Sum of principal ideals in a commutative rng

abstract-algebracommutative-algebraidealsrngs

Let's define the principal ideal of an element $a$ of a commutative ring $R$ with or without identity as $\langle a \rangle = R \cdot a + \mathbb Za$.

It looks like with this definition $\langle a \rangle \cdot \langle b \rangle = \langle a \cdot b \rangle$ in any commutative ring/rng:

  • $r \cdot (ab) + n(ab) = a \cdot (r \cdot b + nb) = (r \cdot a + na) \cdot b \implies \langle a \cdot b \rangle \subseteq \langle a \rangle \cdot \langle b \rangle$;
  • $(x \cdot a + na) \cdot (y \cdot b + mb) = (xy + mx + ny) \cdot (ab) + mn(ab) \implies \langle a \rangle \cdot \langle b \rangle \subseteq \langle a \cdot b \rangle$.

I am trying to check the properties for the sum of principal ideals:

  • $r \cdot (a + b) + n(a + b) = (r \cdot a + na) + (r \cdot b + nb) \implies \langle a + b \rangle \subseteq \langle a \rangle + \langle b \rangle$.

Question:

In which types of rings/rngs $\langle a \rangle + \langle b \rangle \subseteq \langle a + b \rangle$ (and, therefore, $\langle a \rangle + \langle b \rangle = \langle a + b \rangle$)?

Best Answer

With the help from Rob Arthan:

$\langle a \rangle + \langle b \rangle = \langle a + b \rangle$ for any $a$ and $b$ of a commutative ring/rng $R$ if and only if $R = \{ 0 \}$.

Proof:

  • If $R = \{ 0 \}$, then $\langle 0 \rangle$ is the only principal ideal of $R$;
    $\langle 0 \rangle + \langle 0 \rangle = \langle 0 + 0 \rangle = \langle 0 \rangle$.
  • Assuming $\langle a \rangle + \langle b \rangle = \langle a + b \rangle$ for any elements $a$ and $b$ of $R$;
    Then, for an element $x$: $\langle x \rangle = \langle x \rangle + \langle -x \rangle = \langle x + (-x) \rangle = \langle 0 \rangle$;
    $x$ is an element of $\langle x \rangle$, however the only element of $\langle 0 \rangle$ is $0$;
    Therefore, $x = 0$ for an arbitrary element $x$ of $R$.
Related Question