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17

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.


16

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

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 ...


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

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 $\phi(a/b) = |a|$.


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

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 $n\lt40$ ...


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.


11

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

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

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

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

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

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 ...


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

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 $b|a$.


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

Let $F$ be a free $R$-module, where $R$ is a PID, and $U$ be a submodule. Then $U$ is also free (and the rank is at most the rank of $F$). Here is a hint for the proof. Let $\{e_i\}_{i \in I}$ be a basis of $F$. Choose a well-ordering $\leq$ on $I$ (this requires the Axiom of Choice). Let $p_i : F \to R$ be the projection on the $i$th coordinate. Let $F_i$ ...


8

We do this in a few steps. Let $R$ be our integral domain, and we are supposing that every prime ideal in $R$ is principal. Let $S$ be the set of non-principal ideals of $R$, and suppose that $S$ is nonempty. Throw an inclusion partial order on $S$ (or note that it has one, but I like to be physical with my math sometimes). WRT this partial order, let ...


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

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 ...


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 ...


8

This thread (including comments from both Jeremy and Alex) was very useful to me in PhD Quals preparation. I would like to give a detailed proof using Jeremy's and Alex's outline, but without mention of Spec(R), in case the reader is less algebraic-geometry-inclined. Corrections welcome. Thanks. Proposition. Let R be a UFD in which every nonzero prime ...


7

Ok, so it seems to me what you're saying is that you've proven that under the condition that every prime ideal is maximal that every maximal ideal is principal. The problem then follows from a nice condition you might want to remember: Theorem: Let $R$ be an integral domain. Then, $R$ is a PID if and only if every element of $\text{Spec}(R)$ is ...


7

The ring $\mathbf Z[w]$ is isomorphic to the quotient ring $A=\mathbf Z[x]/(x^2+x+1)$. To show $2$ is prime, you have to show $A/2A$ is an integral domain. We have: $$A/2A \simeq \mathbf Z[x]/(2,x^2+x+1)\simeq \mathbf Z/2\mathbf Z[x]/(x^2+x+1)$$ so it is enough to show the polynomial $x^2+x+1$ is irreducible over $\mathbf Z/2\mathbf Z$, which means it has ...


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

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

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 ...



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