New answers tagged

-1

A ring in the mathematical sense is a set S together with two binary operators + and * (commonly interpreted as addition and multiplication, respectively) satisfying the following conditions: Additive associativity: For all $a,b,c$ in $S$: $(a+b)+c=a+(b+c)$, Additive commutativity: For all $a,b$ in $S$: $a+b=b+a$, Additive identity: There exists an element ...


0

We want to calculate the length of the longest ideal of finite length in the ring$$R_\mathfrak{p}/IR_\mathfrak{p}.$$For us, $\mathfrak{p} = \langle x, y\rangle$ and $I = \langle y^2, xy\rangle$. Since $R_\mathfrak{p}$ is a local ring, the obvious choice of an idea of maximal length is $\langle x, y\rangle$. This ideal is a vector space of dimension one over $...


1

Hint : try to identify $C_0/\mathrm{Im} f = C_0/\langle v_1-v_0, v_2-v_1, v_0-v_2\rangle$... Edit : quotienting by $\mathrm{Im} f$ means making $v_1-v_0 = v_2-v_1= v_0-v_2=0$, i.e. $v_0=v_1=v_2$.


1

You may be thinking of the fact that composition of functions is not commutative. But in this case, the operation is pointwise multiplication, not composition: pointwise multiplication is commutative.


3

If you had asked the question over any field and not specifically $\mathbb{K}$ (which I read as $\mathbb{R}$ or $\mathbb{C}$) I would have the following answer: Take $R= \mathbb{Q}[X,Y]$, $I=(X^2-2)$, $J=(Y^2-3)$. Then in $R/(I+J)=\mathbb{Q}[\sqrt 2, \sqrt 3]$ we have $(Y-X)(Y+X)=Y^2-X^2=1$. It's impossible (in $R/(I+J)$) to write $Y-X$ as the product of ...


2

Compute the quotient rings: $A / P_1 \cong {\mathbb Z}[X,Y]/(Y^2 - 6X^2, X, Y, 5) = {\mathbb Z}[X,Y]/(X, Y, 5) \cong {\mathbb Z}_5$. $A / P_2 \cong {\mathbb Z}[X,Y]/(Y^2 - 6X^2, X - Y, 5) \cong {\mathbb Z}[X](X^2 - 6X^2, 5) = {\mathbb Z}[X]/(5) \cong {\mathbb Z}_5[X]$. $B / Q_1 \cong {\mathbb Z}[X, T]/(T^2 - 6, X, T+1) \cong {\mathbb Z}[T]/(-5, T+1) \cong ...


3

No. The following constructs a counterexample. Let $R$ be the graded ring $\mathbb{R}[x_1, \ldots, x_5]$ and $I = (x_1, \ldots, x_5)$. In any homgeneous degree $d$, define the "pure" polynomials to be the the products of $d$ linear polynomials, and let the rank of a homogeneous polynomial $f$ be the minimum number of terms needed to express $f$ as a sum of ...


1

It means the greatest common divisor of $m$ and $n$.


0

Hint $\ $ Apply the Lemma below for $\,{\rm D} = K[y].$ Lemma $ $ Let $\rm \,D\,$ be a domain, $\rm \,\color{#c00}{0\ne }c\in D.\,$ If $\rm \ (c,x) = (f)\ $ in $\rm \,D[x]\,$ then $\rm \,c\,$ is a unit. $\rm\ \ f\ \in\ (c,x)\ \Rightarrow\ f\ =\ c\, g_1 + x\,h_1.\, $ Eval at $\rm\: x = 0\ \Rightarrow\ f(0)\ =\ cg_1(0)\ =\ cd,\:$ $\rm \,d\in D$ $\rm\ \ c\ ...


0

Hint: It is well known that, if $a$ and $b$ are coprime, and $\;ua+vb=1$ is a Bézout's relation between $a$ and $b$, the solutions to the linear dipohantine equation $$xa+yb=c\quad (c\in\mathbf Z)$$ is given by $$x=uc+kb,\quad y=vc-ka\quad (k\in\mathbf Z).$$


0

Merely from looking at it, it’s pretty clear that the kernel should be just the submodule generated by $(3,0)$ and $(0,2)$. But what’s a rigorous argument for that and how can one proceed generally? Well, there’s the structure theorem which shows that you are looking for a rank two submodule $M$ of $ℤ^2$ with elementary divisors $2$ and $3$ (because $ℤ/6ℤ \...


1

Hint: if $R$ is a domain and not a field, take $r\in R$ such that $r\ne0$ and $r$ is not invertible. Then the ideal $(r,x)$ in $R[x]$ is not principal.


0

Consider the function $\phi\colon K[x,y]\to K$ defined by $\phi(f) = f(0,0)$ (that is evaluation at $(0,0)$). Check that it is a homomorphism. So its kernel consists of polynomials of two variables vanishing at origin; that is they have no constant terms. This is not a principal ideal (in fact all maximal ideals of this ring are non principal).


2

$2a + 3b\in 6 \mathbb Z \subseteq 3 \mathbb Z$ implies $a \in 3 \mathbb Z$. $2a + 3b\in 6 \mathbb Z \subseteq 2 \mathbb Z$ implies $b \in 2 \mathbb Z$. Thus, $2a + 3b\in 6 \mathbb Z $ iff $a \in 3 \mathbb Z$ and $b \in 2 \mathbb Z$ and the kernel is $ 3 \mathbb Z \times 2 \mathbb Z$.


1

When $b$ is odd $3b$ will also be odd and so adding an even number $2a$ won't make $2a+3b$ a multiple of 6. When $b$ is even $3b$ will be a multiple of $6$ and hence, for $2a+3b$ to be multiple of $6$ we need $2a$ to be a multiple of $6$ which is the same as $a$ being a multiple of $3$. So pairs of numbers of the form $(a,b)=(3x, 2y)$ form the kernel ...


2

Let $J:=(I,x)=I+\langle x \rangle =(c)$ and $(I:x)=(d)$ . Let $i\in I \subseteq J$ , so that $i=uc$ for some $u \in R$ , then $urc \in I , \forall r \in R$ , so $u(c)=uJ \subseteq I$ , so $ux \in I$ , so $u \in (I:x)=(d)$ , so $u=vd$ , thus $i=uc=vdc \in (cd)$ , so $I \subseteq (cd)$ . For the reverse inclusion , $d \in (I:x) \implies dx \in I$ and also $dI ...


2

If $k$ is any infinite field then the ring $R=k[x]/(x^2)$ is a counterexample, since the ring $R$ has only the ideals 0, $(x)$, and $R$.


3

First of all, the quotient $M/\ker f$ may not be free -- consider $\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$. The criterion relating injectivity of $f$ to linear independence of the columns of $A$ works in the case where both source and target are free, but not necessarily otherwise. If the quotient $M/\ker f$ is free, the matrix will not necessarily be the ...


0

1) $a+bi$ is a root of $x^2-2ax+(a^2+b^2)=0$. 2) For the converse, consider the quadratic $x^2+bx+c$, where $b$ and $c$ are integers. If the roots are rational, they are divisors of $c$. Now we deal with the case the roots are non-real. From the quadratic formula, if the roots are non-real and in $\mathbb{Q}(i)$, they are of the shape $\frac{s\pm ti}{2}$, ...


11

Let $\rm\ R = \mathbb Q + x\:\mathbb R[x],\ $ i.e. the ring of real polynomials having rational constant coefficient. Then $\,x\,$ is irreducible but not prime, since $\,x\mid (\sqrt 2 x)^2\,$ but $\,x\nmid \sqrt 2 x,\,$ by $\sqrt 2\not\in \Bbb Q$


10

This is impossible: any polynomial ring over a field is a U.F.D. In such domains, irreducible elements are prime. The simplest example is the ring of quadratic integers $\;\mathbf Z[i\sqrt 5]$, which is not a U.F.D.. In this ring, we have $$2 \cdot 3=(1+i\sqrt 5)(1-i\sqrt 5),$$ so that $2$ divides the product $\;(1+i\sqrt 5)(1-i\sqrt 5)$, but doesn't divide ...


4

No polynomial can satisfy a linear recurrence except in trivial cases because the solution of linear recurrences is a combination of geometric progressions. For instance, $f(n+2)-2f(n+1)=f(n)$ for all $n \in \mathbb N$ implies that $$f(n) = c_1 (1-\sqrt2)^n+c_2 (1+\sqrt2)^n$$ where $c_1, c_2$ are parameters determined by $f(0)$ and $f(1)$. Indeed, $f(n) \...


-1

This holds good in any commutative ring. Say, we write $I+(x)=(a)$ and that $(I:x)=(b)$ for suitable $a, b \in R$. Since $(I: I + (x)) = (I:x)$, we have that $(I:a) = (b)$ as well. We claim that $I=(ab)$. Proof of Claim: Let $i \in I$. Then, we have that $i=ra$ for some $r \in R$. Since $ra \in I$, we note that $r \in (I:a)=(b)$. In particular, $r=sb$ ...


3

What would happen if your polynomial factored as a product $fg$? Let's look at the $x$-degrees of $f$ and $g$. If one of their degrees as a polynomial in $x$ is zero, then that polynomial must divide $4$, and is thus constant. So there are no non-trivial factorizations like this. The only remaining possibility is that $f$ and $g$ are both linear in $x$. ...


3

The if is false. Counter-example: Let $K$ be a field. consider the ring $R=\bigl(K[X,Y]/(XY)\bigr)_{(X,Y)}$. This ring is local by construction, has dimension $1$, and its zero divisors is the union of its two minimal prime ideals, generated by (the images of) $X$ and $Y$ respectively. Added (thanks to an idea of @rschwieb): The only if part is false ...


2

Your question is how to compute the matrix exponential $\exp(a)$ where $a$ is an element of a finite-dimensional real associative algebra $A$. (Your phrasing is not more general: you're implicitly working in the algebra $M_n(A)$.) This reduces to the case of real matrices, because the left regular representation $$A \ni a \mapsto \left( L_a : b \mapsto ab \...


1

I also was stuck on this Jacobson exercise (2.17-5) and rschwieb's proof for a commutative rng helped me out. Start by proving a statement that looks just like exercise 4 in 2.17, but it's actually reversed and I don't see why it matters whether the integer is positive. Claim: R has no zero divisors and $a,r \in R$ both non-zero. $k$ is an integer. If $ra +...


2

For $n \in \mathbb{N}$, $n$ is the norm of a Guassian integer iff $k = x^2 + y^2$ for some integers $x, y$. This is a classic question and known to be true exactly when $n$ has an even number of copies of each prime $p \equiv 3 \pmod{4}$. See this math fun fact. Primes $p \equiv 3 \pmod{4}$ are particularly significant in the Gaussian integers $\mathbb{Z}[...


1

The conjecture is false. Here is a counterexample. Suppose $R$ is a ring with the property that every $r \in R^\times$ satisfies $r^2 = 1$. Then both $(R,+)$ and $(R^\times,\cdot)$ are abelian groups with the property that every element has exponent $2$ — that is, they are vector spaces over $\mathbf{F}_2$. If $B$ is a basis for a vector space $V$ ...


1

We show the implications (a) $\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ (a). Assume (a) and let $e$ be the unit of $R$. Let $S$ be any ring such that $R\subseteq S$ is an ideal. For any $s\in S$, we have $se, es\in R$ and in fact they are equal: $$ es = e(es) = (es)e = e(se) = (se)e = se, $$ where for the second and fourth equality we have used that $...


11

In this answer (and with more detail in this answer) I gave an example, due to Sundaresan, of a compact Hausdorff space $X$ such that if $Y$ and $Z$ are the results of adding one and two isolated points, respectively, to $X$, then $X\sim Z\not\sim Y$, where $\sim$ denotes homeomorphism. Thus, $X\sqcup X\sim X\sqcup Z\sim Y\sqcup Y$, even though $X\not\sim Y$....


1

To have $R=I\oplus J$, we need to know that $I+J = R$ and $I\cap J = (0)$. $eR + (e-1)R=R$ is true for any $e\in R$, since $e-(e-1)=1$. $eR\cap (e-1)R=(0)$ is where we use idempotence. If $x=ae=b(e-1)$, then $ex = ae^2=ae=x$, but $ex=b(e-1)e=b(e^2-e)=b(e-e)=0$, and therefore $x=0$.


0

This is the Pierce decomposition, which says that for an idempotent $e\in R$ we have $$ R=eRe \oplus eR(1-e) \oplus (1-e)Re \oplus (1-e)R(1-e). $$ For commutative rings it simplifies to $R=Re\oplus R(1-e)$. One uses $e^2=e$ several times in the proof, see the above link, e.g., $e(1-e)=e-e^2=0$.


1

$Re\cap R(1-e)=\{0\}$: Indeed if $x$ lies in the intersection, it can be written both as $x=ae=b(1-e)$ for some $a, b\in R$. But then $$x=ae=ae^2=b(1-e)e=b(e-e^2)=b\cdot0=0.$$


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=...


0

Example of division in $\Bbb Z[i]$: $$\frac{11+4i}{2+5i}=\frac{42}{29}-\frac{47}{29}i$$ The nearest Gaussian integer is $2-2i$, so this is the Eculidean quotient: $$11+4i=(2-2i)(2+5i)+(-3-2i)$$ The remainder is $(-3-2i)$. Of course, this does not prove that $\Bbb Z[i]$ is an Euclidean domain, but you can see how the division is. You can try to prove ...


2

I'll answer to the first question. Since $p(x)$ and $q(x)$ are monic, they're also non-zero divisors, and the ideals $p(x)R[x]$ and $q(x)R[x]$ are both isomorphic to $R[x]$.


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 ...


4

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]...


0

A simple counterexample to (2) is the ring of upper triangular $2\times 2$ matrices over a finite field $\mathbb{F}_q$. The element $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ generates a right ideal of index $q$ but a left ideal of index $q^2$, and there is an involutory anti-isomorphism $$\begin{pmatrix}a&b\\0&c\end{pmatrix}\mapsto\begin{pmatrix}...


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 ...


4

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.


1

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

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 $...


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 ...


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

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.



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