All Questions
Tagged with ideals principal-ideal-domains
174
questions
3
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Question about an example on ring theory from Dummit and Foote
Background
Example: If $p$ is a prime, the ring $\Bbb{Z}[x]/p\Bbb{Z}[x]$ obtained by reducing $\Bbb{Z}[x]$ modulo the prime ideal $(p)$ is a Principal Ideal Domain, since the coeffiencets lie in the ...
- 3,683
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Mistake in Proof "Every unique factorization domain is a principal ideal domain"
While doing my Algebra HW, I "proved" that every unique factorization domain (UFD) is a principal ideal domain (PID). I know that this is not true, however I fail to see where exactly is ...
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Prove that $R / J$ is a field and determine $|R / J|$ with $J=(1-2 i) \subseteq \mathbb{Z}[i] .$
Exam question: In the ring of Gaussian integers $\mathbb{Z}[i]$, consider the ideal
$J=(1-2 i) \subseteq \mathbb{Z}[i] .$
(a) Prove that $R / J$ is a field and determine $|R / J|$.
(Check: The order $|...
- 317
2
votes
2
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An easier example of a non-PID where every finitely generated ideal is principal
Say that an integral domain $\mathcal{X}$ is an almost-PID if $\mathcal{X}$ is not a PID but every finitely generated ideal of $\mathcal{X}$ is principal. The question of whether almost-PIDs exist ...
- 249k
1
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2
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The polynomial ideal associated to a prime ideal is prime
Let $A$ be a ring, $I \subset A$ a prime ideal and $\Phi : A \rightarrow A[x]$ the canonical inclusion homomorphism. Is $I[x]= \{ \sum_{k=0}^n a_k x^k : n\in \Bbb{N}, (a_k)_{k=0}^n \in I^n \}$ also ...
- 731
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Let $R$ be a P.I.D. and let $a \neq 0$ be an element in $R$. Prove that for a prime element $p$, $p(R/(a)) = ((p) + (a))/(a)$.
Let $R$ be a P.I.D. and let $a \neq 0$ be an element in $R$. Prove that for a prime element $p$, $p(R/(a)) = ((p) + (a))/(a)$.
This is something that came up while reading Dummit and Foote's textbook ...
- 3,800
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1
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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 ...
- 2,067
0
votes
1
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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 ...
user965463
2
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0
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108
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Exhibit all the ideals in the ring $F[x]/(p(x))$, where $F$ is a field and $p(x)∈F[x]$ (describe them in terms of the factorization of $p(x)$)
This is Exercise 9.2.5 in Dummit and Foote's Abstract Algebra
Exhibit all the ideals in the ring $F[x]/(p(x))$, where $F$ is a field and $p(x)$ is a polynomial in $F[x]$ (describe them in terms of ...
- 116
0
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Let $R$ be a factorial ring in which every ideal generated by two elements is a principal ideal. Show that $R$ is a principal ideal ring. [duplicate]
I want to check if my solutions for this problem are right.
Let $R$ be a factorial ring in which every ideal generated by two elements is a principal ideal.
Show that $R$ is a principal ideal ring.
...
- 900
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Find a basis for an ideal $I$ of $k[X]$ viewed as a $k$-subspace of the vector space $k[X]$.
I am looking for an answer to the following question.
If $k$ is a characteristic zero field and $I=(X^3+X^2,X^6)k[X]$, find a basis for the $k$-subspace $I$ of the vector space $k[X]$.
It is clear ...
- 3,619
0
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1
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119
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Ring that is not a principal ideal domain has at least 6 ideals
I am stuck on the proof for this. This is what I manage to conclude so far: we have two trivial ideals, (0) and (1). Since it is not a PID, the ring R cannot be a field, so it must have some element ...
- 3
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0
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31
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If $P\cap B$ is maximal in $B$, then $P$ is maximal in $B[y]$
As part of a much larger proof, I have encountered the following situation. I have that $B$ is a PID and $P$ is a prime ideal of the polynomial ring $B[y]$. I know that the intersection of $P$ and $B$ ...
- 2,067
3
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2
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196
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$P$ is a prime ideal $\iff$ for $r, s \in R$ such that $rRs \subset P$, then $r \in P$ or $s \in P$.
If $P$ is an ideal in a not necessarily commutative ring $R$, then the following conditions are equivalent:
a) $P$ is a prime ideal,
b) If $r,s\in R$ such that $rRs\subset P$ then $r\in P$ or $s\in P$...
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Let $\psi:\mathbb Z[x] \rightarrow \mathbb R$ be the homomorphism defined by $\psi(p(x))=p(\sqrt3)$.
Let $\psi:\mathbb Z[x] \rightarrow \mathbb R$ be the homomorphism defined by $\psi(p(x))=p(\sqrt3)$.
a) Prove that the kernel of $\psi$ is a principal ideal.
b) Find the subring $S$ of $\mathbb R$ ...
