Let $R$ be some integral domain and consider the following property: For all $a,x,y\in R$, if $a\mid xy$, then there exist $b\mid x$ and $c\mid y$ with $a=bc$.  Question: Is there a 'simple' classification of integral domains with this property?    Own work: Note that the property implies that irreducible elements are prime, so if $R$ satisfies the ascending chain condition on principal ideals, it must be a UFD. Since the property clearly holds in any UFD, the only 'interesting' $R$ are those that satisfy the property, but not the ascending chain condition on principal ideals.   Perhaps one can prove that the property is equivalent with irreducible elements are prime, in which case it holds precisely when $R$ is a GCD domain. This would be the type of 'simple' classification I'm looking for.
122398
In a , if $a|bc$, then $a$ can be written as $a = a_1a_2$ so that $a_1|b$ and $a_2|c$.  Is this property of a commutative ring strictly weaker than the property of being a UFD?  If so, does it have a name, or is it equivalent to some other well-known property?
112917
I wanted to ask if someone had a proof for the following claim:     Given and integral domain $A$ and a finitely generated $A$-algebra $B$, show that there exists elements $x_1,...,x_n \in B$ algebraically independent over $A$ and a non-zero element $a \in A$ such that $B[\frac{1}{a}]$ is integral over $A[\frac{1}{a}][x_1,...x_n]$.   Thank you!
9489
Let $R$ be an integrally closed domain. Let $P$ be a prime ideal of $R[x]$ with the property that $P\cap R=\left\{ 0\right\} $ and $P$ contains a monic polynomial. Prove that $P$ is principal.   Help me a hint to prove.  Thank for any insight.
87201
The book I am reading mentioned that the greatest common divisor of $a_1, \cdots a_n$ in an integral domain $R$, whenever it exists, is unique up to arbitrary invertible factors. What does "up to arbitrary invertible factors" mean?  Definition: Let $a_1, \cdots, a_n$ be nonzero elements of the ring $R$. An element $d \in R$ is a greates comon divisor of $a_1, \cdots a_n$ if it possesses the properties:   $d \vert a_i$, for $i=1, c\dots n$ $c \vert a_i$, for $i=1, \cdots n$ implies $c \vert d$.
98919
Is there a proof for $\Bbb Z$ being an integral domain? Or is it an axiom that if $a*b=0$ then $a=0$ or $b=0$ where $a$, $b$ belong to $\Bbb Z$. Is it so obvious?
50159
This problem goes as follows:     Prove that $\mathbb{Q}[x,y]/\langle x^2+y^2-1\rangle$ is an integral domain and that its field of fractions is isomorphic to the ring of rational functions $\mathbb{Q}(t)$.    The first part is straightforward, $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$, and  hence a prime ideal, as $\mathbb[Q][x,y]$ is a euclidean and hence a UFD. The quotient by a prime ideal is an integral domain.   How do I do the second part? I guess, I have to set a isomorphism by making a substitution such that only one variable remains. But I don't see how $x^2+y^2-1$ is going to help me make a choice.
7759
My lecture notes state that an 'easy' result is     If $R$ is an integral domain then an irreducible element of $R$ remains irreducible in $R[x]$, and the units in $R$ and in $R[x]$ are the same.   I can't seem to get my head around why this is the case, and what a unit in $R[x]$ means intuitively because I don't see how the units can be the same if $R$ is the coefficients of the polynomials in $R[x]$. I.e. for an unit say $\alpha \in R$ then what is the 'corresponding' unit in $R[x]$? I is it $\alpha x$ or $\alpha x^2$... or am I getting the wrong end of the stick here?
89381
Let $R$ be an integral domain and let $S$ be a ring with $R \le S \le \text{Frac}(R)$ (fraction field).   Question: Is there a multiplicatively closed subset $U \subseteq R\setminus \{0\}$ such that $S=R[U^{-1}]$ ?
50083
Disclaimer: this is a homework question. I'm looking for direction, not an answer.     Given a field $F$, show that $F[x,x^{-1}]$ is a principal ideal domain.   I'm unsure how to proceed. Would it be better to prove this directly? (ie, let $I$ be an ideal, show that $I = (f)$ for some $f \in F[x,x^{-1}]$). The proof that polynomials are a PID would involve (I imagine) division by remainder and use of degree, both of which don't seem to have obvious parallels for Laurent polynomials. Should I try and devise some parallels and mimic the proof for polynomials? Or is this overkill? (or just wrong?)  edit 1: I guess the approach I mentioned above amounts to showing that Laurent polynomials are a euclidean domain (we know that euclidean domain => principle ideal domain, so this would be sufficient) Are they a euclidean domain though? (it seems like this would have been the question if they were, instead of asking if they were a PID).  edit 2: I've spat this out, think most parts of it are correct, though it seems kind of ugly/cumbersome (but that might just be me trying to spell things out more than is needed):     Given a Laurent polynomial $f \in F[x,x^{-1}]$, define its "negative   degree" $\deg^-(f)$ to be the largest power of $x^{-1}$ that appears   in $f$.      Let $I$ be an ideal of $F[x,x^{-1}]$. Note that $\{x^{-\deg^-(f)}f \mid f \in I\} \subseteq F[x]$. Let $J$ be the   ideal in $F[x]$ generated by this set. $F[x]$ is a principal ideal   domain, so $J$ is a principal ideal and we have $J = (j)$ for some $j \in F[x]$.      We claim $I = (j)$ (now meaning an ideal of $F[x,x^{-1}]$).      Let $f \in I$. Then $x^{-\deg^-(f)}f \in J$, meaning $f = x^{\deg^-(f)}g$, where $g = x^{-\deg^-(f)}f$ is in $J$. Because $g \in J = (j)$, there exists $g' \in F[x]$ such that $g = g'j$, and thus $f = (x^{\deg^-(f)}g')j$ is a multiple of $j$, so $f \in (j)$.      Let $f \in (j)$. Then $f = gj$ for some $g \in F[x,x^{-1}]$. But note   that $j = x^{-\deg^-(f')}f'$ for some $f' \in I$, so $f = gx^{-\deg^-(f')}f'$, so $f$ is a multiple of an element of an ideal   $I$, so $f$ itself is in $I$.      This shows $I = (j)$. So an arbitrary ideal of $F[x,x^{-1}]$ is   principal, so $F[x,x^{-1}]$ is a principal ideal domain.   I kind of feel like I still don't "get" the proof (I more-or-less see how each part works with the others but I'm having trouble seeing the bigger picture), though this may be due to a poor handle on ideals in general.
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.
70246
If $R$ is a Noetherian normal domain, then it is equal to the intersection of its localizations at height one primes.      What is an example of a non-normal domain that is strictly contained in such an intersection?    I'd prefer an example that is a finitely generated $\mathbb{C}$-algebra, and all I want is a candidate suggested. I'd prefer to verify it myself.
7186
I am trying to understand the proof for the following theorem:     Let $R$ be an integral domain. If $x \in R$ is prime, then $x$ is irreducible.   Here is the proof:    I typed this a while ago and I don't understand the part where if $x | bc$, then $x=bc$? Is something wrong at this step?  Also, is the definition of prime elements where $p$ is prime if whenever $p|ab$, then either $p|a$ or $p|b$? Now I don't feel so sure. This could be the reason why I am not understanding the proof...
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I've managed [with help from the wonderful people lurking on this site :)] to prove that for an integral domain $A$, if $A$ is integrally closed, then $S^{-1}A$ is integrally closed for all multiplicatively closed subsets $S$ of $A$.  The problem that I want to apply this to is:     $A$ is integrally closed if and only if $A_{P}$ is integrally closed for all maximal ideals $P$ of $A$.   So far:  $(\Rightarrow)$ If $P$ is a maximal ideal of $A$, then $A\setminus P$ is multiplicatively closed, because maximal ideals are prime. So by the result mentioned above, $A_{P} = (A \setminus P)^{-1}A$ is integrally closed.  Update:  Attempting to prove the contrapositive.  Assume there is some $\frac{r}{s}$ in $F$ (the field of fractions of $A$) such that $\frac{r}{s}$ is integral over $A$ but not in $A$.  I feel like I haven't done hardly anything at all with your hint, but I suppose that I would want to somehow get a maximal ideal of $A$ from this $\frac{r}{s}$ such that $A_{P}$ is not maximal, thus proving the converse.
65309
Show that if $ R $ is an integral domain then a polynomial in $ R[X] $ of degree $ d $ can have at most $ d  $ roots.  Thoughts so far:  I feel like I might be missing something here. If $ R $ is an integral domain, then so is $ R[X] $. The factor theorem gives a correspondence between roots and factors. Clearly (is this clear?) a polynomial of degree $ d $ can't have more than $ d $ irreducible factors.
66357
Prove that the quotient ring $\mathbb{C}[x,y]/(x^2+y^2-1)$ is a unique factorization domain.   I am trying to prove first it is a principal ideal domain. However I am really stuck on this problem
107212
I'm wondering  when Bezout's Theorem is valid.  I know when the ring $R$ is a Euclidean Domain or a Principal Ideal Domain there is always $\gcd(a,b)$ and I can find $x,y \in R$ such that $\gcd(a,b) = ax + by.$     I know when the ring $R$ is a Unique Factorization Domain, there is always $\gcd(a,b)$. My question is, in this case can I find $x,y \in R$ such that $\gcd(a,b) = ax + by.$ ?   Now if $R$ is just a commutative ring, if there is $d \in R$ such that $&lt;a&gt;+&lt;b&gt; = &lt;d&gt; $ I know that $d = \gcd(a,b)$. In this case, can I find $x,y \in R$ such that $\gcd(a,b) = ax + by$ ?
96972
Let $B$ be a commutative domain and $A \subseteq B$ a subring. Suppose that the extension $A \to B$ is finite, i.e. $B$ is finitely generated as an $A$-module. If $I \cap A = \lbrace 0 \rbrace$ for some ideal $I$ of $B$, can we conclude that $I = \lbrace 0 \rbrace$?  I know that if $I$ contains a nonzero prime ideal, the statement holds since $A \to B$ is an integral extension, whereby no two comparable prime ideals of $B$ can have the same intersection with $A$. But does the statement still hold if $I$ does not contain a nonzero prime ideal? And what if $A \to B$ is only assumed to be an integral extension?