# Tag Info

6

The following proof is found here. One can actually show that $M_a$ is not generated by countably many elements. Assume the contrary that $M_a$ is generated by $(f_1, f_2, \cdots, f_n, \cdots)$. By normalizing we assume $|f_i(x)| \le 1$ for all $i, x$. Then define $$f(x) = \sum_{i=1}^n \frac{\sqrt{|f_i(x)|}}{2^i}.$$ Then $f \in M_a$ and there is $g_1, ... 5 Yes, the highlighted proposition is false as shown by the following counterexample: Given a field$k$, take$R=k\times k$and$I=P=k\times \{0\}$. Then$IP=I$and yet$I\neq0$. 5 Let$A$be a commutative ring. Given any subset$S\subseteq A$, there is the ideal generated by$S$: $$\langle S\rangle=\bigcap_{\substack{\text{ideals }I\\ \text{with }I\,\supseteq \,S\strut}}\!\!\!I$$ It is comprised of elements of$S$any finite sum of elements of$S$any product of an element of$A$with an element of$S$any finite sum of elements ... 5 Hint: let$R$be your integral domain, and suppose$\alpha \in R$is nonzero. Consider the ideals$\langle \alpha \rangle, \langle \alpha^{2} \rangle, \langle \alpha^{3} \rangle, \ldots$. What must be true if any of these two ideals$\langle \alpha^{k_{1}} \rangle$and$\langle \alpha^{k_{2}} \rangle$are equal? 4 Set$A=\mathbb C[X,Y]/(X^2,XY,Y^2)$. We have$\dim_{\mathbb C}A=3$, a basis is$\{1,x,y\}$, and the proper ideals of$A$are exactly the$\mathbb C$-subspaces of the vector space generated by$x$and$y$. 4 Your confusion comes from mistaking a set of elements with a single element. The product$IJ$is the set of all products of the given form for any possible$n$(including 0 for the zero element), making the definition$IJ=\left\{\sum_{i=0}^n a_i*b_i\mid a_i\in I, b_i\in J, n\ge 0\right\}$So in your example the set contains$2*3$,$4*6$,$2*3+4*6$, ... 4$\mathbb{Z}[x]$, e.g. with$I=(2)$and$J=(x)$, then $$(I+J)(I\cap J)=(2,x)(2x)=(4x,2x^2)\subsetneq (2x)=IJ.$$ 4 If we assume$M_a$is finitely generated, say$M_a = \langle f_1,f_2, \cdots, f_n \rangle$, such that$|f_k(x)| \le 1$for$x \in [0,1]$. Consider,$f(x) = \sum\limits_{k=1}^{n} \sqrt{|f_k(x)|} \in M_a$. Then, there are functions$\{g_k\}_{1 \le k \le n} \in C[0,1]$, such that,$f(x) = \sum\limits_{k=1}^{n} g_k(x)f_k(x)$say,$\max\limits_{k=1}^{n} g_k(x) ...

3

This is essentially a generalisation of the notion of coprimality in the integers. If two integers are coprime, then their gcd is 1 (and by Bézout's identity gcd can be expressed as a linear combination of those two numbers), which is the generator of the entire ring $Z$. Hence the sum of their ideals is $Z$. So while I can't say what to expect in a ring ...

3

Let $\mathfrak{q}$ be a prime of $\mathcal{O}_L$ lying over $\mathfrak{p}$, a prime in $\mathcal{O}_K$. Let $E$ be the inertial group, i.e. $$E = \{\sigma \in G : \sigma (\alpha) \equiv \alpha \mod \mathfrak{q} \}.$$ Let $D$ be the decomposition group, i.e. $$D = \{\sigma \in G : \sigma(\mathfrak{q}) = \mathfrak{q} \}.$$ Moreover, let $L_E$ and $L_D$ ...

3

We have $(I\cap J) I \subset JI = IJ$, and $(I \cap J) J \subset IJ,$ and therefore it follows that $$(I\cap J) (I + J) = (I\cap J)I + (I\cap J) J \subset IJ.$$ In a Dedekind domain, like $\mathbb{Z}$ also the converse containment is true, because intersection of ideals haves like lcm.

3

Note that $J$ is the ideal $(6,X)$: if $f(X)$ has constant term divisible by $6$ then $f(X)=Xg(X)+6k\in (6,X)$, and conversely. Then the quotient is $$\frac{\Bbb Z[X]}{(6,X)}\simeq \Bbb Z_6$$ The isomorphism is induced by the map that sends $f(X)=\sum a_i X^i$ to $\overline {f(0)}\in \Bbb Z_6$. Since $\Bbb Z_6$ is not a domain, the ideal is not prime.

3

Take $R=\mathbb{Z}$, $I=2\mathbb{Z}$ and $r_1=r_2=0$. Note that $I^2=4\mathbb{Z}\subsetneq I$.

3

What you have written is correct: $J/(7,7)J$ is isomorphic to $\Bbb Z/7\Bbb Z$ but what I think you're doing is you're confusing $J/(7,7)J$ with $\Bbb Z^2/(7,7) J$. $(7,7)J$ is maximal in $J$. It is not maximal in $\Bbb Z^2$ as you have demonstrated.

3

If $M$ is maximal, $R/M$ is not necessarily a field. This is true when $R$ is commutative and has an identity. In fact, if $R$ is not commutative, then $R/M$ might not even be a division ring (the non-commutative version of a field). Theorem. For any ideal $M$ in a ring $R$, there is a bijection between the set of ideals of $R$ which contain $M$ and the set ...

2

When you multiply by $\begin{bmatrix} a & b\\0 & 0 \end{bmatrix}$ on the right, you get just a scalar multiple by $a$. So the left ideal generated by $\begin{bmatrix} a^{\prime} &b^{\prime}\\0&0 \end{bmatrix}$ is just the collection of all scalar multiples. This is independent of whether $b^{\prime}$ is $0$ or not. For example: The left ...

2

Let $R$ be your integral domain, and take $a \in R$, $a \ne 0$. We have to prove that $a$ is a unit. Let $\mathcal I$ be the set of ideals of $R$ and consider the map $\mathcal I \to \mathcal I$ given by $I \mapsto aI$. This map is injective (*). Since $\mathcal I$ is finite, the map is surjective. Thus, $R=aI$ for some ideal $I$ and so $1=ai$ for some ...

2

The collection $\mathfrak{R}$ of all nilpotent elements in a commutative ring $A$ is indeed an ideal, called the nilradical of $A$. To prove this, consider $a, b \in \mathfrak{R}$. That is, $a^n = b^m = 0$. Then expanding $(a + b)^{m+n - 1}$ by bionomial theorem, you get a sum whoose terms are $a^p b^q$ where $p + q = m + n - 1$. Clearly one cannot have ...

2

For the first one, let $I=(m_i)$ and $J=(n_i)$, with implied summation over $i$. Then $(I: x_j)=(m_i/\gcd(m_i, x_j))$. So $((I+J): x_j)=(m_i/\gcd(m_i, x_j), n_i/\gcd(n_i, x_j))$ and $$((I: x_j)+(J: x_j))=((m_i/\gcd(x_j)+(n_i/\gcd(n_i,x_j)))=(m_i/\gcd(m_i, x_j), n_i/\gcd(n_i, x_j))$$ so assuming the right side of your equation is the ideal generated by the ...

2

The ring $R=\mathbf Z_3[x]/(x^3+2x^2+2)$ has only two prime ideals $(x+1)R$ and $(x^2+2x+2)R$, since in $\mathbf Z_3[x]$, they are the irreducible factors of $x^3+2x^2+2$. Thus $R$ is a semi-local ring of Krull dimension $0$. Its zero-divisors is the union of the two prime ideals in $\operatorname{Ass}R=\operatorname{Spec}R$.

2

If we have $I_1 \subset I_2$, then the intersections equals $I_1$, which is prime. Otherwise pick $x \in I_1 \setminus I_2$ and $y \in I_2 \setminus I_1$. We have $xy \in I_1 \cap I_2$ but none of the factors is contained in the intersection. In fact the assumption $I_2 \not\subset I_1$ is not needed, since if $I_2 \subset I_1$ holds, the intersection is ...

2

This is not true. For instance, consider the localization $R=\mathbb{Z}_{(p)}$, or more generally the localization of any domain at a height $1$ prime. The only primes in $R$ are $0$ and $(p)$, the nilradical is $0$, but the Jacobson radical is $(p)$.

2

$IJ=(6,2X,3X,X^2)$. Now, since $2$ and $3$ are relatively prime, $(2X,3X)=(X)$. Hence, $IJ=(6,X,X^2)$, and because $X^2\in(X)$, $IJ=(6,X)$.

2

Suppose $x= (i+j)a\in (I+J)\cdot(I\cap J)$, where $i\in I, j\in J$ and $a\in I\cap J$ then $x= (i+j)a=ia+ja=ia+aj\in IJ$ thus $(I+J)\cdot(I\cap J)\subset IJ$

1

The classic example for a ring that is not a polynomial ring is the ideal $(2,1+\sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$. I don't think there's going to be anything simpler if you rule out examples from polynomial rings. But personally I find the conceptually simplest example is the ideal $(x,y)$ in $R[x,y]$, where $R$ can be any non-zero ring at all. It ...

1

Even if you know the prime ideals of a commutative ring $R$ very well, there is nothing substantial you can say about the prime ideals of $R[x]$: if there were algebraic geometers would be out of business! For example it is very easy to determine the prime ideals of $R=\mathbb C[y,z]$ but essentially impossible to classify the height-$2$ prime ideals of ...

1

An ideal has to be a subgroup (of the additive group of the ring) to begin with. So if you prove that all subgroups are already ideals, you are done. PS Just correct the statement every sub-group is $mZ_n$ for $0<m<n$ to every subgroup is of the form $m Z_{n}$ for $m \mid n$.

1

Macaulay2 shows that $x^2,y^2$ is a Gröbner basis for $I$, so $\dim\mathbb C[x,y]/I=\dim\mathbb C[x,y]/(x^2,y^2)=4$.

1

Induction on $n\ge1$. For the case $n=1$ set $c_1=a_1b_1$. Inductively there exist $c_1, \dots, c_{n-1}$ such that $c_i \in (a_1, \dots, a_i) \cap (b_1, \dots ,b_i)$ and $\operatorname{ht}(c_1,\dots,c_i)=i$ for all $i$, $1\le i\le n-1$. The height of ideals $(a_1,\dots,a_n)$ and $(b_1,\dots,b_n)$ is $n$. Thus $(a_1,\dots,a_n)\nsubseteq\mathfrak p$ and ...

1

I don't have the book with me, but I don't think there are any typos. 1. It means each of (only) f_1 to f_s has multidegree $\delta$, not that every polynomial in the ideal has the same degree (only every polynomial in the list). Why? Because we want to say that in the sum $\Sigma c_if_i$ Has smaller degree because we are saying Some cancellation occurs ...

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