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## Hot answers tagged ideals

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 ...

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

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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. 2 How about$\mathbb{F}_p \oplus \mathbb{C}.$2$2$is in the set, but$2x$isn't. 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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