So I am to prove that every $\mathbb C$-subalgebra of the ring $\mathbb C[x]$ is a finitely generated $\mathbb C$-algebra.  So.. what I know is:   $\mathbb C$ is Noetherian and thus $\mathbb C[x]$ is too. Every finitely generated $\mathbb C$-algebra is a quotient of $\mathbb C[x_1,...,x_n]$.   Well, I am not even sure how to use these two things to begin with. Any help?
39281
Suppose $k$ is a field, $A$ is a $k$-subalgebra of the polynomial ring $k[x]$. Must $A$ be a finitely generated $k$-algebra?   Thanks.
14194
Assume that $M$ and $N$ are two finitely generated $R$-modules. Then $\operatorname{Hom}_{R}(M,N)$ is a finitely generated $\mathbb Z$-module and/or $R$-module (in this case, assume that $R$ is commutative)?   Note that $R$ is not Noetherian ring. Is there any counterexample (for any cases)?
64265
Are $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ and $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ isomorphic as $\mathbb{R}$-vector spaces?   I am having a very hard time at digesting tensor products and I do not know how to "compare" tensor products over different rings. My hunch is that they are not isomorphic. It is easy to see that $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ is an $\mathbb{R}$-vector space of dimension $4$. I suspect that $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ is also an $\mathbb{R}$-vector space but of lower dimension, but I have no idea how to show this or if indeed this intuition is correct.  Thank you.
73365
Let $A$ be a commutative Noetherian ring and let $I$ and $J$ be ideals of $A.$  Suppose that $I\subseteq J$ and that $A/I \cong A/J$ as rings.  I want to prove that $I=J.$    Observations so far:  1) If we drop the hypothesis $I\subseteq J,$ then the result is false:  $$\mathbb{Q}[X,Y]/(X)\cong \mathbb{Q}[X,Y]/(Y) \text{ but } (X)\neq (Y).$$  2) If $A/I\cong A/J$ as $A$-modules, then $I=J$:  $$I=\mathrm{Ann}_A(A/I)=\mathrm{Ann}_A(A/J)=J.$$  I'm now a bit stuck as to how to proceed. Any hints would be most helpful!  Many thanks :)
40012
Say I have a module $M$ over a ring $R$ that is   $1.$ finitely-generated, i.e. $\exists$ finite $X \subset M$ such that $M = \sum_{i=1}^{n} Rx_{i}, x_{i}\in X$ (but $X$ is not necessarily free)  $2.$ free, i.e. $\exists$ a basis $Y \subset M$ a.k.a. $Y$ is free and generates M, (but $Y$ is not necessarily finite).  Now I want to show, that there exists a finite basis for $M$. For vector spaces, this would obviously be trivial, but for some  $R$-Modul, I'm not sure how to proceed.   Furthermore, is every minimal generating subset of $M$ a basis?
45169
Can anyone please help me out on how to use the fundamental theorem of finitely generated abelian groups to prove that $({\mathbb{Q}},+)$ is not finitely generated?
93148
Prove that: $\forall a,b\in\mathbb{C} \exists c,d \in \mathbb{C}: c+d = a \land cd = b$   We just learned about the characteristic/minimal polynomial and diagonalization but I am not sure if it has something to do with this question.  I'd be glad for guidance!
31304
Let $A$ be a ring. How can I prove that:     $A$ is an Artinian ring $\Leftrightarrow \exists$ a faithful $A$-module which is of finite length.   I know that if a ring has a faithful $A$-module which is Noetherian, then the ring is Noetherian. Can I use this to prove the result above? I need your help to solve this problem. Thank you.
96369
Is it true that if $x_1, \dots, x_n \in \Bbb C$ are algebraically independent over $\mathbb Q$, then they are algebraically independent over $\overline{\mathbb Q}$?  I know how to prove it for $n=1$, by the contrapositive. If $P(x)=0$ and $P \neq 0$ has coefficients $a_0,...,a_n$ in $\overline{\mathbb Q}$, then $x$ is algebraic over $\mathbb Q(a_0,...,a_n)$ which is itself algebraic over $\mathbb Q$, therefore $x$ is algebraic over $\mathbb Q$, so it satisfies some relation $Q(x)=0$ where $Q \neq 0$ is a polynomial with rational coefficients.  For $n \geq 2$, I tried induction, but I was not sure how to proceed.  Thank you for your help!
68673
Let $A$ be the $\mathbb{C}$-subalgebra of $M_4(\mathbb C)$ consisting of matrices of the form   \begin{pmatrix} * &amp; * &amp; * &amp; *\\ * &amp; * &amp; * &amp;*\\ 0 &amp; 0 &amp; *&amp;0 \\ 0 &amp; 0&amp;* &amp;* \end{pmatrix}   Then how would I find the Jacobson radical $J(A)$ of $A$?    I don't really understand finding the Jacobson radical of $\mathbb{C}$-subalgebras.  Any help on this would be greatly appreciated, thank you!
30695
On Hilton and Stammbach's homological algebra book, end of chap. 2, they wrote in general $F(\mathfrak{C})$ is not a category at all in general. But I don't quite get it. I checked the axioms of a category for the image, and I think they are all satisfied. Am I missing something? Thanks.
19105
Given a collection of sets $\mathcal{C}$ and $E$ an element in the $\sigma$-algebra generated by $\mathcal{C}$, how do I show that $\exists$ a countable subcollection $\mathcal{C_0} \subset \mathcal{C}$ such that $E$ is an element of the $\sigma$-algebra, $\mathcal{A}$ generated by $\mathcal{C_0}$?    The hint says to let $H$ be the union of all $\sigma$-algebras generated by countable subsets of $\mathcal{C}$....although I don't know why.
47042
In commutative rings we have the following  Theorem. $R$ is Noetherian if and only if each prime ideal of $R$ is finitely generated.  From this Theorem I am looking for commutative rings $R$ in which every maximal ideal is finitely generated but $R$ is non Noetherian.  Question: Is there a straightforward example of a commutative ring $R$ so that each maximal ideal is finitely generated, but $R$ is non Noetherian?  Thank You
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?
21174
Let $C(\mathbb{R})=\{ f:\mathbb{R}\to \mathbb{R} \colon \ f \text{ continuous}\}\subseteq \mathbb{R}^{\mathbb{R}} $. How to prove formally that $C(\mathbb{R}) \notin \mathcal{B}(\mathbb{R}^{\mathbb{R}})$ (the product sigma algebra)?  The classic proof said that   $A \in \mathcal{B}(\mathbb{R}^{\mathbb{R}})$ iff exists $J\subseteq \mathbb{R}$ with $|J|\leq \aleph_0$ and $B\in \mathcal{B}(\mathbb{R^J})$ such that $A=B \times \mathbb{R}^{\mathbb{R}\setminus J}=\{f \in \mathbb{R}^\mathbb{R} \colon  \ (f(j) \colon \ t \in J) \in B\}$. In simpler terms,  $A$ only can have countable restrictions but the set of continuous functions have uncountable restrictions since every function on $C(\mathbb{R})$ needs to be continuous in every point $x \in \mathbb{R}$.  The last part of this proof always didn't satisfied me at all, and I don't really know how to do this proof formally.   Any help or references will be appreciated.
103140
I need some help on this exercise from A Course in Ring Theory by Donald S. Passman     Find all finitely generated graded $K[x]$-modules up to abstract isomorphism. Remember, $K[x]$ is a principal ideal domain.   The result is supposedly similar to the well-known  in the non-graded case.  So let $M$ be a finitely generated $K[x]$-module with a minimal generating set of homogeneous elements $\alpha_1, \ldots, \alpha_n$ with $d_i$ the degree of $\alpha_i$ (such a set exists because every element of M is a sum of homogeneous elements). We then get a surjective graded homomorphism $$\phi \colon K[x](d_1) \oplus \ldots \oplus K[x](d_n) \to M,\;e_i \mapsto \alpha_i,$$ where $K[x](d_i)$ denotes the graded module $K[x]$ with its grading shifted upwards by $d_i$ and $e_i = (0,\ldots,1,\ldots,0)$. So the homomorphism theorem for graded modules states that $M$ is - as a graded $K[x]$-module - isomorphic to $$\left(\phi \colon K[x](d_1) \oplus \ldots \oplus K[x](d_n)\right)/\ker(\phi).$$ How do I go on now?  Obviously $\ker(\phi)$ is a graded submodule of something like a free-graded $K[x]$-module. Is there maybe an analogue to the  for the graded case which lets me express this quotient in a nice way ?  Thanks for helping me out!
59572
This is an exercise from Rotman, An Introduction to Homological Algebra, which I've been thinking now and then for a few days and I haven't solved it yet. I've decided to ask here because it is bugging me and I don't have any friends also studying this subject (in case it is of any use, I'm an undergraduate).     Let $R$ be a commutative noetherian ring. If $A$, $B$ are finitely generated $R$-modules, then $\operatorname{Hom}_R(A,B)$ is a finitely generated $R$-module.   Here's what I've thought as of yet (maybe the problem is that I haven't been trying to apply the useful theorems...):  First of all, since $R$ is noetherian, it suffices to inject $\operatorname{Hom}_R(A,B)$ into some finitely generated module (in noetherian rings, submodules of f.g. are f.g.).  Now, since $R$ is commutative (or since $R$ is noetherian), a finitely generated $R$-module is a quotient of $R^n$ for some $n$. Then let us write $A\simeq \frac{R^n}{I}$ and $B\simeq \frac{R^m}{J}$. Now we've got:  $\operatorname{Hom}_R(A,B)\simeq \operatorname{Hom}_R\left(\frac{R^n}{I},\frac{R^m}{J}\right)$.  If I had $\operatorname{Hom}_R\left(\frac{R^n}{I},\frac{R^m}{J}\right) \hookrightarrow \operatorname{Hom}_R(R^n, R^m)$ (this is what I tend to think is the wrong direction, but a interesting question-rising path nevertheless) then it would be over since $\operatorname{Hom}_R(R^n, R^m)\simeq R^{nm}$ which is finitely generated.  So I ask myself: can I see $\frac{R^n}{I}$ as a submodule of $R^n$? Because if I could, then doing the same thing for $R^m$ and passing it to the hom, then it would be over. Now, this would be true if the sequence $$0\to I\hookrightarrow R^n\to \frac{R^n}{I} \to 0$$ split. But this (of course) doesn't always happen. But if I somehow had the third module to be projective, or the first one to be injective, then it would happen.  But why would, for example, be $\frac{R^n}{I}$ be projective?  And this is where I got stuck. I'm sorry if this is overly detailed, but I remember reading that it is always good, when asking about a textbook question, to add what you've thought up to that moment. Also, even if my thoughts don't lead to the solution, I would like to know if they are correct!