Let $R$ be a principal ideal domain containing distinct prime elements $p, q$. Let $n, m \geq 1$ be integers. Is there a simple proof for the claim that the elements $p^n$ and $q^m$ of $R$ are relatively prime, i.e. that $(p^n) + (q^m) = R$, or equivalently that the greatest common divisor of $p^n$ and $q^m$ is $1$?
29309
If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$   This seems clear, but I don't know how to prove this..  I was trying to show this by induction such that if $a^{n+1}$ = $rs$ and $b^{n+1}$ = $rt$, then $s,t$ are divisible by $a,b$ respectively, but i think this is a wrong way..
61754
Problem:   If $x^{n-1} \equiv 1 \pmod{n}$, and for all divisors $q$ of $n - 1$, $a^{q} \not\equiv 1 \pmod{n}$, then $n$ is prime. $(n &gt; 1)$   I read the proof in the book and it was very straightforward; however, I wonder is there a way to prove it by just using congruence property?    And another related question about power residue: If we have $a^{n - 1} \equiv 1 \pmod{n}$. Is there any relation between $n - 1$ and $\phi(n)$?  Because I thought of $a^{\phi(n)} \equiv 1 \pmod{n}$, when $(a, n) = 1$. I try to think of away to connect these two ideas, but I still cannot see it. Any idea?    Thanks,
51418
Let $\Bbb{P}$ be the set of prime numbers and $Q=\Bbb{P}\cup\{1\}$. Is it true that every natural numbers ($\neq 0$) is the sum of distinct elements of $Q$? I tried from $1$ to $60$ and it seems true.
75004
Let $I$ be a homogeneous ideal of a graded ring $S$, $I\ne S$. I want to show that there exists a homogeneous prime ideal which contains $I$.    I proved the following:  Let $T$ be the set of all homogeneous ideals which contain $I$. Then, by Zorn's lemma, there exists a maximal element of $T$, say $P$. I will claim that $P$ is prime. Suppose that for homogeneous elements $a,b \in S$, $ab \in P$ but, $a\notin P$. Then $\langle a \rangle + P$ is a homogeneous ideal which contains $I$. It contradicts by maximality of $P$ so, $a\in P$.  Is it right???
97170
Let $R$  be a ring with with no zero divisors. If $a, b \in R$ are such that $a^m = b^m$ and $a^n = b^n$, where $m$ and $n$ are relatively prime positive integers, then show that $a = b$.   My attempt: Let $m&gt;n$. $\therefore a^na^{m-n}= b^nb^{m-n}$ or, $ a^{m-n} = b^{m-n}$. How do I proceed from here?
51487
Let $I \subset R$ be an ideal and $P_i$  $(i=\{1,...,n\})$ prime ideals with $I\subseteq\bigcup_{i=1}^nP_i$. Prove that then $I$ is contained in one $P_i$.   I don't know how to show this because I don't have any approach.  So I am looking for something to start with or something like a sequence of tips (Since I know this is a large proof)  Thanks in advance!
72172
Let $a$ belong to a group and $|a|=m$. If $n$ is relatively prime to $m$ show that $a$ can be written as the $n^{th}$ power of some element in the group.   We need to show that if $a\in G$ and  $a^m=e\implies \exists \ b\in G $ such that $b^n=a$ i.e.$b^{nm}=e=({b^m})^n$ i.e. $\exists$ an element of order $mn$ in $G$.   How to show this?
105949
Let $R \subset S$ be an integral extension of domains and $\mathfrak p \subset R$ a prime ideal. Can it be the case that there are infinitely many distinct primes ${\cal P} \subset S$ such that ${\cal P} \cap R=\mathfrak p$?    Certainly this is impossible if $S$ is a Dedekind domain, because the primes lying over $\mathfrak p$ are the primes of $S$ occurring in the factorization of $\mathfrak p$ over $S$.  I don't have much of an intuition for integral ring extensions that aren't number fields, so past this I'm not particularly sure.
47406
If $|G|$ and $|{\rm Aut}(G)|$ are relatively prime, then first, it can be shown that $G$ must be abelian: this is clear, since if $G$ would non-abelian then a prime divisor of $|G/Z(G)|$ (which divides $|G|$) is necessarily a divisor of $|{\rm Aut}(G)|$.  Therefore, we assume that $G$ is abelian group with $(|G|, |{\rm Aut}(G)|)=1$. It may be possible to deduce something more; I guess that $G$ must be cyclic. Is this true?
80160
In the usual definition, a principal ideal domain $R$ is also assumed to be an integral domain. However, the property that every ideal is generated by a single element does not seem to immediately imply that the ring is integral. Is this correct and if so:  Do there exist rings where every ideal is generated by a single element and has zero divisors?  I am most interested in the case where $R$ is commutative with unity, but don't mind examples where these properties don't hold. Also, assuming there are examples, is there any reason why we make this assumption?
54424
Maybe this question is too obvious to everyone.      In Noetherian ring $R$, ideal $I\subset R$, $P\in\mathrm{Ass}(I)$, then $P=I:a$ for some $a\in R$ where $\mathrm{Ass}(I)$ denotes associated primes of $I$.   This is an exercise 8.29 in Gathmann's Commutative Algebra notes. I am supposed to find this statement obvious. However, up to now, I still did not find this statement obvious. Can someone explain intuitively why noetherianness here allows me removing radical $\sqrt{I:a}$ without going into details?  If $Q$ is a primary ideal of $I$, then certainly $\sqrt{Q:a}=P$ for $a\not\in P$. It seems taking $(p_1,\dots, p_n)=P$, $p_i^{n_i}a\in Q$. Since $Q$ primary, I must have $p_i\in Q$ and $p_i^{n_i}\in Q$. (This argument seems very wrong as this deduces the same conclusion without noetherian property.)
82956
I'm trying to work out the following past paper question and I've got stuck.  $R$ is an integral domain and $S = R[t]$, the polynomial ring in one variable over $R$. We have that $Q$ is a prime ideal strictly contained in another prime ideal $P$ with  $$ P \cap R = Q \cap R  = P_0. $$ We also have $Y = R\setminus P_0$.  I have shown that $P_0S$ is a prime ideal, and that after localisation we have $$ \frac{SY^{-1}}{P_0SY^{-1}} \simeq F[t]    $$ where  $$ F = \frac{RY^{-1}}{P_0Y^{-1}}. $$  This is a field since $P_0 = R\setminus Y$, so $P_0Y^{-1}$ is a maximal ideal of $RY^{-1}$.   What I now have to show is that $P_0S  =  Q$, and the question says that i should do this using the above isomorphism of polynomial rings, but I can't figure it out.  A hint or solution would be great, thanks!
110995
The proof of the following proposition is not completely clear to me. I get everything up until the bold part and I have a feeling some crucial steps are omitted, can anybody help clear this up?     Let $R$ be an integral domain. If every two elements of $R$ have a greatest common divisor, then every irreducible element in $R$ is prime.   Proof:  Consider an irreducible element $p \in R$ and $x,y \in R$ such that $p\vert xy$. Suppose now that $py$ and $xy$ have a greatest common divisor $z$ in R. We want to conclude from this that $p \vert x$ or $p \vert y$. This is obvious if $xy = 0$, so we may assume that $xy \neq 0$. Then $z \neq 0$. As both $p$ and $y$ divide each of $py$ and $xy$, we have that $z = pu = yv$ for certain $u,v \in R$. Using the cancellation law with $\boldsymbol z \boldsymbol \neq \boldsymbol 0$, we obtain that $\boldsymbol v \boldsymbol \vert \boldsymbol p$. As $p$ is irreducible, either $v \in R^\times$ (i.e. the set of invertible elements of $R$) or $v \sim p$ (i.e. $Rv = Rp$). If $v \sim p$, then $p \vert x$. If $v \in R^\times$, then $p$ divides $v^{-1} pu = v^{-1} z=y$.
91132
Let $f: R \rightarrow S$ be a ring homomorphism, with $R$ and $S$ commutative. If $P$ is a prime ideal of $S$, show that the preimage $f^{-1}(P)$ is a prime ideal of $R$.    Define $g: S \rightarrow S/P$ with kernel $s$. Let $h = g \circ f: R \rightarrow S/P$. Since $h$ is a ring homomorphism, the kernel is an ideal of $R$. Also, from the first isomorphism theorem, we know that $R/\ker(h) \cong S/P$. Since $P$ is a prime ideal of $S$, we know that $S/P$ is an integral domain. Since $R/\ker(h)$ is isomorphic to $S/P$, it must also be an integral domain, which implies that the kernel of $h$ (which is the preimage of $P$) is a prime ideal of $R$.  Do you think my answer is correct? The reason why I was a bit skeptical of my answer is because I did not use the fact that $R$ and $S$ are commutative. So I'm wondering if I missed something $\dots$  Thank you in advance.
116224
Show that if $P$ is a prime ideal of a  commutative ring $R$ with unity then $P[x]+\langle x\rangle $ is a prime ideal of $R[x]$. Here $P[x]$ consists of all polynomials whose coefficients are in $P$.  I tried to show it using the fact that if $f(x)g(x)\in P[x]+\langle x\rangle $ then either $f(x)\in P[x]+\langle x\rangle $ or $g(x)\in P[x]+\langle x\rangle $. But I don't know how to show it. Please help me out.
106906
I am studying ring theory and I am having a hard time understanding a part of a proof of a certain theorem. Note that in my lecture notes, a ring is by definition commutative with $1$. Let $R$ be a ring, then the set of units of $R$ is denoted by $R^*$. Consider the following theorem:  Let $R$ be a principal ideal domain (PID) and $\pi\in R$ irreducible. For any $a,b\in R$ with $\pi|ab$ we have $\pi|a$ or $\pi|b$.  The following is the first part of the proof from my lecture notes:  Consider the ideal $(a,\pi)$. Since $R$ is a PID this ideal is generated by a single element, which we call g. Note that $g|\pi$. Hence either $g$ is a unit, or $g=\epsilon\pi$ with $\epsilon\in R^*$. In the second case, since $g|a$ we also get $\pi|a$ and our assertion is proven.  The second part of the proof covers the case in which $g$ is a unit; I understand what's going on there (under the assumption that $g$ is indeed a unit) so I've omitted it. Please note that this is the proof word by word, i.e. I have quoted it from my lecture notes. First of all, is it true that, since $(g)=(a,\pi)$ we have $\{rg\ |\ r\in R\}=\{r_1a+r_2\pi\ |\ r_1,r_2\in R\}$ and by choosing $r_1=0$ and $r_2=1$ we have $rg=\pi$ for some $r\in R$, thus $g|\pi$? If this is true, then by similar reasoning one can proof that $g|a$ (which is stated in the last sentence of the proof). Next, the proof says that $g$ is either a unit or $g=\epsilon\pi$ for some unit $\epsilon$. By definition, we see from the factorisation above that either $g$ or $r$ is a unit. The first case corresponds to the statement in the proof that $g$ can be a unit. But if $g$ is not a unit, why is $g=\pi\epsilon$ for some unit $\epsilon$? Lastly, how does $\pi|a$ follow from all that has been said in the proof? EDIT: As requested by a user in the comments, here is the second part of the proof in my lecture notes, along with the first one for easy reading:  Consider the ideal $(a,\pi)$. Since $R$ is a PID this ideal is generated by a single element, which we call g. Note that $g|\pi$. Hence either $g$ is a unit, or $g=\epsilon\pi$ with $\epsilon\in R^*$. In the second case, since $g|a$ we also get $\pi|a$ and our assertion is proven. Now suppose  that $g$ is a unit and hence $(a,\pi)=R$. In this case there exist $x,y\in R$ such that $ax+\pi y=1$. Multiply by $b$ to get $abx+\pi by=b$. Both terms on the left are divisible by $\pi$. Hence the sum $b$ is also divisible by $\pi$ which proves our theorem.
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 .