# Tag Info

11

This case is easy: $$\{(x),(x^2),\ldots,(x^n),\ldots\}$$ The ideals of this set are all different because $x^j\notin (x^k)$ when $k>j$.

6

Yes, it does have infinitely many, basically if $f$ and $g$ have different degrees then $(f)$ is different from $(g)$. So at least one for every $n\in\mathbb N$.

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 $I\ne I^2$.

5

Recall since $R=\mathbb{C}[x,y]$ is a finitely generated $\mathbb{C}$-algebra, every quotient of $\mathbb{C}[x,y]$ is a finitely generated $\mathbb{C}$-algebra. For any prime ideal $\mathfrak{p}\subset R/I$ we have $$\text{height } \mathfrak{p} + \dim (R/I)/\mathfrak{p} = \dim R/I$$ Since $\dim R=2$ we have $\dim R/I \in\{0,1,2\}$. However, it cannot be $2$ ...

4

This answer uses the information, given in OP, that $2, 1 + \sqrt{-5}$ are irreducible elements that are not associated. Suppose $I = (a)$. If $\{2, 1 + \sqrt{-5}\} \subset I$ then $a \mid 2$ and $a \mid 1 + \sqrt{-5}$. Thus as $2, 1 + \sqrt{-5}$ are irreducible: either $a$ is a unit, or $a$ is associated to $2$ and $a$ is associated to $1 + ... 4 Usually an ideal also has to be an additive subgroup of the additive structure of the ring. Groups are never empty (they at least contain an identity), so ideals cannot be empty. 4 It even has an infinite number of prime ideals (for each degree$n$there is an irreducible polynomial of degree$n$). 4 Your thoughts are on the mark. For any ideal$I$in any ring$R$, commutative or not, we have$1_R \in I \Rightarrow I = R, \tag{1}$since in the case$1_R \in I$we have, for any$r \in R$,$r = 1_R r = r1_R \in I. \tag{2}$Now maximal ideals$M$are generally defined to be proper, that is,$M \subsetneq R$; this precludes$1_R \in M$. 4 We have$3+-2=1 \in (3,2)$, hence the generated ideal contains$1$, hence it contains$\mathbb Z$3 Let$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 ...

3

The ideal $R$ is generated by $(1)$. Remember that for this, you are allowed to multiply by any element in $R$. So for example $x^2y$ is in $(1)$ since $x^2y = (x^2y) \cdot 1$. Your statement about $I$ is not correct. For example, $I = (x^2y + x^3y^2)$ is finitely generated but not of the form you claim. To solve the exercise, you want to find some ideal ...

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 ...

3

The idea is correct. But, the details are lacking. You know that $A$ contains a unit (not the unit element $1_R$). Recall that a unit is an element $u$ that has a multiplicative inverse, that is there exists a $v$ such that $uv =1_R$. So, you have such an element $u$. It might be $1_R$ itself, but it does not have to be. And if it is not there is a ...

3

Let $R=k[x_1,\ldots,x_n]$ so I don't have to keep writing it. I'm going to construct this explicitly, although I'm going to do it in two stages. First we need a map $\psi$ from $O_P(\Bbb{A}^n)\to O_P(V)$. Define $\psi(f/g)=\frac{\overline{f}}{\overline{g}}$, where $\overline{f}$ is the image of $f$ under the canonical projection from $R$ to $R/I$ and ...

3

If $z=a+bi$ does not fulfill $2\mid a-b$, then $z-1$ does. Hence the order of the quotient is $2$. (The generator of the ideal is $1+i$).

3

(Below i am assuming that $R$ is the field of real numbers.) The point you are missing is that the set of ideals of a ring is in general a partially ordered set under inclusion. In the case of $R[x]$ you are right that each irreducible element generates a maximal ideal. But these maximal ideals are not comparable in terms of inclusions. Take for example the ...

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.

3

From the way the example is set the ideal $(2^n,2^{n-1}x,2^{n-2}x^2,...,2x^{n-1},x^n)$ is an ideal with $n$ honest generators in $\Bbb Z[x]$ because we cannot generate any of these generators by using the preceding ones.

3

The statement you want to prove is wrong, think for example of $\phi \colon \mathbf Z \to \mathbf Q$ the inklusion and $J = 2\mathbf Z$. Then $\phi^{-1}(J) = 2\mathbf Z \subseteq \mathbf Z$ is an ideal, but $J = 2\mathbf Z\subseteq \mathbf Q$ is not, as $\mathbf Q$ is a field and $2\mathbf Z \ne 0,\mathbf Q$.

3

$\sqrt{0}=\ker\pi$ is the intersection of all the prime ideals of $R/I$. Therefore, let $\mathfrak{p}\in \text{Spec}(R/I)$. Consider the projection $\pi:R/I\to R/\sqrt{I}$. You know that $\pi^{-1}(\pi(\mathfrak{p}))=\mathfrak{p}$, and therefore that $\pi(\mathfrak{p})\in\text{Spec}(R/\sqrt{I})$. You said you already knew that this implies ...

2

Let $d := \text{gcd}(x,y)$. Then there are $a,b\in R$ with $ax+by = d$, and hence $\frac dy = a\frac xy + b \in S$. If $I \subseteq S$ is an ideal, it should be easy to check that $I\cap R$ is an ideal of $R$. Is there a particular axiom you're having trouble with? Certainly, once we know this, then $I \cap R = zR$ for some $z\in R$, and we claim that $I = ... 2 You're right that a maximal ideal of$A_f$corresponds to an ideal in$A$not containing any power of$f$, but this ideal need not be maximal in$A$. A distinguished open set$D(f)$is nonempty if and only if$f$is not nilpotent (since the nilradical is the intersection of all prime ideals). Showing that the closed points are dense amounts to showing that ... 2 The comment of rschwieb made me think it's time put everything together. The statement A) The set of zero divisors of a commutative ring is the union of prime ideals. requires the existence of prime ideals in any commutative ring. According to this MO answer, prime existence is equivalent to the boolean prime ideal theorem (BPI for short). Since there ... 2 Hint. Let$K:=ker f$. Then$R/ K \cong S$. So maximal ideals of$S$correspond to maximal ideals of$R/ K$. On the other hand maximal ideals of$R/K$correspond to those maximal ideals of$R$which contain$K$. 2 You have to prove that if$r\equiv s \mod I$, then$rm=sm$for any$m\in $M$. That is equivalent to $(r-s)m=0$, which is by definition since $r\equiv s\mod I\iff r-s\in I\subseteq\operatorname{Ann}_AM$.

2

The converse is not true. Let $\phi:\mathbb{Z}\rightarrow \mathbb{Q}$ be the inclusion homomorphism. Then $\phi^{-1}(\mathbb{Z})= \mathbb{Z}$ is an ideal in $\mathbb{Z}$ but $\mathbb{Z}$ is not an ideal in $\mathbb{Q}$. Ideals are kernels of ring homomorphisms. Viewing them like this might help understand why this only works one way: the pre-image along ...

2

A $\sigma$-ideal is an ideal (with additional conditions) in the representation of the $\sigma$-algebra $\Sigma$ as a Boolean ring. Every Boolean algebra can become a Boolean ring by taking the ring addition to be $A\oplus B=(A\sqcap \overline B)\sqcup(\overline A\sqcap B)$ and the ring multiplication to be $A\otimes B=A\sqcap B$. In case of a subalgebra ...

2

I just thought that I would recast ajotatxe’s argument in a slightly different manner. Observe that we have a multiplicative function $N: \Bbb{Z}[\sqrt{-5}] \to \Bbb{N}_{0}$ defined by $$\forall (a,b) \in \Bbb{Z}^{2}: \quad N(a + b \sqrt{-5}) \stackrel{\text{df}}{=} a^{2} + 5 b^{2}.$$ Suppose that $\{ 2,1 + \sqrt{-5} \} \subseteq \langle x \rangle$ ...

2

See augmentation ideal. Obviously, augmentation map $\varepsilon$ is an epimorphism $R\to S$. Then $I=\ker(\epsilon)$ and $R/I\cong S$. So if $G$ commutative, then $I$ is a prime ideal of a commutative ring $R$. If $G$ nonabelian, then $R$ noncommutative and I don't know, what is the name for such ideals. From this clear, that $I$ maximal iff $S$ is a field ...

2

Note that $\ker(\phi) \subset R$ is an ideal, and $I +\ker(\phi)\subset$ is also an ideal in $R$. Since $\phi(\ker(\phi)) = 0\in S$, we see that $\phi(I +\ker(\phi)) = \phi(I)$. The question wants you to see that $\phi(I) = S,$ so the image of $I$ is the whode codomain iff $I +\ker(\phi) =R$.

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