstract ...


16

In "concrete" algebra one works with things like integers, rational numbers, real numbers, complex numbers, matrices, quaternions, permutations, polynomials, geometric transformations (e.g. isometries, similarities, reflections, inversions, projectivities, etc.), etc., subject to operations like addition, multiplication, and composition. In "abstract" ...


6

Mathematics has to do with sets. There is no definition about what the set is, but we all know that set is made up by elements. Many problems in nature can be represented by sets, and the relations of the elements on these sets. The mathematical discipline that studies THE RELATIONS OF ELEMENTS on a given set is called algebra. There are many good properties ...


2

There won't be any condition, since $\phi : A \to B$ and $A/\mathrm{ker}(\phi) \to B$ have the same image.


0

For every idempotent $e\notin \{0,1\} $ , $e\in Jac(R) $ so $1-e$ is invertible. Then from $e(1-e)=0$ we conclude that $e=0$, a contradiction.


2

For any finite group $G$ and finite-dimensional $\mathbb{F}_p[G]$-module $M$, I'll denote by $\Omega M$ the kernel of an $\mathbb{F}_p[G]$-projective cover $P(M)\to M$. Let $F(M)$ be a free $\mathbb{Z}[G]$-module with a surjection $F(M)\to M$, and complete to a short exact sequence $$0\to K(M)\to F(M)\to M\to0$$ of $\mathbb{Z}[G]$-modules. Then $K(M)$ is ...



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