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15

If $\Bbb Z[X]$ were a principal ideal domain, then its quotient by the ideal generated by$~X$, an element that is obviously irreducible, would have to be a field. But it is clear that $\Bbb Z[X]/(X)\cong\Bbb Z$ which is not a field.


13

Yes, below is a sketch a proof that $\rm\: \mathbb Z[w],\ w = (1 + \sqrt{-19})/2\ $ is a non-Euclidean PID, based on remarks of Hendrik W. Lenstra. The standard proof usually employs the Dedekind-Hasse criterion to prove it is a PID, and the universal side divisor criterion to prove it is not Euclidean, e.g. see Dummit and Foote. The latter criterion is ...


12

No; for example, $(x,y)$ is a maximal ideal which is not principal in $F[x,y]$. And also, $F[x,y]/(x^2-y^2)$ is not an integral domain since $(x-y)(x+y)=x^2-y^2$. On the other hand, the polynomial $y^2-x$ is irreducible and hence $F[x,y]/(x-y^2)$ is an integral domain.


11

Consider a finite generating set and find a common denominator for its elements. Now, look at what you've got... Alternatively, if $M$ is such a submodule, it has no torsion and we know from the structure theorem for f.g. modules over a PID that it must be free. Its rank can be computed by first tensoring with $K$ over $R$ and computing the dimension over ...


11

In P. Samuel, Anneaux factoriels, pages 36-37, it's proved that $A=\mathbb R[X,Y]/(X^2+Y^2+1)$ is a UFD. In the following we denote by $x,y$ the residue classes of $X,Y$ modulo $(X^2+Y^2+1)$. Thus $A=\mathbb R[x,y]$ with $x^2+y^2+1=0$. Lemma. The prime ideals of $A$ are of the form $(ax+by+c)$ with $(a,b)\neq (0,0)$. Proof. It's not difficult ...


11

If $I$ is an ideal of $R$, then you can verify that that the set of numerators of $I$, $\{r\in A: \frac{r}{s}\in I\text{ for some $s\in S$}\}$, is an ideal in $A$, thus a principal ideal, generated by some element $t$. Now show that the ideal generated by $t=\frac{t}{1}$ in $R$ is the ideal you started out with.


10

Hint: Consider the ideal $(2, x)$. Show that it's not principal. Suppose $(2, x) = (p(x))$ for some polynomial $p(x) \in \mathbb Z[x]$. Since $2 \in (p(x))$, then $2 = p(x) q(x)$ for some polynomial $q(x)\in \mathbb Z[x]$. Since $\mathbb Z$ is an integral domain, we have $\operatorname{degree} p(x)q(x) = \operatorname{degree}p(x) + ...


10

We will prove this using the PID $R:=\mathbb{Z}[\omega]$, where $\omega=\frac{1+i\sqrt{163}}{2}$, with the norm $N(a+b\omega)=|a+b\omega|^2=a^2+ab+41b^2$. Lemma 1: If $p$ is prime in $\mathbb{Z}$ such that $\exists n\in\mathbb{Z}$ such that $p|n^2+n+41$, then $p$ is reducible in $R$. Proof: Let $p|n^2+n+41$ and assume $p$ is irreducible. Then $p$ is also ...


10

Since $4(n^2+n+41)=(2n+1)^2+163$, one way would be to methodically show that $-163$ is a quadratic non-residue mod all (odd) primes less than $41$. Added later: The OP has asked for further details, so here goes. If $p\mid n^2+n+41$, then $(2n+1)^2+163\equiv0$ mod $p$, which means that $-163$ is a square mod $p$. Now the values of $n^2+n+41$ for $n\lt40$ ...


10

Theorem $\ $ The polynomial $\rm\ f(x)\ =\ (x\!-\!\alpha)\,(x\!-\!\alpha')\ =\ x^2 + x + k\ $ assumes only prime values for $\rm\ 0\ \le\ x\ \le\ k-2 \ \iff\ \mathbb Z[\alpha]\ $ is a PID. Hint $\ (\Rightarrow)\ $ Show all primes $\rm\ p \le \sqrt{n},\; n = 1-4k\ $ satisfy $\rm\ (n/p) = -1\ $ so no primes split/ramify. For proofs, see e.g. Cohn, Advanced ...


10

Here is a proof followed by conceptual elaboration, from my 2008/11/9 Ask an Algebraist post. Let R be an integral domain. Let every prime ideal in R be principal. Prove that R is a principal ideal domain (PID) Below I present a simpler way to view the proof, and some references. First let's recall one well-known proof, as presented by P.L. Clark ...


10

Yes. Let $\phi(a/b) = |a|$ where $a/b$ is written in lowest terms. To see that this is a Euclidean function, let $a/b,c/d\in \mathbb{Z}_X$ be nonzero and in lowest terms and write $$\frac{a}{b}=\frac{nd}{b}\cdot \frac{c}{d}+\frac{s}{t}$$ which means that $\phi(s/t)=\phi((a-nc)/b)\leq |a-nc|$ which for a suitable value of $n$ is less than $\phi(a/b) = |a|$.


10

Broadly speaking, the information you gave in your question is up-to-date: ever since Gauss, number theorists have believed there should be infinitely many real quadratic fields of class number one, but we are not any closer to being able to prove this than Gauss was, to the best of my knowledge. Moreover you have identified the key problem: the fact that a ...


9

According to the theorem about finitely generated modules, any finitely generated module over a PID $R$ can be expressed as $$ R^r \oplus R/(p_1^{k_1}) \oplus R/(p_2^{k_2}) \oplus \cdots \oplus R/(p_n^{k_n}) $$ where $p_1^{k_1},\ldots,p_n^{k_n}$ are prime powers in $R$. The $R^r$ is called the free part, and the other summands are elementary divisors. ...


9

The problem is that the set of all composite integers does not form an ideal. For example if you add $21 - 4$ you get $17$ which is a prime and thus not composite. That's why there's no contradiction. Remember that for a subset of a ring to be an ideal it must be closed under addition and under taking multiples by elements of the ring, and in this case the ...


9

Your statement is equivalent to proving (why?) that there are finitely many ideals $(b)$ such that $(a) \subset (b)$. But, $(a) \subset (b)$ iff $b|a$. Now factor $a$ into a finite product of irreducibles and use the fact that a P.I.D. is a U.F.D. to show that there can be only finitely many possibilities for $b$ such that $b|a$.


8

There are various ways to interpret how class groups measure (non)unique factorization. For example, Carlitz (1960) showed that the class group has order at most $2$ iff all factorizations of a nonzero nonunit into irreducibles have the same number of factors. Narkiewicz posed the problem of generalizing this, i.e. devising arithmetical characterizations of ...


8

Suppose $\def\ZZ{\mathbb Z}I\subseteq\ZZ\times\ZZ$ is an ideal. If $(x,y)\in I$, then $(x,0)=(1,0)(x,y)$ and $(0,y)=(0,1)(x,y)$ are also in $I$. If we write $I_1=\{x\in\ZZ:(x,0)\in I\}$ and $I_2=\{y\in\ZZ:(0,y)\in I\}$. then it follows easily from this that $I=I_1\times I_2$. Now $I_1$ and $I_2$ are ideals of $\ZZ$, so that there are $a$, $b\in\ZZ$ such ...


7

Here is a general result: If $D$ is a domain, then $D[X]$ is a PID iff $D$ is a field.


7

We know ideals of $R/I$ are of the form "ideal of $R$ containing $I$" mod $I$. So a descending chain of ideals looks like $I_1/I \supseteq I_2/I \supseteq I_3/I \supseteq \cdots$ where $I_j$ are ideals of $R$ such that $I \subseteq I_j$. Next, $R$ is a PID so there exists $a_j \in R$ such that $I_j=(a_j)$ and $a \in R$ such that $I=(a)$. Don't forget $a ...


7

For your third bullet, if there's only one element to the class group, then unique factorization holds because then all the fractional ideals are principal, and in particular the ring of integers is a principal ideal domain, which is equivalent to unique factorization for Dedekind rings. As to the first two, I'll only state that the class group of ...


7

Let me complete YACP's excellent answer by describing the units of $S=\mathbb{C}[X,Y]/(X^2+Y^2-1)=\mathbb C[x,y]$. We will use the isomorphism $\mathbb C[U,U^{-1}]\cong \mathbb C[x,y]:U\mapsto x+iy$ First note that the units of $\mathbb C[U,U^{-1}]$ constitute the set $(\mathbb C[U,U^{-1}])^{\times}=\bigcup _{n\in \mathbb Z} \mathbb C^*U^n$ of non zero ...


7

There is another way to think about this problem. Since $R:= \mathbb R[x,y]/(x^2 +y^2 -1 )$ is a smooth affine curve, it is a normal ring (i.e. integrally closed in its fraction field), and so it is factorial if and only if it has trivial class group. Here and below I will use ideas discussed in Hartshorne, Ch.II.6, in the subsection on Weil divisors. We ...


7

Given a UFD which is not a field, it has a prime element. It generates a non-zero prime ideal, so that we have a chain of prime ideals of length 1 and therefore the dimension is at least 1. It remains to decide if PIDs are UFD. The usual proof uses the axiom of dependent choice, which is a weak form of the axiom of choice. In fact, it turns out that it is ...


7

Here's what I think is a nice way to find a few PIDs that aren't Euclidean. It's not quite elementary, but if you know a bit about number fields I think it's a lot easier and nicer than the normal drudge. Let $K=\mathbb{Q}(\sqrt{-d})$ for $d>3$ squarefree, with ring of integers $\mathcal{O}_K$. Then the only units in $\mathcal{O}_K$ are $\pm1$. Suppose ...


7

Hint: What about the ideal $\langle2,x\rangle$? To answer your comment, a principal ideal is just a "set of multiples" of the generating element. Thus every element of a principal ideal has the generator as a divisor, and usually has similar properties. For example, in $\mathbb{Z}[x]$, the ideal $\langle2\rangle$ is the set of polynomials where every ...


7

Different prime ideals are coprime because in a Dedekind domain every nonzero prime ideal is maximal. If $P$ and $Q$ are nonzero prime ideals then $P+Q$ is an ideal containing both $P$ and $Q$ and so must be the whole ring if $P\ne Q$.


7

The claimed result is false: For $R/I$ to be a principal ideal domain, it must be a domain. But $R/I$ is a domain if and only if $I$ is a prime ideal. In particular, $\mathbb{Z}/6\mathbb{Z}$ is a quotient of a PID that is not a domain, hence not a PID. What is true is that a quotient of a PID will be a principal ideal ring: every ideal of $R/I$ is ...



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