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9

Is the product of prime numbers prime?


6

The answer is no, consider $\mathbb{Z}$ for example.


6

1) The simplest example is the zero ring. It has no maximal ideal (recall that maximal ideals are required to be proper ideals). For a more interesting example, consider the ring $(\mathbb{Q},+,0)$, where $+$ is the usual addition and $0$ is the zero multiplication. It is known that $(\mathbb{Q},+)$ has no maximal subgroups, which implies that ...


5

In the ring $\mathbb{Z}[X,Y]$, what happens if we take $U=V=(X,Y)$? Does $X^2 + Y^2$ lie in $U.V$?


4

The original question suggests that you want to ask the following: Is the product of two prime ideals radical? Well, $\sqrt{\mathfrak p\mathfrak q}$ equals the intersection of all (minimal) primes containing $\mathfrak p\mathfrak q$, and therefore $\sqrt{\mathfrak p\mathfrak q}=\mathfrak p\cap \mathfrak q$. So the answer is: $\mathfrak p\mathfrak ...


4

HINT: $k[x,y]/(xy-1)$ is naturally isomorphic to the ring of fractions $k[x][\frac{1}{x}] = S^{-1}\ k[x]$, where $S= \{1,x,x^2, \ldots\}$.


3

Assume $x\in \langle g(x,y)\rangle$. That means that $x=f(x,y)\cdot g(x,y)$. Since the polynomial $x$ is irreducible in $F[x,y]$, we don't have much options left: either $g(x,y)=c$ or $g(x,y)=cx$ for some constant $c\in F$. In the former case $\langle g(x,y)\rangle=F[x,y]$, in the latter case $\langle g(x,y)\rangle=\langle x\rangle$.


3

"To divide is to contain." Thus, if $P$ divides $(I,a)$ then $P$ contains $(I,a)$ and so $P$ contains $I$. This means that $P$ divides $I$.


3

I assume that, for you, an associated prime $P$ of $R$ is a prime ideal that is the annihilator of some nonzero $m\in M$, and that you want to show that this implies that there is some submodule of $M$ isomorphic to $R/P$. Here is my hint: As part of the condition, we are given a nonzero $m\in M$. Using this, construct a homomorphism of modules $R\to M$, ...


3

Let $\;B\;$ be an ideal of $\;R\;$ containing $\;A\;$, and now define : $$B/A:= \{r+a\in R/A\;\;;\;\;r\in B\}$$ Prove that $\;B/A\;$ as defined above is an ideal of the factor ring $\;R/A\;$ . Important: don't forget that $\;A\le B\;$ ! Now, let $\;\overline B\le R/A\;$ (an ideal of $\;R/A\;$ ), and define $$B:=\{r\in R\;\;:\;\;r+A\in\overline B\}$$ ...


3

Suppose that $$W(x) =x^5 +x+1 =a(x) (x^2 +1 ) +b(x) (x^3 +1) $$ then $$1=W(1) =a(1)\cdot 0 +b(1)\cdot 0 =0$$ in $\mathbb{Z}_2 .$


2

The ideal generated by $f$ is not maximal. It is properly contained in the maximal ideal $(x_1,x_2,x_3)$.


2

For a general collection of ideals $\{A_i\}$, no. The problem is that the tuple $(r+A_i)_{i\in I}$ is only a member of $\oplus_{i\in I} R/A_i$ when all but finitely many of the elements are zero. In other words, we can define such a map exactly when our collection of ideals has the following property: for all $r\in R$, $r\in A_i$ for all but finitely many ...


2

No need any assumption on $I$ like being invertible. $I/mI$ is an $R/m$-vector space. (If $ax=0$ in a $K$-vector space, then $a=0$ or $x=0$.)


2

The zero element (which is always unique) of the quotient ring $R/I$ is the ideal $I$. In present question, $I$ consists of all elements $(3-i)x$ with $x\in\mathbb{Z}[i]$.


2

The product $\mathfrak p\cdot \mathfrak q$ of two prime ideals $\mathfrak p\cdot \mathfrak q\subset A$ in a commutative ring $A$ can only be prime if $\mathfrak p= \mathfrak q$ and $\mathfrak p= \mathfrak p^2$. If $\mathfrak p$ is finitely generated (for example if $A$ is noetherian) Nakayama then implies that $\mathfrak p=(e)$ is principal, generated by an ...


2

Let $I=(XY,(X-Y)Z)$. As you observed, $I$ has three minimal primes, one of them being $(X,Z)$. We have $I:(X,Z)=(XY,Y^2,(X-Y)Z)$. Using this we get $$I=(XY,Y^2,(X-Y)Z)\cap(X,Z).$$ Set $J=(XY,Y^2,(X-Y)Z)$. Note that $J$ has two minimal primes, namely $(Y,Z)$ and $(X,Y)$. We have $J:(Y,Z)=(X-Y,XY,Y^2)=(X-Y,Y^2)$, and therefore $J=(Y,Z)\cap(X-Y,Y^2)$. Finally ...


2

Since you mention $\quot(R_i)$, the $R_i$'s must be domains or equivalently the $I_i$'s must be prime, which I now assume. The keys to your question are then: a) The transcendence degree of $K_i=\quot(R_i)$ over $\mathbb C$ is precisely the dimension of the variety $V_i=\text {Spec}(R_i)$ defined by $R_i$. b) The $\mathbb C$-algebra $R_3$ is the tensor ...


2

Let $P$ be a prime ideal in $k[x,y]$. If $xy-1,x\in P$ then $1\in P$, contradiction. If $xy-1,x-1\in P$ then $y-1\in P$, so we can take $P=(x-1,y-1)$. (Note that $xy-1\in(x-1,y-1)$.)


2

I'll assume $1\in S$, which is known to be not restrictive. The map can be more properly written $$ P \mapsto S^{-1}P $$ where $$ S^{-1}P=\left\{\frac{x}{s}:x\in P\right\} $$ It is easy to show that if $P$ is any ideal in $R$, then $S^{-1}P$ is an ideal in $S^{-1}R$. Now, let's show that if $P$ is prime in $R$, disjoint from $S$, then $S^{-1}P$ is prime in ...


2

Prove for two and then use induction: $$I+J=A\implies \forall\,n,m\in\Bbb N\;,\;\;I^n+J^m=A$$ Because $$1=i+j\;,\;\;i\in I\;,\;\;j\in J\implies 1=(i+j)^{n+m-1}=\sum_{k=0}^{n+m-1}\binom{n+m-1}k i^kj^{n+m-1-k}$$ Observe that in the rightmost expression, we have that $$k<n\implies n+m-1-k>n+m-1-n=m-1$$ so either $\;i^k\in I^n\;$ , or ...


2

If $I+J=R$, then $I^2+J \supseteq I^2+IJ+JI+J^2=(I+J)^2=R^2=R$ and hence $I^2+J=R$. By induction we get $I^{2^n}+J=R$ and hence $I^n+J=R$. Exchanging $I$ and $J$ implies $I^n+J^m=R$.


2

In the ring $K[X,Y]$, where $K$ is a field and $X$ and $Y$ are indeterminates, we have $$\Big((X)\cap(Y)\Big)\Big((X)+(Y)\Big)\subsetneq(X)(Y).$$ (Note that $(X)\cap(Y)=(XY)$, and therefore $(XY)(X,Y)\subsetneq(XY)$; otherwise $1\in(X,Y)$, a contradiction.)


2

This is a possible answer that I have come up with. It can possibly be trimmed and edited for clarity. Let $R$ be a commutative domain and suppose that $A \subseteq R$ is an ideal maximal with respect to the property that $A^{-1} \not\subseteq R$. We show that $A$ must be prime. Take $r,s$ in $R$. Suppose that that their product $rs\in A$, but that ...


2

By hypotheses $R=I_{i}+I_{j}$, for all $i\ne j$ whence $$1 = a_j + b_j \ \ \ \ \ j = 1, \ldots n-1 $$ with $a_j \in I_n $ for all $j$ , and $b_j \in I_j $. Thus $$1 = \prod_{j=1}^{n-1} (a_j + b_j) = b_1 b_2 \cdots b_{n-1} + r $$ with $r \in I_n $ and $b_1 b_2 \cdots b_{n-1} \in I_1I_2..I_{n-1} $. This implies $$R=I_{n}+I_{1}I_{2}...I_{n-1}$$


2

If $R\ne I_{n}+I_{1}I_{2}\cdots I_{n-1}$, then there is a maximal ideal $M$ containing $I_{n}+I_{1}I_{2}\cdots I_{n-1}$, so $I_n\subset M$ and $I_{1}I_{2}\cdots I_{n-1}\subset M$. From $I_{1}I_{2}\cdots I_{n-1}\subset M$ there is $I_j\subset M$ with $1\le j\le n-1$, so $R=I_n+I_j\subset M$, a contradiction.


1

First off, these are finite rings, and a prime ideal is maximal in a finite ring. (Proof: if $R/P$ is a domain, it's a finite domain, hence a field by Wedderburn's little theorem. Thus $P$ is maximal.) So it suffices to find the maximal ideals. The maximal ideals of $\Bbb Z/p^2q\Bbb Z$ are those maximal ideals of $\Bbb Z$ containing $p^2q\Bbb Z$. You ...



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