I'm trying to solve Exercise 8.2.4 in Dummit &amp; Foote   Let $R$ be an integral domain. Prove that if the following conditions hold then R is a Principal domain (1) any two nonzero elements $a$ and $b$ in $R$ have a greatest common divisor which can be written in the form $ra+sb$ for some $r,s$ $\in R$, and (2) if $a_1, a_2 , \cdots ,a_n, \cdots $ are nonzero elements of $R$ such that $a_{i+1} | a_i $, then there is a positive integer $N$ such that $a_n $ is a unit times $a_N$ for all $n \ge N$  I tried to understand the solution in  But I am stuck from the outset. In the solution I put a link above, it claims then for any nonzero ideal $I$ in the ring $R$ of the problem, when we put partial order on $I$ as divisibility relation like  $$x \leq y \; \text{if} \; y|x $$  then each chain $C$ in $I$ with such partial order is a countable sequence $\lbrace C_i \rbrace$ with $C_{i+1} | C_i$ But why should this hold? Does divisibility partial order on (an ideal of) an integral domain make each chain be countable? This seems highly nontrivial and even seems wrong.. Any comment on the claim in the above link or on the problem itself will be a great help to me. Thanks in advance.
121367
This is just a textbook problem from Dummit and Foote, but the issue is that our class barely touched on PIDs and the preceding material, so I don't really know or understand much.  Anyway,     Let $R$ be an integral domain.  If $R$ satisfies the following:           i. any two nonzero $a,b \in R$ have a gcd in the form of $d=ra+sb, r,s \in R$          ii. If $a_1, a_2,...$ are nonzero in $R$ such that $a_{i+1}$ divides $a_i$ for every $i$ then $\exists N \in \mathbb{Z}$ such that $a_n$ is an unit times $a_N$  $\forall n \ge N$,          prove that $R$ is a principal ideal domain.   From what I've gathered, the second condition means that $(a_1) \subset (a_2) \subset...$ and $(a_N) = (a_n)$ $\forall n \ge N$.  However, I'm not sure what I should do now. I'm worried that I'm lacking the necessary context here especially since we used Artin pretty much all semester.  In short, my question is: how should I proceed with this proof given what I know?
21686
I like to be thorough, but if you feel confident you can skip the first paragraph.  Review: A ring is a set $R$ endowed with two operations of + and $\cdot$ such that $(G,+)$ is an additive abelian group, multiplication is associative, $R$ contains the multiplicative identity (denoted with 1), and the distributive law holds.  If multiplication is also commutative, we say $R$ is a commutative ring. A ring that has no zero divisors (non-zero elements whose product is zero) is called an integral domain, or just a domain.  We want to show that for a domain, the equation $x^2 = 1$ has at most 2 solutions in $R$ (one of which is the trivial solution 1).    Here's what I did:   For simplicity let $1,a,b$ and $c$ be distinct non-zero elements in $R$.  Assume $a^2 = 1$.  We want to show that letting $b^2 = 1$ as well will lead to a contradiction.  So suppose $b^2 = 1$, then it follows that $a^2b^2 = (ab)^2 = 1$, so $ab$ is a solution as well, but is it a new solution?  If $ab = 1$, then $abb = 1b \Rightarrow a = b$ which is a contradiction.  If $ab = a$, then $aab = aa \Rightarrow b = 1$ which is also a contradiction.  Similarly, $ab = b$ won't work either.  So it must be that $ab = c$.  So by "admitting" $b$ as a solution, we're forced to admit $c$ as well.  So far we have $a^2 = b^2 = c^2 = 1$ and $ab = c$.  We can procede as before as say that $(abc)^2 = 1$, so $abc$ is a solution, but once again we should check if it is a new solution.  From $ab = c$, we get $a = cb$ and $b = ac$, so $abc = (cb)(ac)(ab) = (abc)^2 = 1$.  So $abc$ is not a new solution; it's just one.  At this point I'm stuck.  I've shown that it is in fact possible to have a ring with 4 distinct elements, namely $1,a,b$ and $c$ such that each satisfies the equation $x^2 = 1$ and $abc = 1$.  What am I missing?
43929
How would one prove that $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ is a principal ideal domain (PID)? It isn't a Euclidean domain according to .
15896
Here is the theorem and proof in my textbook.     Thm.  Let $P$ be an ideal of a commutative ring $R$ with identity $1$.  Then $P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain.   Proof.   Suppose that $P$ is a prime ideal of a commutative ring $R$ with $1$.  Then $P$≠$R$ implies $1+P≠0+P$. Hence $R/P$ is a commutative ring $R$ with identity.  Assume that $(a+P)(b+P)=0+P$.  Then $ab+P=0+P$ and $ab∈P$.  By the definition of a prime ideal P we get $a∈P$ or $b∈P$.  That is, $a+P=0+P$ or $b+P=0+P$.  Thus $R/P$ is an integral domain.   Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$ and $R/P$ is a commutative ring $R$ which has no zero divisors.  Hence $P≠R$.  Assume $ab∈P$.  Then $ab+P=0+P$ and $(a+P)(b+P)=0+P$.  Since $R/P$ is an integral domain, we get $a+P=0+P$ or $b+P=0+P$.  So $a∈P$ or $b∈P$.  Thus $P$ is a prime ideal.    I am having problems understanding proof with "Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$.".  I know that $R/P$ has an identity. (cause it is ID)  But, why doesn't "$P$" have an identity?  Isn't there a case $P=R$?  Please help me understanding. I think $P≠R$ should be with condition of Thm..  Thank you in advance.
54559
If $F$ is a field, then $F[x]$ is a principal ideal domain.  Conversely, if $F[x]$ is a principal domain does $F$ have to be necessarily a field?  My Thoughts: Suppose instead of $F$, we take the set of polynomials $R[x]$ over a commutative ring $R$ with unity. Then, suppose $I$ is an ideal of $R[x]$. Let $g(x) \in I$ such that $g(x)$ is the polynomial of the lowest degree in $I$. Then: $ \langle g(x) \rangle \subseteq I .......... (1)$ Let $f(x) \in I$. Then $f(x) = p(x)g(x) + r(x)~~|~~p(x),r(x) \in R[x], \deg r(x) &lt; \deg g(x)$ Since, $I$ is an ideal $\implies f(x) - p(x)g(x) = r(x) \in I$ But, $g(x)$ is of the lowest degree in $I \implies r(x) = 0 \implies f(x) \in \langle g(x)  \rangle \implies I \subseteq \langle g(x)  \rangle ......(2)$ Then from $(1),(2) : I = \langle g(x) \rangle$ Does the Presence of zero divisors in $R[x]$ really make a difference? The only advantage I see is that if there are no zero divisors in $R[x]$ then $I=\{0\} \implies I= \langle 0  \rangle$. But, every ideal contains the zero element. Why is there the condition of a field specifically given in textbooks for $F[x]$ to be a principal ideal domain? Thank you for your help.
90688
Let $F$ be a field, and $F[x,y]$ be a ring of polynomials in two variables. Is $F[x,y]$ a Principal Ideal Domain?   Also show that $F[x,y]/(y^2-x)$ and $F[x,y]/(y^2-x^2)$ are not isomorphic for any field $F$.
14965
Let $R$ be commutative ring with identity that contains a field $K$ as a subring. If$ $R is a finite dimensional vector space over the field $K$, prove that every prime ideal in $R$ will be maximal.   My idea was to prove the integral domain $R/p$ (if $p$ is a prime ideal) was a field as for any ideal $P$, $R/P$ will be a field iff $P$ is maximal in $R$. But how can I use the fact that $R$ is finite dimensional over $K$? I don't understand.
9138
One has the following relations for a domain $R$:  $R$ GCD domain $\Rightarrow$ All irreducible elements are prime  $R$ PID $\Rightarrow$ $(R$ GCD domain $\land$ $R$ statisfies ACCP$)$  $R$ UFD $\Leftrightarrow$ $(R$ GCD domain $\land$ $R$ satisfies ACCP$)$  $R$ satisfies ACCP $\Rightarrow$ $(R$ UFD $\Leftrightarrow$ All irreducible elements are prime$)$  $R$ GCD domain $\Rightarrow$ $(R$ UFD $\Leftrightarrow$ $R$ atomic domain$)$  So what is the difference between a GCD domain and a domain where all irreducible elements are prime? What is the "weakest" predicate $P$ such that  $R$ GCD domain $\Leftrightarrow$ $($All irreducible elements are prime $\land$ $P(R))$ ?
123588
So I know that for it to be an integral domain it has to have the following properties:    Commutative Has multiplicative identity No Zero-Divisors    and if $p(x)$ is reducible it can be written as $p(x) = (x-a)q(x)$ but I can't seem to link it together to prove that it violates one of the three properties.
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 $\gcd(a,b) = ax + by$ ?
34163
Let $R$ be subring of integral domain $S.$ Suppose $R$ is $\text{PID}.$ Let $a\in R$ be a greatest common divisor of $r_1,r_2$ in $R$. ($r_1,r_2 \in R$, not both zero). Could anyone advise me on how to prove $a$ is greatest common divisor of $r_1,r_2$ in $S?$   Hints will suffice, thank you.
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 $c \vert d$.
9368
Theorem: If $R$ is an integral domain, then $R[x]$ is an integral domain as well.  Proof: Let $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$ and $g(x)=b_0+b_1x+b_2x^2+\cdots+b_mx^m$ be two elements of $R[x]$. Let $0\leq i\leq n$ and $0\leq j\leq m$ such that $i$ & $j$ are the smalles numbers such that $a_i\ne 0$ & $b_j\ne 0$. $f(x)\cdot g(x)=\sum_{i=0}^{n+m}{c_ix^i}.$ $c_{i+j}=\overbrace{a_{i+j}b_0+a_{i+j-1}b_1+\cdots +a_{i+1}b_{j-1}}^{=0} +\overbrace{a_ib_j}^{\ne 0}+\overbrace{a_{i-1}b_{j+1}+\cdots+a_0b_{i+j}}^{=0}$ $=a_i+b_j\ne 0\Longrightarrow f(x)g(x)\ne 0$    I don't know why $a_{i+j}b_0+a_{i+j-1}b_1+\cdots +a_{i+1}b_{j-1}=0$ and $a_{i-1}b_{j+1}+\cdots+a_0b_{i+j}=0$.
73097
If $F$ is a field, then $F[x]$ is a principal ideal domain   By a previous theorem, we know that $F[x]$ is an integral domain. Now, let $I$ be an ideal in $F[x]$. If $I = \{0\}$, then $I = \langle 0 \rangle$. If $I \neq \{0\}$, then among all the elements of $I$, let $g(x)$ be one of minimum degree. We will show that $I = \langle g(x) \rangle$. Since $g(x) \in I$, we have $\langle g(x) \rangle \subseteq I$. Now let $f(x) \in I$. Then, by the division algorithm, we may write $f(x) = g(x)q(x) + r(x)$, where $r(x) =0$ or deg $r(x) \lt$ deg $g(x)$. Since $r(x) = f(x) - g(x)q(x) \in I$, the minimality of deg $g(x)$ implies that the latter condition cannot hold. So, $r(x) = 0$ and, therefore, $f(x) \in \langle g(x) \rangle$. This shows that $ I \subseteq g(x)$.  My question is: What is a polynomial of minimum degree in this context?  And how does this minimality imply that deg $r(x) \lt$ deg $g(x)$ cannot hold?
92791
Let $R$ be a principal ideal domain, and $a, b \in R$ with $a^m = b^m$ and $a^n = b^n$ for $m, n \in \mathbb{N}$, and $\gcd(m, n) = 1$. I now want to show that we then already have $a = b$.   I think the statement is easily seen to be true if $a = 0$ (or $b = 0$), since if $a = 0$ and $a^m = b^m$, then $a = b$ since $R$ doesn't contain any nontrivial zero divisors, so $b ≠ 0$ already implicates $b^m = b \cdot b^{m-1} ≠ 0$.  I don't really know how to prove the statement for the general case, where I can't directly apply the fact that $R$ doesn't have any zero divisors except $0$. I most likely have to use the fact at some point that $gcd(a, b) = 1$, but I'm not sure how.
21154
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.
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.