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$R$ 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: $\langle a\rangle \langle b\rangle = \langle ab\rangle$.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Mill May 13 '14 at 15:34
  • $\begingroup$ @Milad In all definitions of PID I've seen, an identity is required. $\endgroup$ – egreg May 13 '14 at 15:45
  • $\begingroup$ 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 $\endgroup$ – Mill May 13 '14 at 16:44
  • $\begingroup$ @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 $1$. $\endgroup$ – egreg May 13 '14 at 16:45
  • $\begingroup$ @Milad Anyway, I've added the argument when $1$ is not required. $\endgroup$ – egreg May 13 '14 at 16:54
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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: $\langle a\rangle \langle b\rangle = \langle ab\rangle$.

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

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