# $R$ is a PID. Show $\langle a\rangle \langle b\rangle =\langle ab\rangle$ and $\langle a\rangle ^n =\langle a^n\rangle$.

$$R$$ is a PID and $$\langle a\rangle$$ and $$\langle b\rangle$$ are ideals of $$R$$ generated by $$a$$ and $$b$$, resp. Show $$\langle a\rangle \langle b\rangle =\langle ab\rangle$$ and $$\langle a\rangle ^n =\langle a^n\rangle$$.

Please check out my following answer and let me know whether it's right, or where the problem is.

For the $$\langle a\rangle \langle b\rangle = \langle ab\rangle$$, here's my solution:

Obviously, $$\forall ab\in \langle a \rangle \langle b\rangle; ab\in \langle ab\rangle$$. So $$\langle a \rangle \langle b\rangle\subseteq \langle ab\rangle$$.
Also $$\forall ab\in \langle ab\rangle; ab\in \langle a \rangle \langle b\rangle$$, because
$$\langle a \rangle \langle b\rangle=\{(mn)(ab)+\cdots |-\}$$ and for $$mn=1$$, $$ab\in \langle a \rangle \langle b\rangle$$
Thus $$\langle ab\rangle\subseteq\langle a \rangle \langle b\rangle$$.
Therefore $$\langle a\rangle \langle b\rangle = \langle ab\rangle \square$$

For $$\langle a\rangle ^n =\langle a^n\rangle$$, here are two solutions. Are both of the valid?

1. By the principle of mathematical induction, for $$n=1$$, $$\langle a\rangle ^1 =\langle a^1\rangle$$. Let it be true for $$n=k$$. The following shows it's also true for $$n=k+1$$.
$$\langle a\rangle ^{k+1}= \langle a\rangle ^k \langle a\rangle=\langle a^k\rangle \langle a\rangle=\langle a^ka\rangle=\langle a^{k+1}\rangle \square$$

2. $$\langle a\rangle ^n=\overbrace{\langle a\rangle\cdots\langle a\rangle}^\text{n times} = \langle a\cdots a\rangle=\langle a^n\rangle \blacksquare$$

In both of these proofs (1 and 2) I have used the equation from the first problem: $$\langle a\rangle \langle b\rangle = \langle ab\rangle$$.

Saying $\forall ab\in\langle a\rangle\langle b\rangle$ doesn't make sense.

Let $x\in\langle a\rangle\langle b\rangle$; then, by definition, $$x=\sum_{i=1}^k (ay_i)(bz_i)$$ for some $y_i,z_i\in R$. But then $$x=\sum_{i=1}^k (ay_i)(bz_i)=ab\sum_{i=1}^k (y_iz_i)$$ so $x\in \langle ab\rangle$. Therefore $\langle ab\rangle\subseteq\langle a\rangle\langle b\rangle$.

Conversely, if $x=ax'\in\langle a\rangle$ and $y=by'\in\langle b\rangle$, then $xy=abx'y'\in\langle ab\rangle$, so $\langle a\rangle\langle b\rangle\subseteq\langle ab\rangle$.

The equality $\langle a^n\rangle=\langle a\rangle^n$ follows by induction on $n$. That part of your proof is correct, although too wordy.

For $n=1$ the assertion is obvious. Suppose it's true for $n$; then $$\langle a^{n+1}\rangle=\langle a^na\rangle=\langle a^n\rangle\langle a\rangle \overset{*}{=}\langle a\rangle^n\langle a\rangle\overset{**}{=}\langle a\rangle^{n+1}$$ where $\overset{*}{=}$ denotes application of the induction hypotheses and $\overset{**}{=}$ denotes the definition of iterated product of ideals.

If you don't assume existence of $1$, an element of $\langle a\rangle$ can be written in the form $na+ra$, with $n\in\mathbb{Z}$ and $r\in R$ (the ring is commutative). Thus an element in $\langle a\rangle\langle b\rangle$ can be written as a sum of elements of the form $$x=\sum_{i=1}^{k}(m_ia+y_ia)(n_ib+z_ib)=pab+rab$$ where $p\in\mathbb{Z}$ and $r\in R$. Also in this case we can conclude that $x\in\langle ab\rangle$.

Similarly for the other inclusion.

• Thanks, especially for teaching how not to be too wordy. :) <br>However, I've learned the definition of a generator to be as follows. <br>$\langle a\rangle = \{ na+ra+as+\sum_{i=1}^m x_iay_i | r,s,x_i,y_i\in R, n\in\mathbb{Z}, m\in\mathbb{N} \}$ <br>Your definition holds when $R$ is a ring with identity. – Mill May 13 '14 at 15:34
• @Milad In all definitions of PID I've seen, an identity is required. – egreg May 13 '14 at 15:45
• Am I being stupid? Can I trust Wikipedia's definition and say that it's not a ring with identity? Wiki: "In abstract algebra, a principal ideal domain, or PID, is an integral domain in which every ideal is principal, i.e., can be generated by a single element." link – Mill May 13 '14 at 16:44
• @Milad I wouldn't consider Wikipedia as the most authoritative source. However, the “Integral domain” article says the author follows the convention that integral domains have $1$. – egreg May 13 '14 at 16:45
• @Milad Anyway, I've added the argument when $1$ is not required. – egreg May 13 '14 at 16:54

For the $$\langle a\rangle \langle b\rangle = \langle ab\rangle$$, here's my solution:

Obviously, $$\forall ab\in \langle a \rangle \langle b\rangle; ab\in \langle ab\rangle$$. So $$\langle a \rangle \langle b\rangle\subseteq \langle ab\rangle$$.
Also $$\forall ab\in \langle ab\rangle; ab\in \langle a \rangle \langle b\rangle$$, because
$$\langle a \rangle \langle b\rangle=\{(mn)(ab)+\cdots |-\}$$ and for $$mn=1$$, $$ab\in \langle a \rangle \langle b\rangle$$
Thus $$\langle ab\rangle\subseteq\langle a \rangle \langle b\rangle$$.
Therefore $$\langle a\rangle \langle b\rangle = \langle ab\rangle \square$$

For $$\langle a\rangle ^n =\langle a^n\rangle$$, here are two solutions. Are both of the valid?

1. By the principle of mathematical induction, for $$n=1$$, $$\langle a\rangle ^1 =\langle a^1\rangle$$. Let it be true for $$n=k$$. The following shows it's also true for $$n=k+1$$.
$$\langle a\rangle ^{k+1}= \langle a\rangle ^k \langle a\rangle=\langle a^k\rangle \langle a\rangle=\langle a^ka\rangle=\langle a^{k+1}\rangle \square$$

2. $$\langle a\rangle ^n=\overbrace{\langle a\rangle\cdots\langle a\rangle}^\text{n times} = \langle a\cdots a\rangle=\langle a^n\rangle \blacksquare$$

In both of these proofs (1 and 2) I have used the equation from the first problem: $$\langle a\rangle \langle b\rangle = \langle ab\rangle$$.

• I believe you should have your reasoning in the question, rather than an answer, because you're asking about the proof. – egreg May 13 '14 at 15:05
• I put the solution in the description box as well. – Mill May 13 '14 at 15:13