I am trying to show that the polynomials in $\mathbb C[x,y]$ that contain no terms of the form $c y^m$ where $m&gt;0$ and $c \in \mathbb C^*$ form a non-Noetherian subring.   I know that I need to find an ideal that isn't finitely generated but I am not sure how.
123701
Here is the problem I am stuck on: Fix a field $K$ and consider the subring $A \leq K[x,y]$ generated by $K \cup \{x,xy,\dots,\}$. Show that $A$ is not Noetherian.  I figure that taking ideals $I_n = (x,xy,\dots,xy^n)$ should give an infinite strictly ascending chain, which would establish that $A$ is not Noetherian, but I cannot figure out how to show that each $I_n$ is a proper subset of $I_{n+1}$. Any help here?
117217
$d \in \mathbb{Z}$ is a square-free integer ($d \ne 1$, and $d$ has no factors of the form $c^2$ except $c = \pm 1$), and let $R=\mathbb{Z}[\sqrt{d}]= \{ a+b\sqrt{d} \mid a,b \in \mathbb{Z} \}$. Prove that every nonzero prime ideal $P \subset R$ is a maximal ideal.   I have a possible outline which I think is good enough to follow.  I think that we need to first prove that every ideal $I \subset R$ is finitely generated.   So if $I$ is non-zero, then $I \cap \mathbb{Z}$ is a non-zero ideal in $\mathbb{Z}$.   Then I need to find $I \cap \mathbb{Z} = \{ xa \mid a \in \mathbb{Z} \}$ for some $x \in \mathbb{Z}$. That way if I let $J$ be the set of all integers $b$ such that $a+b\sqrt{d} \in I$ for some $a\in \mathbb{Z}$, then if there exists a integer $y$ such that $J=\{ yt \mid t\in \mathbb{Z} \}$, then there must exist $s \in \mathbb{Z}$ such that $s+y\sqrt{d} \in I$.   Then all I need to show is that $I = ( x,s+y\sqrt{d} )$.    Now I need to derive that the factor ring $R / P$ is a finite ring without zero divisors, also finite, then since every finite integral domain is a field, every prime ideal $P \subset R$ is a maximal ideal, then I'll be done.
79013
Let $R = \mathbb{C}[t]$ be a ring of polynomials in variable $t$ with coefficients in the field of complex numbers $\mathbb{C}$ and let $$N = R[x]/(tx-t).$$  I claim that $N$ is not a flat $R$- module. If we consider the exact sequence $$0 \rightarrow(t) \rightarrow R$$ such that the ideal $(t)$ is viewed as an $R$-module.   We know that the above sequence exact iff the map $(t) \rightarrow R$ is injective. Let us assume that N is flat then this means that $N \otimes  (t) \rightarrow N \otimes R$ has to be injective. How do I go from here?   Thanks.
123329
I have a polynomial ring $R = K[x_1,\dots,x_n]$ and $N \subset \{x_1,\dots,x_n\}$. Define $R_N = K[x_i:x_i\notin N]$. If $f\not= 0$, show $(f) \cap R_N = (0) \iff f \notin R_N$.  So my entire argument hinges that $f \in R_{N^c}$ and clearly $f\in (f)$, but $f \not= 0$, so the intersection must be $(0)$ since $0 \in R_N$. I m not sure how to do the other direction. They look similar.
71263
Let $R$ be a principal ideal domain and let $P$ be a nonprincipal prime ideal of $R[x]$. I'm having trouble seeing why $P$ must contain two elements with no common divisor.    Can anyone help me? Thanks.
48195
It is known that   Now consider a ring $R$ with exactly three ideals. Take an element $x$ such that $(x)\neq (1)$. For $y\in R$ and $y\notin (x)$, there are two possible cases:   $(y)=(1)$ $(y)\neq(1)$. Then $(y)=(x)$, that is $x=ys=xts$ for some $s,t\in R$. It follows that $$x(1-ts)=0$$ which implies that $x$ is a zero divisor.   However, if there is no such a $y$ of the second case, how to prove $x$ is a zero divisor?
24791
My question is as in the title: is there an example of a (unital but not necessarily commutative) ring $R$ and a left $R$-module $M$ with nonzero submodule $N$, such that $M \simeq M/N$?  What if $M$ and $N$ are finitely-generated? What if $M$ is free? My intuition is that if $N$ is a submodule of $R^n$, then $R^n/N \simeq R^n$ implies $N=0$. It seems like $N\neq 0$ implies $R^n/N$ has nontrivial relations, so $R^n/N$ can't be free.  If $R^n/N \simeq R^n$, we'd have an exact sequence  $0 \rightarrow N \hookrightarrow R^n \twoheadrightarrow R^n/N \simeq R^n \rightarrow 0$  which splits since $R^n$ is free, so $R^n \simeq R^n \oplus N$. Does this imply $N=0$? What if we assume $R$ is commutative, or even local? Maybe Nakayama can come in handy.  I'm interested in noncommutative examples too. Thanks!
66930
Let $\mathbb M$ be one of the following rings: $\mathbb{R}, \mathbb{Q},  \mathbb{F_9}, \mathbb{C} $. Let $I$ be the ideal generated by $x^4+2x-2$. Is the ring $\mathbb M[x]/I$ a field for some of the above choices for $R$?   I have been trying to work my way around some problems involving ideals, but I get really confused when they get mixed with rings and fields. I'm stuck and don't know how to proceed.
117131
Suppose $R$ is a commutative ring. Show that every ideal of $M_n(R)$ is of the form $M_n(I)$  where $I$ is an ideal of $R$.   I have spent 30 minutes on this question and I still got nowhere. Can anyone give some hints ?
55731
I've been trying to show that   $$ K[X,Y]/(Y-X^2)\cong K[X] $$  where $K$ is a field, $K[X]$ and $K[X,Y]$ are the obvious polynomial rings over the indeterminates $X$ and $Y$ and $(Y-X^2)$ is the ideal generated by the polynomial $Y-X^2$.  Though I'm sure there's a fairly easy way to find an explicit isomorphism between the two rings, the following argument jumped out at me:   If we substitute in a value for $X$ - $x$, say, then the ideal $(Y-x^2)$ is a maximal ideal of $K[Y]$.  So the quotient $K[Y]/(Y-x^2)$ is a field; in fact, the homomorphism $K[Y]\to K:P(Y)\mapsto P(x^2)$ and is clearly surjective, so the quotient is isomorphic to $K$.    I'd like to be able to deduce from this that $K[X,Y]/(Y-X^2)\cong F[X]$, but I can't see a nice way to do it.  I know that the 'substitution' maps $P(X,Y)\mapsto P(x,Y)$ are homomorphisms, but I can't see a nice way of pulling all these homomorphisms back to the polynomial ring in two variables.    Or maybe I'm completely wrong and there is no way to turn this into a valid argument.  Can anyone help me?
23345
Lately I've been thinking to annihilator of modules and I've conjectured  a proposition I can't prove, so I'll expose my claim.     Let $A$ be a noetherian reduced (commutative) ring and let $M$ be a faithful finitely generated module on $A$. Is it true that there exists an element $m \in M$ such that $\operatorname{Ann}(m)=0$?
2364
I shall assume all rings to be commutative in this question. The impatient can scroll down to the "blockquote" to read the actual question.  Whenever we have a polynomial over a ring, it defines a function from the ring to itself by evaluation. It's reasonable to ask when two different polynomials define the same function.  From the factor theorem it follows that an $n^\text{th}$ degree polynomial over an integral domain has at most $n$ roots. Then it's easy to show this:  Theorem. Let $R$ be an infinite integral domain and let $f \in R[X]$ such that $f(a)=0$ for all $a \in R$, then $f = 0$. Proof. $f$ has infinitely many roots, so it must be the zero polynomial. $\quad\square$  For finite rings a kind of opposite situation occurs:  Theorem. For any finite ring $R$ there are polynomials over $R$ that are different but agree on all elements. Proof. There are only finitely many functions from $R$ to itself, but $R[X]$ is infinite. $\quad\square$  If we make further assumptions it's of course possible to prove more, as Pete L. Clark wrote in this post: []  Then there is the question of infinite rings that are not integral domains. It's relatively easy to come up with examples of a ring $R$ with positive characteristic and a nonzero polynomial that evaluates to the zero function, e.g.: $$ R := \bigoplus_{n=1}^\infty \mathbb{Z}/6\mathbb{Z} \quad\text{and}\quad f(X) := X^3-X.$$  The Question:     This leaves open the case alluded to in this post's title: Is there a commutative ring of characteristic $0$ (hence infinite) such that a nonzero polynomial evaluates to the zero function?
52324
I've tried to construct examples of rings that match all except one of the properties in the definition of a Dedekind domain. (This is an old number theory qual question from Berkeleys MGSA website).  The only starting point that I can think of would be  $R:=K[X_1,X_2,\ldots]$  for $K$ a field. This is clearly integrally closed as any polynomial relation is contained in $K(X_1,\ldots,X_n)$ for some large enough $n$ and $K[X_1,\ldots,X_n]$ is integrally closed being a UFD.  The problem is then to get rid of enough ideals from this ring $R$, so that its dimension would be 1, but I can't seem to figure out what to do.  Anyone have any other ideas for rings that could work as a starting point? The polynomial ring with an infinite number of variables is pretty much the only example that I know of for how to construct a non-noetherian ring that's still a domain.
20324
I'm answering this question:     Let $M$  be a module. Show that if  $M$ is not Noetherian then $M$ has a submodule $N$ such that $N$ is not finitely generated but $A$ is finitely generated whenever $N &lt; A ≤ M$.   The solution starts as follows (I'm following a given solution): Let $Y$ be the set of non-finitely-generated submodules of $M$. If $(N_α)$ is a chain in $Y$ then $Q= \bigcup_αN_α $ is in $Y$, else $Q$ is f.g., which implies $Q=N_α$ for some $α$, whence $N_α$  is f.g., contradiction.  I don't understand why if $Q$ is f.g., then $Q=N_α$ for some $α$? Am I missing some fact?  Thanks!
17207
The question is this:      Let $f:\mathbb C[x,y]\rightarrow\mathbb C[t]$ be the homomorphism that sends $x\mapsto t+1$ and $y\mapsto t^3-1.$ Determine the kernel $K$ of $f$, and prove that every ideal $I$ of $\mathbb C[x,y]$ that contains $K$ can be generated by two elements.   Solution:  I have shown that $((x-1)^3-y-1)=K.$ However, I am having trouble showing the second part. Any hints??
21262
Let $X\in\mathbb{A}^n$ and $Y\in\mathbb{A}^m$ be affine varieties (irreducible algebraic sets). Then, if we denote by $i_X$ and $i_Y$ their ideals in their respective affine spaces, we define their product in $\mathbb{A}^{n+m}$ as the set $$ X\times Y = \{(x,y):x\in X\text{ and } y\in Y\} = \{(x,y):x\in Z(i_X)\text{ and }y\in Z(i_Y)\}. $$ Using the second characterization it's easy to show that in fact $X\times Y$ is closed in $\mathbb{A}^{n+m}$ and particularly that $X\times Y = Z(I_X+I_Y)$, where $I_X$ and $I_Y$ are the vanishing ideals of $X\times\mathbb{A}^m$ and $\mathbb{A}^n\times Y$ (these ideals clearly are in bijective correspondence with $i_X$ and $i_Y$, respectively).  Then, by the Nullstellensatz, we get $I(X\times Y) = \sqrt{I_X+I_Y}$. The problem is that the references I've seen claim that in fact $I(X\times Y) = I_X+I_Y$ (, example 2.3.9 in page 25, and Georges' answer in , specially the last display).   This would only be true if $I_X+I_Y$ is radical, which I haven't been able to show for this case and know isn't true in general (see ). Are my references wrong, or what am I missing?
41674
In self studying abstract algebra and I've come upon the following problem which I could not solve directly. For any $d\in \mathbb{Z}$ we are asked to show that $\mathbb{Z}[\sqrt d]=\{a+b\sqrt{d} ~\colon a,b\in \mathbb{Z} \}$ is an integral domain. This reduces to showing that the ring has no zero divisors which I tried to do through algebraic manipulation.  I have seen two solutions to this problem that use higher level techniques. One defines conjugate and norm and proves the claim using simple consequences of these definitions. The other says that since the given ring is a subring of the field $\mathbb{C}$ it can have no zero divisors.  I don't like either approach because the first uses ideas that haven't been introduced in the text and isn't a natural place to start (for me). The second, while both true and simple, is not satisfying to me. That $\mathbb{C}$ is a field has not been established and assuming it just makes the problem trivial.   My question is whether or not naive tactics like algebraic manipulation are capable of demonstrating the result. Am I lacking in cleverness or is such an approach impossible?