Multiindices under Addition form Commutative Monoid

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Theorem

Let $Z$ be the set of multiindices.

Let $+$ denote the addition of multiindices.


Then $\left({Z, +}\right)$ is a commutative monoid.


Proof

We check each of the axioms in turn.

Let $k = \left \langle {k_j}\right \rangle_{j \in J}$, $\ell = \left \langle {\ell_j}\right \rangle_{j \in J}$ and $m = \left \langle {m_j}\right \rangle_{j \in J}$ be multiindices.


Closure

Trivially we have that:

$j \mapsto \left({ k_j + \ell_j }\right)$

is a sequence of integers indexed by $J$.

We know that finitely many of the $k_j$ are non-zero, and finitely many of the $\ell_j$ are non-zero.

Therefore finitely many of the $k_j + \ell_j$ are non-zero.

This shows that $k + \ell$ is a multiindex and so $+$ is closed.

$\Box$


Associativity

For all $j \in J$, we have:

\(\ds \left(\left(\ell + k\right) + m \right)_j\) \(=\) \(\ds \left(\ell + k\right)_j + m_j\)
\(\ds \) \(=\) \(\ds \left(\ell_j + k_j\right) + m_j\)
\(\ds \) \(=\) \(\ds \ell_j + \left(k_j + m_j\right)\) by Integer Addition is Associative
\(\ds \) \(=\) \(\ds \ell_j + \left(k + m\right)_j\)
\(\ds \) \(=\) \(\ds \left(\ell + \left(k + m\right) \right)_j\)

Thus $+$ is shown to be associative.

$\Box$


Commutativity

For all $j \in J$, we have:

\(\ds \left(\ell + k\right)_j\) \(=\) \(\ds \ell_j + k_j\)
\(\ds \) \(=\) \(\ds k_j + \ell_j\) By Integer Addition is Commutative
\(\ds \) \(=\) \(\ds \left(k + \ell\right)_j\)

Thus $+$ is shown to be commutative.

$\Box$


Identity Element

Let $0_Z$ be the multiindex defined by:

$\left(0_Z\right)_j = 0$

for all $j \in J$.

Then we have, for all $j \in J$:

\(\ds \left(0_Z + \ell\right)_j\) \(=\) \(\ds 0 + \ell_j\)
\(\ds \) \(=\) \(\ds \ell_j\)

Since we have seen that $\left({Z, +}\right)$ is commutative, this shows that $0_Z$ is an identity element for $Z$.

$\blacksquare$

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