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How Should I show that these $k$-algebras are not Isomorphic?

Question Show that the $k$-algebras $k[x,y]/\langle xy \rangle$ and $k[x,y]/\langle xy-1 \rangle$ are not isomorphic. Attempt At first, I thought $xy=0$. This would mean both $x$ and $y$ are zero ...
Mr Prof's user avatar
  • 451
1 vote
0 answers
66 views

$R = \mathbb R[X,Y]/(XY - 1)$ and $I$ be the ideal of $R$ generated by the image of the element $X - Y$ in $R$. Describe $R/I$

Let $R = \mathbb R[X,Y]/(XY - 1)$ ($\mathbb R$ is the set of real numbers) and I be the ideal of R generated by the image of the element X - Y in R. I want to find a way to describe R/I, i.e. find a ...
Jishnu's user avatar
  • 21
2 votes
0 answers
93 views

Is $(I(R:_{Q(R)} I))^n$ generated by $(fI)^n$ as $f$ varies over $(R:_{Q(R)} I)$?

Let $(R, \mathfrak m)$ be a Noetherian local domain of dimension $1$ which is not a UFD. Let $Q(R)$ be the fraction field of $R$. If $I\subsetneq \mathfrak m$ is a non-zero, non-principal ideal of $R$ ...
Alex's user avatar
  • 433
4 votes
1 answer
125 views

Counterexample of Divisibility of Ideals with Product of Ideals

Given a commutative ring $R$ with unity, we define for $I,J\subseteq R$ ideals $I\ \vert\ J\iff I\supseteq J$ $IJ=\{\sum_i a_ib_i:a_i\in I, b_i\in J\}$ For every commutative unitary ring $R$ it ...
tripaloski's user avatar
1 vote
1 answer
58 views

ID's, PID's, Noetherian rings and valuation rings: implications amongst them

I am trying to establish some implications between being an ID, a PID, a Noetherian ring and a valuation ring. First of all, I know that PID $\Rightarrow$ Noetherian, because in a PID every ideal is ...
kubo's user avatar
  • 2,067
0 votes
1 answer
130 views

If $R$ is a ring such that $\forall x\in R, x^n=x$ for some $n>1$ then when $P$ is prime, why is $R/P$ finite?

I was recently reading a post on MSE which had an argument like: If $P$ is a prime ideal of a ring $R$ all of whose elements satisfy $x^n = x$, then $R/P$ is an integral domain with the same property....
Thomas Finley's user avatar
0 votes
1 answer
58 views

Free submodules of an integral $R.$

I was told by the author of the answer here Showing that the rank of $M$ is exactly $1.$ that: Free submodules of an integral domain $R$ are exactly the principal ideals of $R.$ I am wondering which ...
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1 vote
0 answers
53 views

Is $I^n \not = I^{n+1}$ for all non-zero proper ideals $I$ of an integral domain? [duplicate]

For $R$ a (commutative with 1) integral domain, is it possible to have $I^n = I^{n+1}$ for some non-zero proper ideal $R$? I realise that in the case of a Noetherian domain, we can apply Krull's to ...
George's user avatar
  • 846
0 votes
1 answer
112 views

Does the ring $k[x,y,z,w]/(wx-yz)$ contain nonconstant invertible polynomial?

Problem: Let $k$ be an algebraically closed field, $k[x,y,z,w]$ be polynomial ring, and $(wx-yz)$ be the ideal generated by $wx-yz$. Does $A(X):= k[x,y,z,w]/(wx-yz)$ contain any nonconstant invertible ...
cxh007's user avatar
  • 475
0 votes
1 answer
20 views

Question related to infinitely generated ideals

Let $A$ be a domain and $X \subset A$. Then $(X) = \{a_1x_1 + \cdots + a_nx_n \mid n \in \mathbb{N},\, a_i \in A,\, x_i \in X\}$ What I understand from the above theorem is that if $x\in X$ then ...
ops's user avatar
  • 343
0 votes
1 answer
61 views

Equality of principal domains $(a) = (b)$ does not imply that $a = bu$ for some unit $u$. [duplicate]

I proved that in an integral domain that the implication always holds, but apparently it doesn't necessarily hold in all rings, and I have been looking for counterexamples. But, I'm confused why this ...
user avatar
1 vote
1 answer
85 views

Is it really necessary to work with the fraction field here?

Let $A$ be an integral domain, $\mathfrak{p}$ a prime ideal of $A$. Let $f(X)=a_nX^n+...+a_1X+a_0$ be a primitive and non constant polynomials of $A[X]$. We also suppose that $a_k \in \mathfrak{p}$ ...
Maman's user avatar
  • 3,330
2 votes
2 answers
169 views

r-ideals in commutative rings

I'm learning about r-ideals in commutative rings from a journal by Rostam Mohammadian. "A proper ideal I in a ring R is called an r-ideals (resp., pr-ideal), if ab is an element in I with ann(a)=<...
yash_'s user avatar
  • 23
1 vote
1 answer
909 views

Find the kernel of a homomorphism between polynomial rings

Specifically, I'm trying to solve the following problem: Let $R$ be an integral domain and let $x$, $y$ and $t$ be indeterminates. Let $R[x,y]$ denote the ring of polynomials in $x$ and $y$ over $R$...
Atsina's user avatar
  • 2,134
3 votes
4 answers
527 views

Integral Domains and Maximal Ideals

I need to show that if $R$ is an integral domain (containing three ideals: $\{0\}$, $I$ and $R$), then $a,b\in I\Rightarrow ab=0$. I know that since $R$ is an integral domain, $ab=0\Leftrightarrow a=...
user87938971's user avatar

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