Is $\mathbb{Z}[\sqrt{-7}]$ a UFD or not? I feel like there should be some easy example showing that it is not, but I can not think of one.
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I'm considering the ring $\mathbb{Z}[\sqrt{-n}]$, where $n\ge 3$ and square free. I want to see why it's not a UFD.   I defined a norm for the ring by $|a+b\sqrt{-n}|=a^2+nb^2$. Using this I was able to show that $2$, $\sqrt{-n}$ and $1+\sqrt{-n}$ are all irreducible. Is there someway to conclude that $\mathbb{Z}[\sqrt{-n}]$ is not a UFD based on this? Thanks.
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Let $R$ be a UFD. Then for any two elements $a,b \in R \setminus \{0\}$ $ab \sim \gcd (a,b){\operatorname{lcm} (a,b)}$.   My attempt :  Let $d = \gcd (a,b)$.Then $d|a$ and $d|b$.Then $\exists x,y \in R$ such that $a = dx$ and $b = dy$.Now it is to be shown that $c = dxy$ is the lcm of $a,b$.Which I find difficulty to show.It is clear that $a|c$ and $b|c$. Now if $u$ is any common multiple of $a$ and $b$.Then $dx|u$ and also since $y|b$ we have $y|u$ i.e. $dxy|u^2$ $\implies c|u^2$.But I have to show $c|u$ to complete the proof which I fail to prove.Please help me to complete the proof.  Thank you in advance.
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I know that $\mathbb Z[\sqrt{7}]$ is a UFD, and I can write the equation $(2 + \sqrt{7})(3 - 2\sqrt{7}) = (5 - 2\sqrt{7})(18 + 7\sqrt{7}) $. So clearly these are not all irreducibles. How do I determine which of these aren't irreducible? How do I factor those into irreducibles?  It seems difficult when dealing with $\mathbb Z [\sqrt{d}] $ with $d$ positive, since the norm map isn't as nice (it's not always positive).  Thanks
61503
Problem: Is there an irreducible $f\in \mathbb{Z}[x]$, whose image in every $(\mathbb{Z}/p\mathbb{Z})[x]$ has a root for $p$ prime? If there is, what is the minimal degree possible?   I can only prove that $x^2-c$ is impossible, by quadratic reciprocity and Chinese remainder theorem. Even the case of $a x^2 - c$ is unknown to me.  Meanwhile, if $f$ is not required  to be irreducible, but only have no root in $\mathbb{Z}$, then $(x^2-a)(x^2-b)(x^2-ab)$ is a solution for non-sqaure $a,b,ab$, since if both $a,b$ are not squares in $\mathbb{Z}/p\mathbb{Z}$, then $ab$ is a square. Yet I still don't know if this is of minimal degree.  Also, it is natural to pose generalizations   When $p$ is not necessarily prime (equivalently for all prime powers). When $p$ is odd prime. When $p$ represents sufficiently large primes (equivalently all but a finite number left).
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I was wondering, is the localization of a UFD also a UFD?    How would one go about proving this? It seems like it would be kind of messy to prove if it is true.  If it is not true, what about localizing at a prime? Or what if the UFD is Noetherian?
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I am trying to show that $k[x,y,z]/(xz-y^2)\not\cong k[x,y]$.    The latter is a UFD, so I am trying to show the former is not. Clearly $x$ is not prime, since $x\mid xz$ which implies $x\mid y^2$, but $x\mid y$.      So if I can show that $x$ is irreducible then $k[x,y,z]/(xz-y^2)$ is a not a UFD, because irreducible implies prime in UFDs.    Attempt: Assume $x$ is reduced into $ab$ in $k[x,y,z]/(xz-y^2)$, then we know there exists $c\in k[x,y,z]$ such that $x=ab+c(xz-y^2)$ in $k[x,y,z]$, or more usefully as $x-ab=(xz-y^2)c$. Someone suggested that I write $a,b$ such they are degree at most 1 in $y$, that is, replace every $y^n$ using $xz$. This means the LHS is degree at most 2 in $y$ and the RHS is degree at least two in $y$ in  $x-ab=(xz-y^2)c$. I am not sure where to go from here just by knowing that $c$ has no $y$ terms though. Ultimately I want to show that this forces $x=ab$ in $k[x,y,z]$ which, since $x$ is irreducible there, shows that either $a,b$ is a unit, which means $x$ is irreducible in $k[x,y,z]/(xz-y^2)$.
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I think its silly question but I have nt this in my mind at this time.   Any one can help  How we can see $U(10) \overset{def}= \{1,3,7,9\}$, $\mathbb Z/4\mathbb Z \overset{def}= \{0,1,2,3\}$, $U(5) \overset{def}= \{1,2,3,4\}$ are isomorphic groups where $U(10)$ and $U(5)$ are groups under multiplication?
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Let $R$ be a UFD and $r\in R$ irreducible. If $S=R-(r)$, why is $S^{-1}R$ a PID?   I've finished proving the integral domain part, which was pretty easy.  How to prove it's principal? consider $R \subset S^{-1}R \subset F$, where $F$ is the field of fractions of $R$?  thank you! :p
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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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If a nonzero element $a$ of a ring $\mathbb{D}$ does not have a multiplicative inverse, then $a$ must be a zero divisor.   If $\mathbb{D}$ has finitely many elements, then the statement is true. However, I don't see why, as for the infinite case, 7 isn't a zero divisor only because every element $a &gt; 0$ when multiplied by a $b &gt; 0$ gives you $ab &gt; 0$ which has nothing to do with infinity or finite amounts of elements.
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Clearly, $\mathbb Z$ is not dense in $\mathbb Q$ (and not dense in $\mathbb C$).  But why is the ring of algebraic integers $\overline{\mathbb Z}$ dense in $\overline{\mathbb Q}$?  In particular, if this is so, it is dense in $\mathbb C$.
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Let $R$ be a $\text{UFD}$, and let $a,b,c \in R$ such that $1=\text{gcd}(a,b).$ Suppose $a |c, \ b|c.$ Could anyone advise me on how to prove $ab |c \ ?$ How do I use the fact that every nonzero non unit $r \in R$ can be expressed as finite product of irreducible elements in $R\ ?$ Hints will suffice.
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This theorem seems to hold great potential for an intuitive explanation, yet I'm still struggling for it.  Basically, it is saying that if $y = F(x)$, then y would have distribution $Unif[0,1]$. Intuitively, why?  Many thanks!
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It is a known fact that if $R$ is a UFD, then $R[X_1,X_2,\dots]$ is also a UFD, but there is a subtlety that is making me uncomfortable.  The standard approach essentially goes something along the lines of...let $f$ be a polynomial, so it only involves finitely many variables, say up to $X_1,\dots,X_n$. Then since $R[X_1,\dots,X_n]$ is a UFD, $f$ has a factorization into irreducibles. Then $f$ cannot have a factorization involving indeterminates $X_N$ not in $R[X_1,\dots,X_n]$. If it did, then $f$ would have two factorizations in the UFD $R[X_1,\dots,X_n,\dots, X_N]$, a contradiction.  This seems to assume that the prime factorization in $R[X_1,\dots,X_n]$ is still a prime factorization in $R[X_1,\dots,X_n,\dots,X_N]$. But how can you see that irreducibles in $R[X_1,\dots,X_n]$ are still irreducibles in $R[X_1,\dots,X_m]$ for $m&gt;n$? This does not seem obvious.  Let $R_k=R[X_1,\dots,X_k]$. The explanantion I found is this: Suppose $p\in R[X_1,\dots,X_n]$ is irreducible in $R_n$. Suppose $p=ab$ for $a,b\in R_m$. Evaluating $X_1,\dots,X_n$ at $1$, you get  $$ \bar{p}=\overline{ab}\in R[X_{n+1},\dots,X_m] $$ But $\bar{p}\in R$, which implies that $a$ and $b$ do not involve any variables $X_{n+1},\dots,X_m$. So $a,b\in R_n$, and thus one is a unit, hence a unit in $R_m$.   The part I don't follow is how evaluation at $1$, implies $a$ and $b$ do not have indeterminates other than $X_1,\dots,X_n$. Isn't it possible that some of the higher indexed indeterminates disappear when you evaluate? Suppose for instance $p\in R_1$, and $a\in R_2$ is $a=-X_1X_2+X_2$. Then evaluting at $1$ gives $\bar{a}=0$ so we may not get the desired contradiction since $\overline{ab}=0\in R$?
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Is there any field of characteristic two that is not $\mathbb{Z}/2\mathbb{Z}$?  That is, if a field is of characteristic 2, then does this field have to be $\{0,1\}$?
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An example of a non Noetherian UFD.     I know an example is $$K[x_1,\ldots,x_n,\dots]$$ with $K$ a field, but I don't know why. Can someone give another example or better an explanation?   Is it not Noetherian because it's not finitely generated? And why do you know it is a UFD?? My question now is how to prove this is a UFD.
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We're proving that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain, using the norm function $$\nu (a + b\sqrt{2} ) = |a^2 - 2b^2|$$ and the first part says that since $\nu (a + b\sqrt{2} ) = |(a + b\sqrt{2})(a - b\sqrt{2})|$ it's clear that $\nu (xy) = \nu(x) \nu(y)$? ... Can someone please explain to me how this is clear?