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## Hot answers tagged ring-theory

10

No nontrivial group $(G,\cdot)$ can be extended to a ring structure $(G,+,\cdot)$, because $G$ has no zero element with respect to the group operation $\cdot$. A zero element, also called an absorption element in a monoid, is an element $0\in G$ for which $g\cdot 0=0=0\cdot g$ for all $g\in G$. Cancelling $0$ from both sides of $0=0\cdot g$ would imply every ...

5

Let $K(\alpha)/K$ be an algebraic extension of fields, and suppose $\alpha$ has minimal polynomial $f(x)$ over the base field $K$, with $\deg f(x)=n$ (with $n\ge2$). Let $A\in M_n(K)$ be the companion matrix associated to $f(x)\in K[x]$. Then the there is an isomorphic copy of $K(\alpha)$, call it $L$, sitting inside the matrix algebra $M_n(K)$, which the ...

5

Please see the following paper by Bhargava and Satriano here, which contains the details of this computation and other related results. M. Bhargava and M. Satriano, On a notion of “Galois closure” for extensions of rings, Journal of the European Mathematical Society 16, 1881-1913 (2014). The aim of the authors is to define for commutative ring ...

4

The first step is to show that $$\frac{{\Bbb Z}[x]}{(5,x^3+x+1)}\simeq\frac{{\Bbb F}_5[x]}{(x^3+x+1)}$$ For this you can just define an obvious morphism and show that is injective and surjective ($\Bbb F_q$ denotes the finite field with $q$ elements). Next, you need to show that $x^3+x+1$ is irreducible in ${\Bbb F}_5[x]$. For, note that any decomposition ...

4

If you don't care about mapping $1$ to $1$, then we can embed any field $K$ into the ring $A$ of $2\times 2$ matrices over $K$, by sending $a\in K$ to $\begin{pmatrix}a & 0 \\ 0 & 0\end{pmatrix}$. This preserves addition and multiplication, and $K$ clearly this does not lie in the center of $A$. If you care about mapping $1$ to $1$, then we have to ...

3

Let $R$ be a finite nonzero ring, not necessarily commutative, not necessarily having identity, but having the property that $ab=0$ implies that one of $a,b$ is zero. Let $a$ be nonzero in $R$. Then left multiplication by $a$ is a bijection of $R$ and there exists $b$ such that $ab=a$. It follows that $ab^2=ab$ and $b^2=b\neq0$ so that $b$ is a nonzero ...

3

The best approach is to recall that $\mathbb{Z}(i)$ is a PID (Principal Ideal Domain), which is in fact an Euclidean domain with respect its usual norm. Once you notice this, you will realize that your approach using the Chinese Remainder Theorem is the correct one. The only problem is that you are factoring over $\mathbb{Z}$ instead that over $\mathbb{Z}[i]... 3 Hint$ $Note that for principal ideals:$\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$,$ $i.e.$(a)\supset (b)\iff a\mid b,\,$thus$\qquad\quad\begin{eqnarray} R/(p)\,\text{ is a field} &\iff& (p)\,\text{ is maximal} \\ &\iff&\!\!\ (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (a)\\ &\iff&\ p\ \ \text{...

3

Use (or prove) the following facts, in order. In an integral domain, an element is irreducible iff the ideal it generates is maximal among principal ideals; Therefore, in a PID, an element is irreducible iff the ideal it generates is maximal; An ideal in a commutative ring is maximal iff the quotient of the ring by that ideal is a field.

3

The answer is no. Let $\varphi : k[x,y] \to k[x,y], f \mapsto f(x,0)$ and $$A = \{ f \in k[x,y] ~|~ f(x,0) \in k \} = \varphi^{-1}(k).$$ $A$ is well known to be non-noetherian - $(y,xy,x^2y,x^3y, \dotsc)$ is not finitely generated - but it fits in the following cartesian square (the horizontal arrows are inclusions): $$\require{AMScd} \begin{CD} A @>&... 3 Let R be a k-algebra and f\colon k[x]\to R be a k-algebra homomorphism where f(x)=r is invertible. We want to see that there is a unique homomorphism \hat{f}\colon k[x,y]/I\to R, I=(xy-1), such that \hat{f}\circ p=f, where$$ p\colon k[x]\to k[x,y]/I \qquad p(x)=x+(xy-1) $$Define g\colon k[x,y]\to R by g(x)=r and g(y)=r^{-1}. Then$$ g(...

2

As noted in Rostami's answer, you do get that $J(R)$ is nil, hence is nilpotent if it is finitely generated. (Here is an alternative proof, just for fun. Since idempotents are locally 0 or 1, these assumptions imply $J(R) =$nil$(R)$ holds locally, hence globally.) But here is the main point of my answer: an example of a quasilocal ring whose maximal ...

2

Let us first see how will be the elements of $\frac{{\Bbb Z}[x]}{\langle 5, x^3+x+1\rangle}$ look like.$$\frac{{\Bbb Z}[x]}{\langle 5,x^3+x+1\rangle} =\Big\{p(x)+5q(x)+r(x)(x^3+x+1):\ p(x),q(x),r(x)\in \Bbb Z[x] \Big\}.$$ And there is a natural homomorphism between $\frac{{\Bbb Z}[x]}{\langle 5,x^3+x+1\rangle}$ and $\frac{{\Bbb Z_5}[x]}{\langle x^3+x+1\... 2 If$S=\{i_1,\dots,i_n\}$generates$I$, then$R[It]$is generated as an$R$-algebra by the elements$i_1t,\dots,i_nt$(any element of$I^n$is an$R$-linear combination of$n$-fold products of elements of$S$, so any element of$I^nt^n$is an$R$-linear combination of$n$-fold products of elements of$St$). In particular,$R[It]$is a finitely generated$R$-... 2 The converse is indeed true. If$I \subset M \subset R$is a maximal ideal of$R$, then$R/I/M/I \cong R/M$by the third isomorphism theorem for rings, so$M/I$is a maximal ideal of$R/I$. Note that the same argument shows that prime ideals in$R$containing$I$project to prime ideals in the quotient$R/I$. 2 These results are somewhat straightforward with this result: Let$R$be a commutative ring with$1 \neq 0$. The following are equivalent.$R$is a field.$R[X]$is a Euclidean domain.$R[X]$is a P.I.D.$\Rightarrow$2. is a standard result.$\Rightarrow$3. follows from the result that says "Euclidean domains are P.I.D.'s"$\Rightarrow$1.: Suppose$...

2

The first part is easy. We have $R[X]/I=R[X]/\langle X\rangle\cong R$. We are given that $I$ is maximal. So $R[X]/I$ is a field. Hence $R$ is a field. So $R[X]$ is an ED, and hence a PID.

2

Knowing isomorphism classes of ideals of a ring $A$ doesn't give too much information about quotients $A/Ax$. This is the whole point of say the classification of finitely generated modules over PIDs: in a PID every principal ideal is a free $A$-module of rank one (in particular, every two nontrivial ideals of a PID are isomorphic as $A$-modules) but the ...

1

Hint: use the fact that PID are unique factorization domains. Let $N$ be the norm, $N(1+\sqrt-5)=6$, suppose that $\sqrt-5=ab, N(ab)=N(a)N(b)=6$, set $a=u+v\sqrt-5, N(a)=1$ implies $u^2+5v^2=1$ this implies $v=0, u^2=1$, $N(a)=2$ implies also $v=0, u^2=2$ impossible, you cannot have $N(a)=3$ with a similar argument, so $N(a)=1$ this implies that $a=1$ or $a=... 1 If$p(x)=x^d+a_{d-1}x^{d-1}+\dots+a_0$, then the quotient is generated (in fact, freely generated) by$S=\{1,x,\dots,x^{d-1}\}$. Indeed, you can prove by induction that$x^n$is in the submodule generated by$S$for each$n$. For$n<d$this is trivial. For$n\geq d$, you have$x^{n-d}p(x)=0$so$x^n=-a_{d-1}x^{n-1}-\dots-a_0x^{n-d}$, which is generated ... 1 It is infact never true that$\mathfrak{m}[x]\subseteq R[x]$! All of these questions become simple when you consider the following fact. An ideal$\mathfrak{p}\subseteq R$is prime if and only if$R/\mathfrak{p}$is an integral domain. Moreover$\mathfrak{p}$is maximal if and only if$R/\mathfrak{p}$is a field. The first statement is simple, the second ... 1 The one in Wikipedia is the standard definition of a simple ring, but from what you describe, Lang's definition amounts to what is more commonly called a simple Artinian ring. For a simple ring (by which I mean the definition appearing in Wikipedia) the conditions of being right Artinian or left Artinian are equivalent. As a special case of the Artin-... 1 Counterexample to claim one. Let$A$be the abelian group of sequences of integers, let$R$be the endomorphism ring of$A$. Let$X$be the endomorphism that right-shifts a sequence: $$X(a_0, a_1, \ldots) = (0,a_0, a_1, \ldots).$$ Then$YX = 1$, where$Y$is the left-shift operator $$Y(a_0, a_1, \ldots) = (a_1, a_2, \ldots),$$ so the ideal$R \cdot X$is the ... 1 Yes, because$\varprojlim A_i$is isomorphic to one of the$A_i$in the case of a finite directed index set. In a directed set$I$, any two elements must have a maximum, hence$I$has a maximum$i_0$in the case where$I$is finite. So$i_0 \geq i$for all$i \in I$, and we have ring homomorphisms$\phi_i = \phi_{i,i_0}: A_{i_0} \rightarrow A_i$for all$...

1

Theorem: $F[x]$ is an Euclidean domain for every field $F$. the kernel of the map namely K is an ideal of the ring $Q[x]$ .As $Q[x]$ is a PID , $K=(p(x))$ , for any monic polynomial $p(x)$ of lowest degree contained in $K$. By rational root theorem $f(x)$ is irreducible in $Q[x]$. Also it is of lowesr degree monic irreducible polynomial in $k$. Because if $... 1 There is no connection. A semiring is a weakening of "ring" and "semiring of sets" is a weakening of "ring of sets" and that is all. 1 It's the smallest subalgebra of$A$containing$X$, that is, the intersection of all subalgebras containing$X$. It can also be seen as the$\mathbb{F}$-subspace of$A$generated by all products of elements of$X$. 1 If$R$is a finite commutative ring with maximal ideals$I_1,\dots,I_k$then by the Chinese Remainder Theorem there is a surjective homomorphism $$R\to R/I_1\times\dots\times R/I_k,$$ and so$\vert R\vert\geq\prod_{i=1}^k\vert R/I_i\vert$. Since each$R/I_i$has at least two elements, this gives$\log_2(\vert R\vert)$as an upper bound for the number of ... 1$ann(M)$is not trivial, there exists$x\neq 0, x\in ann(M)$, this implies$M\in ann(x)$since$M$is maximal,$M=ann(x)$. Suppose$x\in M$,$x^2=0$. Consider$f:M\rightarrow Rx$defined$f(y)=yx$, the kernel of$f$is$M$, thus$R/M\simeq Rx$. Impossible since a field does not have nilpotent elements. If$N$is another maximal element, there exists$m\in M$... 1 An obvious thing to try is to define$\psi:k[x]_x \rightarrow k[x,y]/(xy-1)$,$\psi(\frac{f(x)}{x^n}) = \overline{f(x)y^n}$(the RHS is an equivalence class in the quotient ring). You will have to prove that$\psi$is well-defined, a homomorphism, and it is the inverse of$\phi\$.

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