If an integral domain $R$ doesn't satisfy the ascending chain condition for principal ideals, then factorization of an element of $R$ (into irreducible elements) might fail: if you choose the wrong elements, the factorization process might not terminate. However, can it be the case, that in some such domain, a factorization still exists for all elements (or at least for some element where the factorization can fail)? It could be that maybe if you choose the correct divisors then the factorization terminates. The only example i know of a integral domain that doesn't satisfy ACCPI is $P/I$ where $P$ is the ring of polynomials of countable many variables $(x_k)$ over an integral domain $R$ and $I$ is the ideal generated by the relations $x_{k+1}^2 = x^k$. I thought of maybe trying to adjoin some new elements $y_k$ such that $x_k = y_k^{p(k)}$ where $p(k)$ is the kth prime starting at $3$ and hope those $y_k$'s turn out to be irreducible and that the result still doesn't satisfy ACCPI, as then an element $x_i$ would maybe have non-terminating factorization if you choose $x_k$'s to factorize it but would terminate if you factor by $y_i$. However this ring seems very complicated to work with and i don't have strong reasons to believe it will work.
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I recently just completed an exercise which states an atomic domain $A$ is a UFD $\iff$ it satisfies the ascending chain condition on principal ideals. This is all well and good, but because this is an if and only if statement, that seems to suggest there are atomic domains that DON'T satisfy the ACCP. Does anyone know of such an example? I would imagine you need something like the algebraic numbers but I am unsure.
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I'm working on the following problem...     Suppose that $A$ is a Dedekind domain with fraction field $K$.  $L/K$ is a finite separable extension of $A$ of degree $n$, and $B$ is the integral closure of $A$ in $L$ (so that $B$ is also a Dedekind domain).  Suppose that $\mathfrak{p}$ is a nonzero prime ideal of $A$, and $B\mathfrak{p}=\mathfrak{q}^{e_1}\cdots \mathfrak{q}_r^{e_r}$ is the prime factorization of $B\mathfrak{p}$.  Show that $$e_1f_1+\cdots +e_rf_r=n$$ where $f_i=[B/\mathfrak{q}_i:A/\mathfrak{p}]$.    My first question is, why does integrality over $A$ imply that $B$ is also a Dedekind domain? My second question is, well, how to do the problem.  I need an outline so that I know what I'm looking for. What results will be needed here?
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Fraleigh Section 23 12.Give an example to show that a factor ring of an integral domain may be a field 13.Give an example to show that a factor ring of an integral domain may have divisors of $0$ 14.Give an example to show that a factor ring of a ring with divisors of 0 may be an integral domain.  For 12, $Z/2Z$ works For 13, $Z/4Z$ works For 14, $4Z/8Z$ ( isomorphic to Z$_2$?) works. -&gt; I just figured out 4Z is not a ring with divisors of 0... I'm unsure about my answer for 14. Is it correct? And any other examples for 12,13,14?
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Possible Duplicate:          I am trying to see whether the ideal generated by irreducible element in a principal ideal domain (PID) is maximal ideal.  Suppose r is irreducible in a PID say D  Let I be an ideal of D containing (r) the ideal generated by r   Since  D is a principal ideal domain, there exist s in D such that I=(s), therefore (r) is a subset of (s).  So,  r=st , for some t in  D   but r is irreducible, this implies that s or t is a unit.  If s is a unit then I= (s)= D .  If t is a unit then (r)= I=(s). But I am not sure this is true, because I do not have reason for saying (r)= I=(s), so that I can conclude and say (r) is maximal.  I need a little help for this. Thanks
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Let $S$ be a subring of $\mathbb{Q}$ containing $\mathbb{Z}$. Prove that it is a principal ideal domain.  So here is what I tried. Take any ideal $I\subset S$. Take any two elements, say $a=p/q, b=r/s$ in $I$. Rationalize denominators (we can do that since $\mathbb{Z}\subset S$. Now we have $aqs=p=K(r)=K(bqs)+R$, that is, just apply the Euclidean Algorithm. Now, $a-Kb=R/qs\in I$, so take the set of all such remainders $R/qs$.   There are two possibilities. Either this set has a smallest positive element, or it does not.   Suppose it has a smallest such element, call it $x=m/n$. Let $a=p/q\in I$. We show $a=lx$. For suppose not. Rationalizing denominators with $n, q&gt;1$, $p=Lm+Z$, so then $Z&lt;m$ so $Z/qn&lt;m/n=x$, giving a contradiction.   Now though, I am stuck on the case where there is no such smallest element :(  Any thoughts?  EDIT: For those marking it as a duplicate: The first answer in the linked question, as mentioned in the comments, feels somewhat unnatural and requires all this extra machinery. On the other hand, the second answer given is incomplete: The poster says "Now show that the ideal generated by $t=\frac{t}{1}$ in R is the ideal you started out with." But if $\frac{r}{s}\in I$, then there exists $k$ such that $r=kt$, so $\frac{r}{s}=\frac{k}{s}\cdot \frac{t}{1}$, but $\frac{k}{s}$ is not necessarily in $S$.
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In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal  Attempt: Let $R$ be the principal ideal domain. A principal ideal domain $R$ is an integral domain in which every ideal $A$ is of the form $\langle a \rangle = \{ar~~ | ~~r \in R\}$  Let $ \langle a \rangle $ be a prime ideal $\implies R/A $ is an integral domain.  A finite integral domain is a field . Hence, if we prove that $R/A$ is finite, then $R/A$ is a field $\implies A$ is a maximal ideal.  Now, $\langle a \rangle = \{ar~~ | ~~r \in R\}$  Since, R is an integral domain, there are no zero divisors and cancellation is allowed $\implies ar_1 = ar_2 \implies r_1=r_2 \implies ar_i$ maps to a different member of $R$ for each different $r_i \implies \langle a \rangle$ represents the elements of $R$ in some random order.  $\implies \langle a \rangle = R$ and hence, $R/A \approx {0}$ is finite and hence $A$ is maximal.  Is my attempt correct?
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In Milne's , he mentions     Remark 3.14   The local ring $\mathcal{O}_{V,P}$ is an integral domain if $P$ lies on a single irreducible component of $V$...       ... if $P$ lies on more than one irreducible component of $V$, then $\mathcal{O}_P$ contains more than one minimal prime ideal, and so the ideal $(0)$ can't be prime.   Why does having more than one minimal prime ideal mean $(0)$ cannot be prime? I'm probably missing something obvious...    Thanks in advance ^_^
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I am looking for an integral domain in which we have an infinitely ascending chain of ideals.   Clearly, this can't be a PID. Also, I am looking for examples other than infinite dimensional fields, integral domains like $k[x_1,x_2,\dots]$.
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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.
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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
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I need to find an example of a ring that is not a PID but every ideal is principal. I know that $\mathbb Z\times\mathbb Z$ is not an integral domain, so certainly is not a PID, but here every ideal is principal. I already proved that if R and S are ring every ideal in R x S is I x J with ideals in the original ring. But I cant follow from that that $\mathbb Z\times\mathbb Z$ has only principal ideals.      Explicitly, if $I$ is an ideal $(a)\times(b)$ which would be its generator $(c,d)$ in $\mathbb Z\times\mathbb Z$?    Thanks.
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There is the next characterization for valuation domains:  Let $D$ be an integral domain and $K$ its field of fractions. The following are equivalent:   For every nonzero $x$ in $K$, either $x$ in $D$ or $x^{−1}$ in $D$. The ideals of $D$ are totally ordered by inclusion. The principal ideals of $D$ are totally ordered by inclusion (i.e., the elements in D are totally ordered by divisibility.) There is a totally ordered abelian group $Γ$ (called the value group) and a surjective group homomorphism (called the valuation) $v :$ $K^×$ $→ Γ$ with $D = \{ x ∈$ $K^×$ | $v(x)$ ≥ $0 \}$ $∪$ {0}.   The equivalence between 1, 2 and 3 are clear, and I understand that the fourth can be found in Krull's article (1936), but I don't speak german and I can't find a proof anywhere. Any link or hint will be much appreciated.
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Let $D$ be a domain which is not a field. If there exists $b \in D - \{0\}$ such that for all $a \in D - \{0\}$, $a|b$, then is it true that $b = 0$?   This is certainly true if one has a UFD with infinitely many irreducibles (that are not associates) or a Jacobson domain which is not a field (or more generally a domain where the intersection of all non-zero prime ideals is the zero ideal), but I can't seem to be able to prove this for a domain in general or find a counterexample.  If this statement is false for an arbitrary domain, then is there some additional weak hypothesis on the domain which makes the above statement true?   I was actually trying to answer a question asked yesterday on MSE, which was the following:  If $D$ is a domain which is not a field and $Q = Frac(D)$, then $Hom_D(Q,D) = \{0\}$. Note that if $\varphi$ is any such $D$-linear map, then for all $a \in D- \{0\}$, $a\varphi(1/a) = \varphi(a/a) = \varphi(1)$. Thus, I get for all $a \in D - \{0\}$, $a|\varphi(1)$, and I want to conclude that $\varphi(1) = 0$, but cannot.
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The proof of the following proposition is not completely clear to me. I get everything up until the bold part and I have a feeling some crucial steps are omitted, can anybody help clear this up?     Let $R$ be an integral domain. If every two elements of $R$ have a greatest common divisor, then every irreducible element in $R$ is prime.   Proof:  Consider an irreducible element $p \in R$ and $x,y \in R$ such that $p\vert xy$. Suppose now that $py$ and $xy$ have a greatest common divisor $z$ in R. We want to conclude from this that $p \vert x$ or $p \vert y$. This is obvious if $xy = 0$, so we may assume that $xy \neq 0$. Then $z \neq 0$. As both $p$ and $y$ divide each of $py$ and $xy$, we have that $z = pu = yv$ for certain $u,v \in R$. Using the cancellation law with $\boldsymbol z \boldsymbol \neq \boldsymbol 0$, we obtain that $\boldsymbol v \boldsymbol \vert \boldsymbol p$. As $p$ is irreducible, either $v \in R^\times$ (i.e. the set of invertible elements of $R$) or $v \sim p$ (i.e. $Rv = Rp$). If $v \sim p$, then $p \vert x$. If $v \in R^\times$, then $p$ divides $v^{-1} pu = v^{-1} z=y$.
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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?
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If R was a unique factorization domain, can we deduce that for a nonzero element d in R, d has a finite number of divisors?   I need this in solving this question  "If R is a unique factorization domain then there are only finite number of distinct principal ideals that contain the ideal (d)."  My idea to solve it: if a divides d then (d) is contained in (a), so if we have finite divisors then we have finite principal ideals. Am I wrong?
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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.