Is $R=\mathbb Z [(1+\sqrt{-19})/2]$  a Euclidean domain?    Its Voronoi region seems relatively small, but its hard to have intuition about division with remainder. I predict it is not since the norm of $(1+\sqrt{-19})/2$ is $\sqrt5&gt;1$.   Here is my attempt at a proof:  I show that an Euclidean Domain $D$ (with not every nonzero element a unit) has a non-unit element $p$ such that $\forall x\in D$ $p|{x}$ or $p|{x-u}$ for some unit $u\in D$. I now want to reason by trying to show the conditions of the contrapositive are true for $R$. In other words, that $\nexists p\in R$ such that $p|{x}$ or $p|{x-u}\ \ \forall x\in R$ for $p$ nonunital and for some unit $u\in D$.   I'm having difficulty showing this. Any help appreciated.
29843
Let $\alpha = \frac{1+\sqrt{-19}}{2}$. Let $A = \mathbb Z[\alpha]$. Let's assume that we know that its invertibles are $\{1,-1\}$. During an exercise we proved that:     Lemma: If $(D,g)$ is a Euclidean domain such that its invertibles are $\{1,-1\}$, and $x$ is an element of minimal degree among the elements that are not invertible, then $D/(x)$ is isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.   Now the exercise asks:     Prove that $A$ is not a Euclidean Domain.   Everything hints to an argument by contradiction: let $(A, d)$ be a ED and $x$ an element of minimal degree among the non invertibles we'd like to show that $A/(x)$ is not isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.  How do we do that? My problem is that, since I don't know what this degree function looks like, I don't know how to choose this $x$!  I know that the elements of $A/(x)$ are of the form $a+(x)$, with $a$ of degree less than $x$ or zero. By minimality of $x$ this means that $a\in \{0, 1, -1\}$. Now I'm lost: how do we derive a contradiction from this?
117217
$d \in \mathbb{Z}$ is a square-free integer ($d \ne 1$, and $d$ has no factors of the form $c^2$ except $c = \pm 1$), and let $R=\mathbb{Z}[\sqrt{d}]= \{ a+b\sqrt{d} \mid a,b \in \mathbb{Z} \}$. Prove that every nonzero prime ideal $P \subset R$ is a maximal ideal.   I have a possible outline which I think is good enough to follow.  I think that we need to first prove that every ideal $I \subset R$ is finitely generated.   So if $I$ is non-zero, then $I \cap \mathbb{Z}$ is a non-zero ideal in $\mathbb{Z}$.   Then I need to find $I \cap \mathbb{Z} = \{ xa \mid a \in \mathbb{Z} \}$ for some $x \in \mathbb{Z}$. That way if I let $J$ be the set of all integers $b$ such that $a+b\sqrt{d} \in I$ for some $a\in \mathbb{Z}$, then if there exists a integer $y$ such that $J=\{ yt \mid t\in \mathbb{Z} \}$, then there must exist $s \in \mathbb{Z}$ such that $s+y\sqrt{d} \in I$.   Then all I need to show is that $I = ( x,s+y\sqrt{d} )$.    Now I need to derive that the factor ring $R / P$ is a finite ring without zero divisors, also finite, then since every finite integral domain is a field, every prime ideal $P \subset R$ is a maximal ideal, then I'll be done.
111431
Letting $\mathbb{Q}$ be equipped with the Euclidean metric.  What I can work out is that it is bounded as its contained in the closed ball of radius ${\sqrt2}/{2}$ centred at ${\sqrt2}/{2} $.  Its not compact as it can be expressed as union of the two disjoint open sets $[0,{\sqrt2}/{2}) $and$ ({\sqrt2}/{2}, \sqrt2)$ (though I'm not sure if this makes it not compact in $\mathbb{Q}$ or just $\mathbb{R}$).  And its not closed as the sequence of truncations of $\sqrt2$ in $\mathbb{Q}$ converges to $\sqrt2$
63936
Let $R$ be a Euclidean domain in which the quotient and remainder are always unique. Does it follow that the ring $R$ is either a field or a polynomial ring $F[X]$ for some field $F$?
75407
Exhibit an integral domain $R$ and a non-zero non-unit element of $R$ that is not a product of irreducibles.   My thoughts so far: I don't really have a clue. Could anyone direct me on how to think about this? I'm struggling to get my head round irreducibles.  Thanks.
91504
In Herstein’s Topics in Algebra, p.147, there is a sentence  If $\pi$ in $R$ is a prime element of $R$ and $a \in R$, then either $\pi \mid a$ or $(\pi,a) = 1$, for, in particular, $(\pi,a)$ is a divisor of $\pi$ so it must be $\pi$ or $1$ (or any unit).  I’m having a bit of difficulty seeing why the last part of this sentence is true (“for, in particular…”). Note:  Herstein’s definition of a prime element in a Euclidean ring $R$ is a nonunit $\pi$ such that whenever $\pi = ab$, where $a, b$ are in $R$, then one of $a$ or $b$ is a unit in $R$. This result comes before the Unique Factorization Theorem, so that can’t be used. However, we have proved that every element of a Euclidean ring $R$ is either a unit or can be written as a product of finitely many primes. (I don’t know if this is useful, but it’s what I tried to use, to no avail.)  My attempt: I tried supposing to the contrary, so that $(\pi,a)$ is neither $\pi$ nor $1$ nor a unit. Since $(\pi,a)$ divides $\pi$ we can write $\pi = (\pi,a)k$ for some $k \in R$. Then by definition of $\pi$ being prime, $k$ must be a unit, since $(\pi,a)$ isn’t. So $\pi$ and $(\pi,a)$ are associates. This is where I tried a few other things and got stuck each time.
61080
Why is $(2, 1+\sqrt{-5})$ not principal in $\mathbb{Z}[\sqrt{-5}]$?   Say $(2,1+\sqrt{-5})=(\alpha)$, then since $2\in(2,1+\sqrt{-5})$ we have $2\in (\alpha)$, so $\alpha\mid2$ in $\mathbb Z[\sqrt{-5}]$. Writing $2=\alpha\beta$ in $\mathbb Z[\sqrt{-5}]$ and taking norms, $4=N(\alpha)N(\beta)$ in $\mathbb Z$, so $N(\alpha)\mid4$ in $\mathbb Z$. Similarly, since $\sqrt{-5}\in(\alpha)$ we get $N(\alpha)\mid5$, thus $N(\alpha)$ is a common divisor of $4$ and $5$, therefore $N(\alpha)=(1)$, so $\alpha$ is a unit. But that means $1\in (2,1+\sqrt{-5})=(\alpha)$, so it must be the whole ring, but it cannot to reach a contradiction, so how can I find an element, which is not in $(2, 1+\sqrt{-5})$.  Or is there a simpler method ? (Maybe not UFD would imply not PID)
17993
Let $p$ be a prime number. I could already show, that $\mathbb{Z}[\sqrt{-2}]$ is an euclidean domain and that $$p\text{ reducible }\Leftrightarrow\text{ }p=a^2+2b^2\text{ for some $a,b\in\mathbb{Z}$.}$$  Now I wonder how to prove  $$p\equiv 1,2,3\text{ mod }8 \Rightarrow p=a^2+2b^2\text{ for some $a,b\in\mathbb{Z}$.}$$  Maybe there is a possibility to show  $$p\equiv 1,2,3\text{ mod }8 \Rightarrow p\text{ reducible }?$$
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 $S=R[U^{-1}]$ ?
37204
As stated in the question, I am looking for a Noetherian ring over which the Euclidean algorithm is not valid.  I am trying to construct non-trivial examples of Noetherian rings.  Thank you.
108839
While trying to look up examples of PIDs that are not Euclidean domains, I found a statement (without reference) on the  page of Wikipedia that  $$\mathbb{R}[X,Y]/(X^2+Y^2+1)$$  is such a ring. After a good deal of searching, I have not been able to find any other (online) reference to this ring.  Can anyone confirm this result? Is there a reference for it (paper, textbook or website)?
69736
Show that if $a+bi$ is irreducible over $\mathbb{Z} [i]$ then $a^2 + b^2$ is prime over $\mathbb{Z} $ . (a,b are non-zero integers)  I know that $\mathbb{Z} [i]$ is a Euclidean domain and so  $a+bi$ is prime in it. Can I say that as the norm of  $a+bi$ is $a^2 + b^2$ that $a^2 + b^2$ must be prime over $\mathbb{Z} $? This doesn't seem correct to me but it's the best try I have been able to come up with.
79588
I was recently reading about integral ring extensions. One of the first examples given is that $\mathbb{Z}$ is integrally closed in its quotient field $\mathbb{Q}$. Another is that $\mathbb{Z}[\sqrt{5}]$ is not integrally closed in $\mathbb{Q}(\sqrt{5})$ since for example $(1+\sqrt{5})/2\in\mathbb{Q}[\sqrt{5}]$ is integral over $\mathbb{Z}$ as a root of $X^2-X-1$, but $(1+\sqrt{5})/2\notin\mathbb{Z}[\sqrt{5}]$.     Now I'm  curious, can we find what are all integers $n$ such that $\mathbb{Z}[\sqrt{n}]$ is integrally closed (equal to its integral closure in its quotient field)?   One thing I do know is that unique factorization domains are integrally closed, so I think rings like $\mathbb{Z}[\sqrt{-1}]$, $\mathbb{Z}[\sqrt{-2}]$, $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{3}]$ are integrally closed, as they are Euclidean domains, and thus are UFDs. But can we say what all integers $n$ are such that $\mathbb{Z}[\sqrt{n}]$ is integrally closed? Thanks!
69614
Which of the following sets are compact:   $\{(x,y,z)\in \Bbb R^3:x^2+y^2+z^2=1\}$ in the Euclidean topology. $\{(z_1,z_2,z_3)\in \Bbb C^3:{z_1}^2+{z_2}^2+{z_3}^2=1\}$ in the Euclidean topology. $\prod_{n=1}^\infty A_n$ with the product topology where $A_n=\{0,1\}$ has discrete topology. $\{z\in \Bbb C:|\operatorname{Re} z |\leq a \}$ for some fixed positive real number $a$ in the Euclidean topology.   $1$ is closed and bounded and hence compact,$2$ is closed but not bounded and hence not compact.  $3$ is compact by Tychonoff Theorem and $4$ is not bounded and hence not compact.  Are these correct?
41674
In self studying abstract algebra and I've come upon the following problem which I could not solve directly. For any $d\in \mathbb{Z}$ we are asked to show that $\mathbb{Z}[\sqrt d]=\{a+b\sqrt{d} ~\colon a,b\in \mathbb{Z} \}$ is an integral domain. This reduces to showing that the ring has no zero divisors which I tried to do through algebraic manipulation.  I have seen two solutions to this problem that use higher level techniques. One defines conjugate and norm and proves the claim using simple consequences of these definitions. The other says that since the given ring is a subring of the field $\mathbb{C}$ it can have no zero divisors.  I don't like either approach because the first uses ideas that haven't been introduced in the text and isn't a natural place to start (for me). The second, while both true and simple, is not satisfying to me. That $\mathbb{C}$ is a field has not been established and assuming it just makes the problem trivial.   My question is whether or not naive tactics like algebraic manipulation are capable of demonstrating the result. Am I lacking in cleverness or is such an approach impossible?
3301
The following is a problem from Dummit &amp; Foote.   Let $R$ be an integral domain. Prove that if the following two conditions are true, then $R$ is a principal ideal domain.    Any two non-zero elements $a$ and $b$ in $R$ have a greatest common divisor that can be written as $d=ra+sb$, $r,s \in R$ If $a_1,a_2,\dots$ are non-zero elements of $R$ such that $a_{i+1}\mid a_i$ for all $i$, then there exists a positive integer $N$ such that $a_n$ is a unit times $a_N$ for all $n \ge N$.   I am a bit confused really. Isn't the first condition sufficient? For e.g. an ideal generated by two elements $a,b$ would be a principal ideal, as any two elements have a gcd. This can then be extended to ideals generated by an arbitrary number of elements by induction.   Obviously, there is something wrong in what I am doing. What is the second condition for? I can only think of defining principal ideals generated by the elements in a chain, and then finding a maximal ideal with respect to divisibility. I have no idea.
46957
In  in these notes, it is said that  Since $x − a$ has leading coefficient $1$, which is a unit, we may use the Division Algorithm...  Why is this true? I thought that the Division Algorithm is only guaranteed to work in Euclidean domains not any integral domain.