# Tagged Questions

227 views

### Equivalence Of Definitions Of Prime Ideal In Ring Without $1$

Let $R$ be a rng, so that $1\not\in R$. I am trying to show that following are equivalence of definition of prime ideal $P$; i) $AB\subseteq P$ with $A,B\subseteq R$ implies $A\subseteq P$ or ...
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### No nonzero proper ideals of $K$-algebra $A$ implies the ring $A$ has no nonzero proper ideals

This is from Seth Warner's Classical Modern Algebra. The problem is: If $A$ is a nontrivial $K$-algebra possessing no nonzero proper ideals, then there are no nonzero proper ideals of the ring ...
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### Where in the proof did Herstein use the fact that $A$ is a two-sided ideal of $R$?

I'm reading Noncommutative Rings by I. N. Herstein. The theorem I'm having trouble with is 1.2.5, on page 16 of the book. Some definition 1. Regular ideal An ideal $\rho \subset R$ is called a ...
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### In a PID without unit an ideal is maximal iff it is prime.

As the titles says, I need to show that in a PID $R$ an ideal is maximal iff it is prime. This is easy to do if $R$ has a multiplicative identity. I can not do it if $R$ does not have an identity. It ...
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### Is there any non-monoid ring which has no maximal ideal?

Is there any non-monoid ring which has no maximal ideal? We know that every commutative ring has at least one maximal ideals -from Advanced Algebra 1 when we are study Modules that makes it as a very ...
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### Why is $S/Z$ a domain for the ideal $Z=\{z\in S\mid za=0,\;\forall a\in R\}$ in $S$?

Suppose $R$ is a rng with no zero divisors, not necessarily commutative. I know $R$ can be embedded into a ring $S:=\mathbb{Z}\times R$ by identifying $r\in R$ with $(0,r)\in S$. The operations on $S$ ...
A maximal ideal is always a prime ideal, and the quotient ring is always a field. In general, not all prime ideals are maximal. 1 In $2\mathbb{Z}$, $4 \mathbb{Z}$ is a maximal ideal. ...
### $I+J=R$, where $R$ is a commutative rng, prove that $IJ=I\cap J$.
So I basically have to prove what is on the title. Given $R$ a commutative rng (a ring that might not contain a $1$), with the property that $I+J=R$, (where $I$ and $J$ are ideals) we have to prove ...