Ring Theory ideals

Question

If $R$ is a ring and $a,b \in R$. How do I show that $(a+b)^n -a^n \in \langle \,b\,\rangle$ for all natural numbers

$R$ need not be commutative or unital

I know to use induction. Just can’t figure how to use the assumption for n=k to get true for n=k+1 — Partey5, Oct 20 '19 at 23:21
Think about this. $\overline{a+b}=\overline{a}$ in $R:/\langle b\rangle$, so $(\overline a)^n=(\overline{a+b})^n$ in $R:/\langle b\rangle$, so $(a+b)^n-a^n\in\langle b\rangle$. — Rushabh Mehta, Oct 20 '19 at 23:28
If you do want to use induction, try writing $(a+b)^{k+1}$ as $(a+b)(a+b)^k$. Then look at doing some expanding and factorising so you can use your inductive hypothesis — Dave, Oct 21 '19 at 00:37
It doesn't matter whether $R$ is commutative or not. The expansion of $(a+b)^n$ has all words that have $k$ $a$s and $\ell$ $b$s, where $k+\ell=n$. If $R$ is not commutative, then the order matters. But in any case, every term except for $a^n$ contains a $b$, and hence lies in $\langle b\rangle$. — Arturo Magidin, Oct 21 '19 at 00:39

score 0 · Accepted Answer · answered Oct 21 '19 at 06:35

0

This can be simply proved by induction. It trivially holds for $n = 1$:

$(a + b) - a = b \in \langle b \rangle; \tag 1$

if

$(a + b)^k - a^k \in \langle b \rangle, \tag 2$

then since $\langle b \rangle \subset R$ is an ideal,

$(a + b)((a + b)^k - a^k) \in \langle b \rangle; \tag 3$

we have

$(a + b)((a + b)^k - a^k) = (a + b)^{k + 1} - (a +b)a^k = (a + b)^{k + 1} - a^{k + 1} - ba^k; \tag 4$

thus

$(a + b)^{k + 1} - a^{k + 1} - ba^k \in \langle b \rangle; \tag 5$

now since

$ba^k \in \langle b \rangle, \tag 6$

we conclude that

$(a + b)^{k + 1} - a^{k + 1} \in \langle b \rangle \tag 7$

as well.

answered Oct 21 '19 at 06:35

Robert Lewis

71,180

1

How does $ba^k \in $ when elements in this ideal are of the form $rbs$ where $r,s \in R$ ? – Partey5 Oct 21 '19 at 22:07
There is more to $\langle b \rangle$ than $RbR$; $\langle b \rangle = RbR + bR + Rb + \Bbb Z b$. Seen? – Robert Lewis Oct 21 '19 at 22:15

score 0 · Answer 2 · answered Oct 21 '19 at 08:01

Let $A = \{a,b\}$, considered as a finite alphabet. Then, in a non-commutative setting, $$ (a+b)^n = \sum_{u \in A^n} u $$ Here $A^n$ denotes the set of all words of length $n$. Apart from $a^n$, all these words contain at least one occurrence of $b$, and thus define an element belonging to the ideal generated by $b$. It follows that $(a+b)^n -a^n$ belongs to this ideal.

Ring Theory ideals

2 Answers2