Discussion between parisa and Thomas

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18:51

Suppose that $R$ be ring and $J$ is ideal of $R$ and $I$ is ideal of $J$. if $I$ have identity element show that $I$ is ideal of $R$. I don't have any idea to solve it. please help me with a hint.

Thomas

If $I$ contains the (multiplicative) identity, then $RI = R$?

parisa

@Thomas:$R$=$\mathbb Z\times\mathbb Z$ and $I$=$\mathbb Z\times 0$

Thomas

So you mean the additive identity?

parisa

@Thomas: no.since ideal is addaitive group must have additive identity

Thomas

But $\mathbb{Z}\times 0$ doesn't contain the multiplicative identity from $\mathbb{Z}\times\mathbb{Z}$.

parisa

18:51

@Thomas: yes mulitplicative identity is (1,0) and distinct from mulitiplicative identity of $\mathbb Z\times Z$

Thomas

Ok, I thought something was strange with your question. My second question is: In your definition of a ring, do you have a multiplicative identity?

parisa

I don't know R has identity

@Thomas

Thomas

This all depends a bit on whether or not your rings have a multiplicative identity. If you say that $I$ is an ideal in $J$, that makes it sounds like $J$ is a ring. If a ring has a $1$, then $J= R$, so I am guessing that your rings do not have a mutiplicative identity.

If your definition of a ring does not require the existence of a $1$ (multiplicative identity), then an ideal is actually a subring.

parisa

yes,if identity $R$ in $I$ then $I$=$R$. you say $R$ hasnot 1 but I has 1?

@user 1:dorost migam?

Thomas

My questions is: What does it mean that $I$ is an ideal of $J$?

parisa

19:00

@Thomas:may be we say that when Ideal has transitive law

@user 1:komak

Thomas

What book are you using?

parisa

this is question of champion

Thomas

I don't know that book.

I posted an answer for you.

parisa

thanks a lot

parisa

19:22

@Thomas:Iranian Math Olympiad for University Students question.

user 1

این سوال برا المپیاد ضایع است. به نظرم کامنت را پاک کنید بهتره. هرجور صلاحه

parisa

پاک نمشه:(

شما فارغ تحصیل ریاضی هستید یا دکتری میخونید؟خوش به حالتون پایتون خیلی خوبه

نه اشتباه میکنید.اطلاعاتتون خیلی خوبه.موفق باشید.راستی توماس بازم کامنت گذاشته

@Thomas.what do you means $0\in I$

@Thomas:@Thomas.what do you means $0\in I$

Thomas

19:41

With these problems you have to be attentive to your definitions. Now a ring $R$ is an Abelian group $(R, +)$ with one more operation (that we usually denote by multiplication). Because of this we say that $0 \in R$ to point out that $R$ has an additive identity element.

Now an ideal of a ring $R$ is a subgroup $I$ and because of this you will always have $0\in I$ for any ideal $I$.

parisa

yes.every ideal and sub ring of $R$ has 0

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