Suppose that $R$ is a commutative ring and $I$ is a finitely generated ideal of $R$ such that $I^2=I$. Show that $I=Re$ for some idempotent element $e$.  I think I should use induction on the number of generators of $I$. I have proved that it's true for $n=1$:  Suppose $I=\langle a\rangle$ for some $a \in I$. Since $I^2=I$ we have $\langle a^2\rangle = \langle a\rangle$ and therefore $\exists r \in R: a=ra^2$. Multiplying both sides by $r$ and setting $e=ra$ we see that $e^2=e$ and $I=\langle e\rangle=\langle ra\rangle\subseteq \langle a\rangle =I$. Therefore, $I=Re$.  Now for $n=2$, things look difficult. Suppose $I=\langle a_1,a_2\rangle$.   Again, we have $\langle a_1^2,a_1a_2,a_2^2\rangle=\langle a_1,a_2\rangle$ therefore there exists some coefficients in $R$ such that we have:  $$a_1 = r_{11}a_1^2 + r_{12}a_1a_2 + r_{22}a_2^2$$ $$a_2 = s_{11}a_1^2 + s_{12}a_1a_2 + s_{22}a_2^2$$  After factorization:  $$a_1 = a_1(r_{11}a_1 + r_{12}a_2) + r_{22}a_2^2$$ $$a_2 = s_{11}a_1^2 + (s_{12}a_1 + s_{22}a_2)a_2$$  It isn't easy to see an idempotent element in these relations, let alone finding one that generates $I$.
118269
I am trying to understand Nakayama's lemma. It looks like some "fixed point theorem". Using Nakayama's lemma , I can easily solve the following question. I want another proof. Thanks.     Let $A$ be a commutative ring with identity, $I$ be a finitely generated ideal of $A$, such that $I^2=I$. Show that there exists an element $e\in I$ with $e^2=e$ and $eA=I$.   You can use anything (geometric interpretation are welcomed) except Nakayama's Lemma.
30362
Denote by $M_{n \times n}(k)$ the ring of $n$ by $n$ matrices with coefficients in the field $k$.  Then why does this ring not contain any two-sided ideal?    Thanks for any clarification, and this is an exercise from the notes of Commutative Algebra by Pete L Clark, of which I thought as simple but I cannot figure it out now.
43828
Let $I$ and $J$ be two ideals of a commutative ring $R$ with $1.$ Give a necessary and sufficient condition so that    $$R/IJ\cong R/I\times R/J.$$   Prove your claim. Then decide whether the following ring isomorphisms are true or not:   $$\mathbb{Q}[x]/\langle x^2-1\rangle\cong \mathbb{Q}[x]/\langle x-1\rangle\times \mathbb{Q}[x]/\langle x+1\rangle,$$   $$\mathbb{Z}[x]/\langle x^2-1\rangle\cong \mathbb{Z}[x]/\langle x-1\rangle\times \mathbb{Z}[x]/\langle x+1\rangle.$$   It has been a while since I took Abstract Algebra and I am preparing for the prelims. I am not sure how to tackle this one. Any help/suggestion/hint will be much obliged. This is what I have so far:     This is a simplified version of the chinese remainder theorem. The map   $R\to R/I\times R/J$ defined by $r\mapsto (r+I,r+J)$ is a ring homomorphism with kernel $I\cap J.$ If the ideals $I$ and $J$ are comaximal, then this map is surjective and $I\cap J=I\cdot J$, so    $$R/(I\cdot J)=R/(I\cap J)\cong R/I\times R/J.$$   My question is about the second part of the question, do I have to check whether: $\langle x-1\rangle$ and $\langle x+1\rangle$ are comaximal and $\mathbb{Q}[x]$ and/or $\mathbb{Z}[x]$ are ring homomorphisms? I believe that these ideals are comaximal in both cases but forget how to prove it.
91403
I have been having a lot of trouble teaching myself rings, so much so that even "simple" proofs are really difficult for me. I think I am finally starting to get it, but just to be sure could some one please check this proof that $\mathbb Z[i]/\langle 1 - i \rangle$ is a field. Thank you.  Proof: Notice that $$\langle 1 - i \rangle\\  \Rightarrow 1 = i\\  \Rightarrow 2 = 0.$$ Thus all elements of the form $a+ bi + \langle 1 - i \rangle$ can be rewritten as $a+ b + \langle 1 - i \rangle$. But since $2=0$ this implies that the elements that are left can be written as $1 + \langle 1 - i \rangle$ or $0 + \langle 1 - i \rangle$. Thus  $$ \mathbb Z[i]/ \langle 1 - i \rangle = \{ 0+ \langle 1 - i \rangle , 1 + \langle 1 - i \rangle\}. $$  This is obviously a commutative ring with unity and no zero-divisors, thus it is a finite integral domain, and hence is a field. $\square$
18653
I'm trying to show that $\langle1-i\rangle$ is a maximal ideal of $\mathbb{Z}[i]$.  I started by assuming there is some ideal $A$ that properly contains $\langle1-i\rangle$, and then I want to show that $1 \in A$, and then I will have that $A= \mathbb{Z}[i] $, so $\langle1-i\rangle$ is maximal.  Is this the right approach?  Thanks  Second Attempt: Because $\forall a+bi \in \mathbb{Z}[i]$, we have $a+bi = (a+b) -b(1-i)$, then we have $a+bi - (a+b) = -b(1-i) \in \langle1-i\rangle$, which implies $a+bi +\langle1-i\rangle = a+b +\langle1-i\rangle$.  So every coset is of the form $c +\langle1-i\rangle$, where $c\in \mathbb{R}$.  Furthermore, $-i(1-i)^2=2 \in \langle1-i\rangle$. Write $c = 2q + r$, where $q,r\in \mathbb{R}$ and $r=0,1$.  Then because $2 \in \langle1-i\rangle$, we have $2q\in \langle1-i\rangle$ and therefore $2q= 2q+r-r = c-r\in \langle1-i\rangle$, so $c+\langle1-i\rangle = r+\langle1-i\rangle$, and every coset is of the form $0 + \langle1-i\rangle$ or $1 +\langle1-i\rangle$. Thus, $1 +\langle1-i\rangle$ is the only nonzero element, and is invertible with itself as its inverse.  Thus, $\mathbb{Z}[i]/\langle1-i\rangle$ is a field with two elements, and by a theorem, $\langle1-i\rangle$ must be maximal.  Is this correct?
96702
This is a question related to :  "...if $R$ is a commutative ring and $I$ a nontrivial ideal of it, then $R/I$ is never free. Thus the only commutative rings $R$ for which every finitely generated module is free are those with no nontrivial ideals, and this is one of several equivalent ways to define a field."  Question 1: How do I show that if $R$ is a commutative ring and $I$ a nontrivial ideal of it, then $R/I$ is not free? Is this really true? I think the following is a counterexample:  $R = \mathbb Z \times \mathbb Z , I = \langle (1, 0) \rangle $. Define $(a,b) + (c,d) = (a+c, b+d)$ and $(a,b) \cdot (c,d) = (ac, bd)$. Then $R/I = \langle (0,1 ) \rangle$ and the basis of $R/I$ is $\{(0,1)\}$.  Question 2: Can you give me an example of a finitely generated module that doesn't have a basis?  Question 3: What's the relationship between $R$ modules and ideals $I$ of $R$? Are all finitely generated $R$-modules of the form $R/I$? Are there no others?
82371
This question is related to .  Another exercise of Dummit-Foote is to show that     Let $I = (2, x)$ be the ideal generated by $2$ and $x$ in the ring $R = \mathbb{Z}[x]$ . Show that the element $2 \otimes_R 2 + x \otimes_R x$ in $I \otimes_R I$ is not a simple tensor, i.e., cannot be written as $a \otimes_R b$ for some $a, b \in I $.    I know how to show that a tensor is not simple in tensor products like $M \otimes_R N$ where $M$ and $N$ are free $R$-modules, but here we can't use this fact; so how to do it ?
105381
Prove that if $I$ is a radical ideal and $ab\in I$, then $I=\operatorname{rad}(I+(a))\cap \operatorname{rad}(I+(b))$.      Deduce that every radical ideal in a Noetherian ring is a finite intersection of primes.   I've done the first part (showing the sets are equal), but don't know how to do the second.
54581
Let $R$ be a commutative ring and $M$ a finitely generated $R$-module. Let $s$ be the maximum number of linearly independent elements of $M$, while $t$ is the minimum number of a system of generators of $M$. Show that $s\leq t$.  It is immediate in the case of vector spaces, but my first attempt (explicit computation) didn't work here due to the fact that there are coefficients (elements of $R$) without an inverse. I tried in so much other ways spending a lot of time, and I should be very glad if you cold give me a precise, complete and rigorous proof. Thank you in advance.
3951
A very basic ring theory question, which I am not able to solve. How does one show that    $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$. Extending the result: $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$, if $a,b$ are relatively prime.   My attempt was to define a map, $\varphi:\mathbb{Z}[i] \to \mathbb{Z}/10\mathbb{Z}$ and show that the kernel is the ideal generated by $\langle{3-i\rangle}$. But I couldn't think of such a map. Anyhow, any ideas would be helpful.
116806
Given $A$ commutative ring with unity. If we have rings $A_1,\dots, A_n$such that $A\simeq A_1\times \dots \times A_n$ there are $e_1,\dots,e_n$ idempotent elements of $A$ such that $e_1+\dots+e_n = 1$ and $e_i e_j =0$ if $i\neq j$.  I'm not sure how general the proof should be written...  The isomorphic relation suggest that if $f$ defines the isomorphism then there is a $(a_1,\dots, a_n)\in \prod A_i$ such that $f(a_1,\dots,a_n)=1\in A$. Could it work thinking of the idemmpotent terms as $e_i=(0,\dots,0, a_i,0, \dots,0)?$ This should satisfy $e_i e_j =0$ if $i\neq j$. But I have two problems with it (1) I don't see why $e_i$ should idempotent as no assumptions in regard to the existence of idempotent elements in $A_i$ has been made. (2) The sum $\sum e_i=1$ has no reason to be true.
23974
Find all ideals of $\mathbb R[x] / \langle x^2-3x+2\rangle$.   I know that $\langle f(x)\rangle \subseteq \langle x^2-3x+2\rangle$ iff $\langle f(x)\rangle$ divides $\langle x^2-3x+2\rangle$. But $\Bbb R[x] / \langle x^2-3x+2\rangle$ is all the combinations $(x^2-3x+2)q(x), q(x)\in \mathbb R[x]$, so isn't the ideals all of the polynomials in $\Bbb R[x]$? Probably not, but what am I missing here?
72171
Let $R$ be a $k$-subalgebra of $S=k[x_1,x_2,\dots,x_n]$. Let $m\subseteq R$ be the ideal generated by the homogeneous elements of $R$ of positive degree. As $S$ is Noetherian, the ideal $mS$ has a finite set of generators, which may be chosen to be homogeneous elements of $m$.    Why can we choose homogeneous elements of $m$ to be the generators?
57022
Let $R$ be a PID (Principal Ideal Domain) and $x$ is an element R. Prove that the ideal $\langle x\rangle$ is maximal if and only if $x$ is irreducible.   Ok, so I know what an irreducible is. I'm thinking that this problem is asking us to set up a proof by contradiction but I can't see how. No one in my study group has any clue.
107519
If ideal $I$ of domain $R$ is free $R$-module, how to prove $I$ is principal ideal?   Is this right if $R$ is just a commutative ring?  My thought:  $(1)$ for non-zero ideal $I$ and $J$, let  $0\not=a \in I, 0\not=b \in J$, since $R$ is  domain, $0\not=ab\in I \cap J$, so $I \cap J\not =\ \varnothing$.  $(2)$ Suppose $\{e_i\}_{i\in \Lambda}$ is basis of free $R$-module $I$, then $I=\bigoplus_{i\in \Lambda}Re_i$, and every $R{e_i}$ is principal ideal generated by $e_i$. Inaccurately speaking, $I$ is generated by elements in $\bigcap_{i\in \Lambda}Re_i$. If $\Lambda$ is a finite set, then that's right. What if $\Lambda$ is an infinite set?  And what if $R$ is just a commutative ring istead of a domain? Thanks in advance!
75449
Just would like to see if I'm right.  Proof: Let $G$ be generated by $x$ (Let $G=\langle x \rangle$ for some $x$ in $G$.) We wish to show that there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$. First, we will show that for all $x^n$ in $G$, $f(x^n)=f^n(x)$. We proceed by induction.   Base case: $n=1$. $f(x*e_G)=f(x)f(e_G)=f(x)$. Now, assume that $f(x^k)=f^k(x)$ for some arbitrary positive integer k. We show that $f(x^{k+1})=f^{k+1}(x)$. Observe that $f(x^{k+1})=f(x^kx)=f(x^k)f(x)=f^k(x)f(x)=f^{k+1}(x).$   Next we will show that for all $x^{-n}$ in $G$, $f(x^{-n})=f^{-n}(x)$. We proceed by induction. Base case: $n=-1$. $f(x^{-1}*e_G)=f(x^{-1})f(e_G)=f(x)$. Now, assume that $f(x^{-k})=f^{-k}(x)$ for some arbitrary positive integer k. We show that $f(x^{-k-1})=f^{-k-1}(x)$. Observe that $f(x^{-k-1})=f(x^{-k}x)=f(x^{-k})f(x)=f^{-k}(x)f(x)=f^{-k-1}(x).$ Because $f$ is onto, $Im(f)=G'$ and therefore $Im(f)=(...,f^{-2}(x),f^{-1}(x),e_G,f(x),f^2(x),...)=G'$.  Thus, there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$ and the proof is complete by the principle of mathematical induction. Q.E.D.  I have a feeling the proof shouldn't be this long. Any suggestions for a shorter, cleaner proof?
9252
Let $R$ be a ring with unity. Let $ (a_i),i\in I $ be a family of non-zero ideals of $R$. Suppose that $R$ is a direct sum of the family $ (a_i),i\in I $ (i.e the additive group $R$ is a direct sum of the subgroups  $ a_i,i\in I $). I wish to prove that in this case $I$ must be a finite set.   I will explain the argument which I gave until the point of my question:   Take the identity element $ 1\in R $. Since $R$ is the direct sum of the $a_i$, there exists a finite subset $I_0 \subseteq I$ and elements $e_i \in a_i \setminus 0 , i \in I_0$ such that $1 = \sum_{i \in I_0} e_i $. Then, for every $x \in R$ we have $x = x.1 = \sum_{i \in I_0} xe_i \in \sum_{i \in I_0} a_i $ since each $a_i$ is an ideal, and so $R = \sum_{i \in I_0} a_i $.  My idea now is to prove that one must have $I = I_0$, but I have a feeling that I'm missing something trivial. Can anyone help me with this final step if possible? Thank you in advance.  On a final note, I'm assuming that each $a_i$ is an ideal, but as far as I was informed, this is also true for left ideals (or  right ideals).