Showing that a set is an ideal

[Mr Davis 97]

Homework Statement

Show that the collection of all nilpotent elements of a commutative ring $R$ is an ideal.

Homework Equations

The Attempt at a Solution

Showing that something is an ideal is somewhat straightforward, but I am a little confused as to what explicitly I have to show. If we denote $N$ as the set in question, then I know that we have to show that $aN \subseteq N$ and $Nb \subseteq$ for all $a,b \in R$. But what else do I have to show? Do I have to show that N is an additive subgroup?

 

[andrewkirk]

Do I have to show that N is an additive subgroup

Yes

 

[Mr Davis 97]

Yes

Consider $(a + b)^{m+n}$, where $a,b \in N$. In the binomial expansion, each summand contains a term $a^{i}b^{m+n−i}$. Now
either $i \ge m$ so that $a^i$ = 0 or $m+ n − i \ge n$ so that $b^{m+n−i} = 0$. Thus each summand of $(a + b)^{m+n}$
is zero, so $(a + b)^{m+n}$= 0 and $N$ is closed under addition. Also, since $0^1 = 0$, $0 \in N$.
Also $(−a)^m$ is either $a^m$ or $−a^m$, so $(−a)^m = 0$ and $-a \in N$.

Does this show that $N$ is a additive subgroup of $R$?

 

[fresh_42]

Consider $(a + b)^{m+n}$, where $a,b \in N$. In the binomial expansion, each summand contains a term $a^{i}b^{m+n−i}$. Now
either $i \ge m$ so that $a^i$ = 0 or $m+ n − i \ge n$ so that $b^{m+n−i} = 0$. Thus each summand of $(a + b)^{m+n}$
is zero, so $(a + b)^{m+n}$= 0 and $N$ is closed under addition. Also, since $0^1 = 0$, $0 \in N$.
Also $(−a)^m$ is either $a^m$ or $−a^m$, so $(−a)^m = 0$ and $-a \in N$.

Does this show that $N$ is a additive subgroup of $R$?

I think you need the power $n+m+1$. But basically, yes. For an ideal you also need $r \cdot a \in N$.

 

[Mr Davis 97]

I think you need the power $n+m+1$. But basically, yes. For an ideal you also need $r \cdot a \in N$.

In the solution to this problem in the book, it also shows that $N$ is closed under multiplication of elements in $N$. It shows that $(ab)^{mn} = (a^m)^n(b^n)^m = (0)(0) = 0$, so $ab \in N$. Isn't this unnecessary? Don't I only need to show that $N$ is an additive subgroup and that is satisfies the absorption property for ideals? Where would being internally closed under multiplication fit in?

 

[fresh_42]

In the solution to this problem in the book, it also shows that $N$ is closed under multiplication of elements in $N$. It shows that $(ab)^{mn} = (a^m)^n(b^n)^m = (0)(0) = 0$, so $ab \in N$. Isn't this unnecessary? Don't I only need to show that $N$ is an additive subgroup and that is satisfies the absorption property for ideals? Where would being internally closed under multiplication fit in?

Well, it does no harm and is almost obvious in a commutative ring. It shows, that $N$ carries also a ring structure. But if $r \cdot a \in N$ for all $r \in R\, , \,a \in N$, doesn't this imply $a\cdot b \in N$ for all $a,b \in N$?

 

[Mr Davis 97]

Well, it does no harm and is almost obvious in a commutative ring. It shows, that $N$ carries also a ring structure. But if $r \cdot a \in N$ for all $r \in R\, , \,a \in N$, doesn't this imply $a\cdot b \in N$ for all $a,b \in N$?

That's what I mean. It seems superfluous. I just want to make sure I know what is strictly necessary to show that some set is an ideal of a ring.

 

[fresh_42]

That's what I mean. It seems superfluous. I just want to make sure I know what is strictly necessary to show that some set is an ideal of a ring.

Then (minimal) it's (for left ideals)

1. $N \neq \emptyset$
2. $a - b \in N$ for all $a,b \in N$
3. $r \cdot a \in N$ for all $r\in R \, , \,a \in N$

Of course left and right ideals don't have to be distinguished in a commutative ring, but in general they have to be. And the first condition cannot be omitted by the second ($a-a \in N)$, as in case $N$ is empty, the second condition is still true, whereas $0 \notin N$, which is needed. So we can also write $0 \in N$ as first condition.