I have a proposition which says that an ideal $I$ is a prime ideal in $R$ iff its quotient ring $R/I$ is a domain. I know that for an ideal to be a prime ideal, $xy\in I\implies x\in I\lor y\in I$. My thought is this: $x\notin \langle x^2-x\rangle$ and $x-1\notin \langle x^2-x\rangle$, but $x(x-1)=x^2-x\in \langle x^2-x\rangle$, so $\mathbb{Q}[x]/\langle x^2-x\rangle$ is not a domain. Am I thinking this right?
123588
So I know that for it to be an integral domain it has to have the following properties:    Commutative Has multiplicative identity No Zero-Divisors    and if $p(x)$ is reducible it can be written as $p(x) = (x-a)q(x)$ but I can't seem to link it together to prove that it violates one of the three properties.
108734
I can prove that the ideal $(4, 2x, x^2)$ in $\mathbb Z[x]$ is not principal. But I failed to prove that this cannot be generated by two elements.  It's really difficult for me. Would you give me a hint.
54559
If $F$ is a field, then $F[x]$ is a principal ideal domain.  Conversely, if $F[x]$ is a principal domain does $F$ have to be necessarily a field?  My Thoughts: Suppose instead of $F$, we take the set of polynomials $R[x]$ over a commutative ring $R$ with unity. Then, suppose $I$ is an ideal of $R[x]$. Let $g(x) \in I$ such that $g(x)$ is the polynomial of the lowest degree in $I$. Then: $ \langle g(x) \rangle \subseteq I .......... (1)$ Let $f(x) \in I$. Then $f(x) = p(x)g(x) + r(x)~~|~~p(x),r(x) \in R[x], \deg r(x) &lt; \deg g(x)$ Since, $I$ is an ideal $\implies f(x) - p(x)g(x) = r(x) \in I$ But, $g(x)$ is of the lowest degree in $I \implies r(x) = 0 \implies f(x) \in \langle g(x)  \rangle \implies I \subseteq \langle g(x)  \rangle ......(2)$ Then from $(1),(2) : I = \langle g(x) \rangle$ Does the Presence of zero divisors in $R[x]$ really make a difference? The only advantage I see is that if there are no zero divisors in $R[x]$ then $I=\{0\} \implies I= \langle 0  \rangle$. But, every ideal contains the zero element. Why is there the condition of a field specifically given in textbooks for $F[x]$ to be a principal ideal domain? Thank you for your help.
8933
Ok, so an ideal is prime if for every $ ab \in I $, $ a \in I $ or $ b \in I $.  Now the two questions:  1: Is that an inclusive or exclusive or? Can both $a$ and $b$ be in I?  2: Can $ab$ be equal to either $a$ or $b$?
6400
Let $A$ be a commutative ring with $1$. Let $I$ and $J_k$, $k=1,\dots,n$ be ideals of $A$ with $I\subseteq \cup _{k=1}^n J_k$. Then I have obtained the following:  (1) If $J_k$, $k=1,\dots,n$, are prime ideals, then there exists some $j$ such that $I\subseteq J_j$.   (2) If $A$ is a principal ideal ring, then there exists some $j$ such that $I\subseteq J_j$.   I want some counterexamples that for all $k=1,\dots,n$, $I$ is not contained in $J_k$. How to get the counterexamples?
57022
Let $R$ be a PID (Principal Ideal Domain) and $x$ is an element R. Prove that the ideal $\langle x\rangle$ is maximal if and only if $x$ is irreducible.   Ok, so I know what an irreducible is. I'm thinking that this problem is asking us to set up a proof by contradiction but I can't see how. No one in my study group has any clue.
50159
This problem goes as follows:     Prove that $\mathbb{Q}[x,y]/\langle x^2+y^2-1\rangle$ is an integral domain and that its field of fractions is isomorphic to the ring of rational functions $\mathbb{Q}(t)$.    The first part is straightforward, $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$, and  hence a prime ideal, as $\mathbb[Q][x,y]$ is a euclidean and hence a UFD. The quotient by a prime ideal is an integral domain.   How do I do the second part? I guess, I have to set a isomorphism by making a substitution such that only one variable remains. But I don't see how $x^2+y^2-1$ is going to help me make a choice.
117169
Let $R = (\mathbb{Z}/2)[x] $ and $I$ the ideal generated by the polynomial $x^{17} - 1$. Is there a non-zero element of $R/I$ s.t. the square of that element is zero in $R/I$? I know a quotient ring $R/P$ is an integral domain iff $P$ is a prime ideal, and since $R$ is a PID, the prime elements are the irreducible, but $x^{17} - 1 $ is reducible in $R$, hence $R/I$ is not an integral domain. So there are nonzero $a,b \in R/I$ s.t. $ab = 0 $, but $b$ might not be $a$, and that's where I got stuck. I've also tried finding $f \in R$ s.t. $ f^2 = g \cdot (x^{17} - 1)$, but don't know how to attack this methodically.
80299
I want to prove the following result:      Let $R$ be a ring and $M$ a maximal ideal in $R$. If $P$ is a prime ideal in $R[x]$ that (strictly) contains $M[x]$, then $P$ is a maximal ideal in $R[x]$.    I have an idea how to prove that, but I'm not quite sure if the argument is totally valid; maybe someone has a cleaner proof. My argument is like that:   $R[x]/M[x]$ is isomorphic to $(R/M)[x]$, that is a PID (since $R/M$ is a field). So, any prime ideal in this ring is maximal; If $P$ contains $M[x]$, then $P$ corresponds to a (prime??) ideal in $R[x]/M[x]$ (this needs more clarification); The prime ideal (that corresponds to $P$) in $R[x]/M[x]$ is then maximal, and this guarantees that $P$ is maximal in $R[x]$.   I'll appreciate any comment.
41252
Let $F$ be a field and $F[x]$ the univariate polynomial ring over $F$. Is it the case that every quotient ring ${F[x]}/{(p(x))}$, where $p(x) \in F[x]$, is a PID (principal ideal domain)?    This is just a minor issue that's been bothering me for the last couple of hours. I'm aware that when the ideal $(p(x))$ is prime (i.e. $p(x)$ irreducible) then ${F[x]}/{(p(x))}$ is a PID.
109765
For some of you, this question is going to seem extremely basic.  I think I understand what an ideal is. Such as $\langle 2, 1 + \sqrt{-5} \rangle$, it consists of all numbers in this ring of the form $2a + (1 + \sqrt{-5})b$.  But then what is $\langle 2, 1 + \sqrt{-5} \rangle^2$? My first thought was $\langle 4, -4 + 2 \sqrt{-5} \rangle$, but that seems wrong somehow. I also had something in the back of my mind saying $\langle 2 \rangle$, but I'm not sure about that one either.  Then I thought about trying to figure out $\langle 2, 1 + \sqrt{-5} \rangle \langle 2, 1 + \sqrt{-5} \rangle$ when I realized I don't actually understand how to multiply ideals to begin with.  I'm only using $\langle 2, 1 + \sqrt{-5} \rangle$ as an example (though that does draw in one question identified as similar that looks much more relevant than all the questions identified as "Questions that may already have your answer"). In a principal ideal domain, would it be correct to think that $\langle a \rangle \langle b \rangle = \langle ab \rangle$?  Any help would be much appreciated.
49117
I've got to prove that an ideal $Q$ whose radical is a maximal ideal is a primary ideal. That is, I want to prove that if $xy\in Q$, then $x\in Q$ or $y^n\in Q$ for some $n&gt;0$.    I've been trying for a while and I'm not sure where to begin. All I've got is that if $\text{Rad}(Q)$ is maximal and by definition it's the intersection of all prime ideals $P_i$ containing $Q$, then it must be equal to each of these $P_i$. So there is only 1 prime ideal containing $Q$, namely $\text{Rad}(Q)$.  Could anyone point me in the right direction? Thanks for any replies.
55902
Let $d \in \mathbb{Z}$ be a square free integer.  $R=\mathbb{Z}[\sqrt d]$ =$\{a+b\sqrt d | a,b \in \mathbb{Z} \}$. Overall, I'm trying to show that every prime ideal $P \subset R$ is a maximal ideal.     So far I showed that $I \subset R$ is finitely generated.  $I=\{(x,s+y \sqrt d)\}$ And now I'm trying to show that R/P, for some prime ideal is a finite ring with no zero divisors.  From there it would follow that R/P is a field and any prime ideal is maximal.   I know R could also be written as $R= \mathbb{Z}[x]/(x^2-d)$.  How could I show that R/P is a quotient of $\mathbb{Z}/n\mathbb{Z}[x]/(x^2-d)$?
67788
Let $S$ be a multiplicatively closed subset of a ring $R$, and let $I$ be an ideal of $R$ which is maximal among ideals disjoint from $S$. Show that $I$ is prime. If $R$ is an integral domain, explain briefly how one may construct a field $F$ together with a ring homomorphism $R\to F$. Deduce that if $R$ is an arbitrary ring, $I$ an ideal of $R$, and $S$ a multiplicatively closed subset disjoint from $I$, then there exists a ring homomorphism $f\colon R\to F$, where $F$ is a field, such that $f(x)=0$ for all $x\in I$ and $f(y)\neq 0$ for all $y\in S$.   [You may assume that if $T$ is a multiplicatively closed subset of a ring, and $0\notin T$, then there exists an ideal which is maximal among ideals disjoint from $T$.]  Here is a question I need to answer. For the first part I can show if $x, y$ are not in $I$ then nor does $xy$. The second part is just the field of fractions.   For the third part, I think I need to find an ideal $J$ containing $I$ such $J$ is prime, so that $R/J$ is an integral domain, and use the second part. To find $J$ prime, I think as suggested by the hint, I should go for a maximal ideal disjoint from $S$ containing $I$, but how can I do that?
71263
Let $R$ be a principal ideal domain and let $P$ be a nonprincipal prime ideal of $R[x]$. I'm having trouble seeing why $P$ must contain two elements with no common divisor.    Can anyone help me? Thanks.
15896
Here is the theorem and proof in my textbook.     Thm.  Let $P$ be an ideal of a commutative ring $R$ with identity $1$.  Then $P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain.   Proof.   Suppose that $P$ is a prime ideal of a commutative ring $R$ with $1$.  Then $P$≠$R$ implies $1+P≠0+P$. Hence $R/P$ is a commutative ring $R$ with identity.  Assume that $(a+P)(b+P)=0+P$.  Then $ab+P=0+P$ and $ab∈P$.  By the definition of a prime ideal P we get $a∈P$ or $b∈P$.  That is, $a+P=0+P$ or $b+P=0+P$.  Thus $R/P$ is an integral domain.   Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$ and $R/P$ is a commutative ring $R$ which has no zero divisors.  Hence $P≠R$.  Assume $ab∈P$.  Then $ab+P=0+P$ and $(a+P)(b+P)=0+P$.  Since $R/P$ is an integral domain, we get $a+P=0+P$ or $b+P=0+P$.  So $a∈P$ or $b∈P$.  Thus $P$ is a prime ideal.    I am having problems understanding proof with "Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$.".  I know that $R/P$ has an identity. (cause it is ID)  But, why doesn't "$P$" have an identity?  Isn't there a case $P=R$?  Please help me understanding. I think $P≠R$ should be with condition of Thm..  Thank you in advance.
81640
Let R be a commutative ring, and let P be a prime ideal of R. Suppose that P has no nontrivial zero divisors in it. Show that R is an integral domain.   My proof:  Take $r,s,a \in R$ with $ar = as$, and $p \in P$. Then  \begin{align*} par &amp;= pas\\ a p r &amp;= a p s &amp;&amp; \text{R is commutative} \end{align*} and since $P$ is an ideal of $R$, then $pr, ps \in P$. Since $P$ is an integral domain, then $a p r = a p s \implies p r = p s$, so $r = s$.  I'm not using the fact that $P$ is a prime ideal which leads me to believe that I messed up a step.