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 May 25th, 2017, 02:26 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Let $R$ be a commutative ring with unity. Let $R$ be a commutative ring with unity. Consider the following two statements. S1 : If for any $\displaystyle a \in R, a^2=0$ implies $a=0$ then $R$ doesn't have nonzero nilpotent elements. S2: If $A$ and $B$ are two ideals of $R$ with $A+B=R$ then $\displaystyle A \cap B = AB$. I checked and found that S1 is true and S2 is false (not true in all cases). Please let me know if it's wrong thanks
 May 25th, 2017, 02:39 AM #2 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Another question I'm adding another question instead of opening a new thread... The kernel of a ring homomorphism from $R[X]$ to $C$ defined by $\displaystyle f(X) \rightarrow f(3+2i)$ is a) $x^2-6x+13$ b) $x^2+6x+5$ c) $R[X]$ d) {0} My answers are : Option A and D. Please check... thank you. Last edited by skipjack; May 25th, 2017 at 01:24 PM.
 May 26th, 2017, 03:25 PM #3 Member   Joined: May 2017 From: Russia Posts: 34 Thanks: 5 Hi! The kernel of the homomorphism is the principal ideal generated by polynomial $\displaystyle x^2-6x+13$. So, option D is incorrect. The conditions of S2 can be satisfied in some cases. For example, let $\displaystyle R=\mathbb{Z}$ and $\displaystyle A=(2),\ B=(3).$ $\displaystyle A$ and $\displaystyle B$ are principal ideals generated by 2 and 3 correspondingly. Then, $\displaystyle A\cap B = A\cdot B=(6)$. And $\displaystyle A + B = \mathbb{Z}$.
May 26th, 2017, 04:33 PM   #4
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Quote:
 Originally Posted by Lalitha183 I checked and found that S1 is true and S2 is false (not true in all cases).
Can you please show the counterexample you found to determine that S2 is false?

I don't mean to sound preachy (it's just my nature!) but writing down the symbols is a big part of the learning process. It would help you learn better if you would say, "I think such and so is false or true or whatever" and then write down your exact reasons, as formally as you can. That's how you'll learn to do these problems And when you get to grad school that's ALL they'll care about. No multiple choice in grad school. Proofs proofs proofs.

 May 27th, 2017, 11:16 PM #5 Member   Joined: May 2017 From: Russia Posts: 34 Thanks: 5 Sentence. Let R be a commutative Euclidean domain with unity. If $\displaystyle A$ and $\displaystyle B$ are two ideals of $\displaystyle R$ with $\displaystyle A+B=R$ then $\displaystyle A \cap B=AB$. Proof. $\displaystyle R$ is a Euclidean domain. So, $\displaystyle R$ is a unique factorization domain and principal ideal domain. $\displaystyle A=(a)$ and $\displaystyle B=(b)$ for some $\displaystyle a\in A, \ b\in B$. $\displaystyle A+B=R$ and $\displaystyle 1\in R$. So, $\displaystyle a\cdot r_1 + b\cdot r_2 = 1$ for some $\displaystyle r_1,r_2 \in R$. It is possible if and only if $\displaystyle a$ and $\displaystyle b$ are relatively prime. $\displaystyle A\cap B\supset AB$. If $\displaystyle r\in AB$ then $\displaystyle r$ is a multiple of both $\displaystyle a$ and $\displaystyle b$. It implies $\displaystyle A\cap B\subset AB$. $\displaystyle A\cap B = AB$. Last edited by skipjack; May 28th, 2017 at 07:08 PM.
 May 28th, 2017, 07:14 AM #6 Member   Joined: May 2017 From: Russia Posts: 34 Thanks: 5 And an example which doesn't fit S2. Let $\displaystyle R = \mathbb{Z}[x]$. And let $\displaystyle A=(2,\ x)$ --- the ideal generated by $\displaystyle 2$ and $\displaystyle x$. And $\displaystyle B=(3, x)$. Then $\displaystyle R=A+B$. $\displaystyle A$ and $\displaystyle B$ both contain element $\displaystyle x$: $\displaystyle x\in A\cap B.$ But $\displaystyle x\notin A\cdot B.$
 May 28th, 2017, 08:32 AM #7 Member   Joined: May 2017 From: Russia Posts: 34 Thanks: 5 Proposition. Let R be a commutative Euclidean domain with unity. If $\displaystyle A$ and $\displaystyle B$ are two ideals of $\displaystyle R$ with $\displaystyle A+B=R$ then $\displaystyle A \cap B=AB$. Proof. $\displaystyle R$ is a Euclidean domain. So, $\displaystyle R$ is a unique factorization domain and principal ideal domain. $\displaystyle A=(a)$ and $\displaystyle B=(b)$ for some $\displaystyle a\in A, \ b\in B$. $\displaystyle A+B=R$ and $\displaystyle 1\in R$. So, $\displaystyle a\cdot r_1 + b\cdot r_2 = 1$ for some $\displaystyle r_1,r_2 \in R$. It is possible if and only if $\displaystyle a$ and $\displaystyle b$ are relatively prime. $\displaystyle A\cap B\supset AB$. If $\displaystyle r\in AB$ then $\displaystyle r$ is a multiple of both $\displaystyle a$ and $\displaystyle b$. It implies $\displaystyle A\cap B\subset AB$. $\displaystyle A\cap B = AB$.

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