Tag Info

Hot answers tagged

8

Let $R$ be a ring with unity, take $r\in R \setminus \{0\}$, and suppose that $(r) = R$ as ideals. In the commutative case, this means that $r$ is invertible. Why? Because every element of $(r)$ is of the form $ar$, so we have $ar=1$ for some $a\in R$. But in the non-commutative case, if we are looking at two-sided ideals, the elements of $(r)$ include ...


6

By unique prime factorization of integers, the multiplicative group $\newcommand{\Q}{\mathbb{Q}}\Q^\times$ is isomorphic to the additive group of $\newcommand{\Z}{\mathbb{Z}}(\Z/2\Z) \times \bigoplus_{p \text{ prime}} \Z$. The isomorphism comes from the fact that every rational number can be written uniquely as $\pm \prod_{p \text{ prime}} p^{a_p}$ for some ...


5

The answer is negative since $A\subset B$ flat and $B$ regular implies $A$ regular; see Bruns and Herzog, Theorem 2.2.12. But in this case $A\simeq k[a,b,c]/(ac-b^2)$, so $A$ is not regular. Edit. A simpler approach: let $I=(x^2,xy)$ and $A/I\to A/I$ be the multiplication by $y^2$. Since $A/I\simeq k[y^2]$ this is injective, but on $A/I\otimes_AB\to ...


5

If it were a PID, then every nonzero prime ideal would be maximal. But $\mathbb Z[X]/p \mathbb Z[X] = \mathbb (Z/p\mathbb Z)[X]$ is an integral domain which is not a field.


4

I'll follow a third way. Let $I=(f_1,f_2,f_3)$ and let's consider $\mathbb C[t,x]/I$. This is isomorphic to $$\mathbb C[x,x^{-1}]/(x^{-2}+x^2-2,x^{-3}+5x^2+1).$$ But $(x^{-2}+x^2-2,x^{-3}+5x^2+1)=(x^4-2x^2+1,5x^5+x^3+1)=(1)$ (why?), so $\mathbb C[t,x]/I=0$.


4

Algebraic Geometry Approach We claim $f_1,f_2,f_3$ have no common $($complex$)$ zeros. Suppose otherwise, that there exists a common zero $(u,v)$; then $$0 = f_1 - 2f_2 = (u^2+v^2-2) - 2(uv - 1) = (u-v)^2,$$ so $u-v=0$. Now we have $0 = f_2 = u^2-1$, so $u = \pm1$. But $$0 = f_3 = u^3+5uv^2+1 = 6u^3 + 1 = \pm 6 + 1 \ne0,$$ contradiction. By Corollary ...


4

HINT: Consider the map $$ \Bbb C[X]\longrightarrow\Bbb C\times\Bbb C\times\Bbb C $$ given by $q(X)\mapsto(q(0),q(1),q(2))$. Convince yourself that is a homomorphism, that is surjective, and compute the kernel.


4

Recall that: The number of finite subsets of a countable set is countable. Every ideal is finitely generated in this case. From this you should be able to derive the fact that the number of prime ideals, and indeed of all ideals, is countable in this case.


4

This follows from $(1,0)(0,1)=(0,0)$, i.e. $R\times R$ has zero divisors (in an integral domain $0\neq 1$). You only mention a special case, which is also more immediate from the one line argument I gave.


4

Hint. In general, $A[T]/(T-a)\simeq A$ for $a\in A$. (In your case consider $A=R[y]$ and $a=y^2$.)


3

Given a ring $R$ that's Noetherian as a $\Bbb Z$-module, it's automatically a (left- or right- as desired) Noetherian ring, as ideals are by definition closed under sum of elements; so every increasing chain of ideals is an increasing chain of $\Bbb Z$-submodules of $R$, and thus stabilizes. $\Bbb Q$ is indeed a Noetherian ring (it only has two ideals) ...


3

I will assume the characteristic of a non-unital ring is simply the exponent of the additive group, i.e. the smallest $n$ such that $na=0$ for every $a \in A$. Then the argument that the characteristic is prime for integral domains follows the usual way: So assume that $n=kl, \;\;\; 0<k , l <n$ is a composite number. Choose $a \in A$ such that $ra ...


3

If $R$ is a ring, then the polynomial ring $R[X]$ is the ring of all polynomials with coefficients in $R$ $$R[X] = \{a_0 + a_1 X + \ldots + a_n X^n : a_i \in R\ \forall i\}$$ with the usual addition and multiplication of polynomials: $$\sum_{i=0}^na_iX^i+\sum_{i=0}^nb_iX^i=\sum_{i=0}^n(a_i+b_i)X^i \quad\text{(some of the $a_i, b_i$ may be ...


3

If you really need a formal definition, the ring $R[x]$ of polynomials over a (commutative) ring $R$ is defined as the set of all functions $f:\mathbb N \to R$ that have finite support, that is, $f(n)=0$ for all but a finite number of $n$. (Here $\mathbb N$ contains $0$.) That set is the same as the set of all sequences in $R$ that are eventually $0$. The ...


3

An easy example would be the ring $$\mathbb{F}_2[x,y]/(x^2,xy,y^2)=\mathbb{F}_2[\tilde{x},\tilde{y}]=\{0,\,1,\,\tilde{x},\,\tilde{y},\,1+\tilde{x},\,1+\tilde{y},\,\tilde{x}+\tilde{y},\,1+\tilde{x}+\tilde{y}\}$$ where the ideal $$I=(\tilde{x},\tilde{y})=\{0,\tilde{x},\tilde{y},\tilde{x}+\tilde{y}\}$$ cannot be generated by a single element.


3

Look at $k[x]/\langle x^n \rangle$, and consider the principal ideal generated by $x+\langle x^n \rangle$.


2

Because $M$ is a maximal element of $P$, it is in particular an element of $P$, or in symbols, $M\in P$. By definition, $P$ is the collection of ideals that contain $I$. Therefore, $M$ contains $I$.


2

a) $4=\frac{2}{1}\cdot\frac21$ and similarly for $10$. On the other side, $5$ is irreducible, $6=\frac21\cdot\frac31$ and $\frac31$ is invertible in $P$, so $6$ is also irreducible, and similarly for $15$. It remains $9=3^2$ which is invertible in $P$. b) $P$ is a UFD as a ring of fractions of a UFD; see here. We also have that $P[x]$ is a UFD; see here.


2

Here's a simple example: let $\star$ be the operation defined by $ 2 \star 2 = 2 $ $ 2 \star p = p $ $ p \star p = 1 $ $ p \star q = 1 $ $ 2 \star -1 = -1 $ $ p \star -1 = 1$ $-1 \star -1 = 1$ and extended to all nonzero rational numbers by distributivity: e.g. $$\begin{align}6 \star 8 &= (2 \cdot 3) \star (2 \cdot 2 \cdot 2) \\&= (2 \star 2) ...


2

Theorem: If $R$ is a Euclidean Domain and $I, J$ ideals in $R$, then $IJ = I \cap J$ means that $I + J = (1)$. Proof: Since $R$ is a Euclidean Domain and $I, J$ are ideals in $R$, then $I = (x)$ and $J = (y)$ for some $x, y \in R$ (since ED $\implies$ PID). Note that \begin{align*} I + J &= \big( \gcd(x, y) \big)\\ IJ &= (xy) \\ I \cap J &= ...


2

For a counterexample take the first Weyl algebra $\mathbf C[p,q]$ where $pq-qp=1$, which is a subalgebra of ${\rm End}_{\mathbf C}\mathbf C[X]$. Here $p$ acts on polynomials by $p f(X)=f'(X)$ and $q$ acts by $qf(X)=Xf(X)$. It should be checked $q,p$ are not invertible, yet $pq-qp=1$ means $(p)=(q)=1$. The point is that in noncommutative cases, even principal ...


2

By definition, we have $\gcd(p,q)\mid p$ and $\gcd(p,q)\mid q$. For any $r,s\in\mathbb{Z}[i]$, if $r\mid s$ then by definition $s=kr$ for some $k\in\mathbb{Z}[i]$, hence $\mathrm{N}(s)=\mathrm{N}(k)\mathrm{N}(r)$, so we must have $\mathrm{N}(r)\mid \mathrm{N}(s)$. Therefore, $\mathrm{N}(\gcd(p,q))\mid \mathrm{N}(p)$ and $\mathrm{N}(\gcd(p,q))\mid ...


1

Consider the ring of differential operators in x. By this I mean differential operators of the form $$ \sum_{i=0}^N p_i(x) {d^n \over dx^n}, $$ where $p_i$ is a polynomial in $x$. Multiplication is given by composition of operators. As a ring, this is generated by $x$ and $y=d/dx$. This is not a division ring, but can be embedded in one. $x$ and $y$ ...


1

For sure $$\mathbb{Z}[a] \cong \mathbb{Z}[x] / \langle x^3 + x^2 - 2x + 8\rangle \cong_{\text{as } \mathbb{Z} \text{ module}} \langle 1, \alpha, \alpha^2 \rangle$$ If my calculations are correct $(\frac{\alpha + \alpha^2}{2})^2 = \frac{\alpha + \alpha^2}{2} - 2\alpha - 2$. Obviously $\langle 1, \alpha, \alpha^2 \rangle \subseteq \langle 1, \alpha, ...


1

This is a community wiki solution designed to remove this item from the unanswered queue. Edit: The question has received an answer at MO, which looks correct to me -- Todd Trimble.


1

Probably what is throwing you off is that $x\in R/P$ and not $x\in R$. Of course, $x^2=x$ in $R$ without $x$ being $1$. But in the quotient, there are no zero divisors (because you're quotienting by a prime ideal.) So you must examine $x^2-x=0$.


1

A element in an integral domain is irreductible if it is not a unit nor a product of non-units. An element $p$ of a commutative ring $A$ (not necessarily an integral domain) is prime if $A / p A$ is an integral domain, that is, is non-zero (that is $p$ is not a unit) and has no zero-divisor, that is, whenever $ab \in pA$ then $a$ or $b$ is in $p A$ which is ...


1

As Derek Holt hinted, you should consider the polynomial $h=f-g$. By hypothesis $h\ne 0$. Then consider the associated polynomial function of $h$, denoted here by $h_K$. Suppose the contrary that $h_K=0$, that is, $h_K(a)=0$ for all $a\in K$. This means that all elements of $K$ are roots of $h$. But over a field a polynomial can't have more (distinct) roots ...


1

$$\mathbb{Z}[\sqrt{-3}]/(2)\simeq\mathbb Z[X]/(X^2+3,2)$$ But $(X^2+3,2)=(X^2+1,2)$, so $$\mathbb{Z}[\sqrt{-3}]/(2)\simeq\mathbb Z[X]/(X^2+1,2)\simeq(\mathbb Z/2\mathbb Z)[X]/(X^2+1)=(\mathbb Z/2\mathbb Z)[X]/(X+1)^2$$ which is not an integral domain, so $2$ is not prime. (If you don't want to identify the quotient $\mathbb{Z}[\sqrt{-3}]/(2)$, only to show ...


1

The implication you're looking for is easy: if $a+b\sqrt{-k}=3(x+y\sqrt{-k})$, then $$ a=3x,\qquad b=3y $$ by equating the real and imaginary parts. More interesting is asking whether $3$ is irreducible. Suppose $$ (a+b\sqrt{-k})(c+d\sqrt{-k})=3 $$ with both elements in the left hand sides noninvertible. Then, by taking conjugates, $$ (a^2+b^2k)(c^2+d^2k)=9 ...



Only top voted, non community-wiki answers of a minimum length are eligible