I am currently reading up on normed algebra and am a bit confused about the conventions regarding the multiplicative properties of a norm.

In a normed ring the norm is only supposed to be submultiplicative in the sense that $\vert xy \vert \leq \vert x \vert\vert y\vert$. This is important, since for example most matrix-norms are only submultiplicative, and so are the norms of most Banach-algebras.

On the other hand a normed vector space is assumed to have a multiplicative norm in the sense that $\Vert\lambda \cdot v\Vert = \vert \lambda \vert \cdot \Vert v\Vert$. I would assume this will be the same for normed modules.

The problem with these definitions is: a normed ring is not a normed module over itself. To be fair, I am not sure how big of a problem it is. Most normed rings we consider modules over (like $\mathbb{Z,R,C},\mathbb{Z}_p,\mathbb{Q}_p, \mathbb{C}_p$) do have multiplicative norms. And I have not yet seen modules over general Banach-algebras.

So what is the correct thing to do here? To only consider normed modules over multiplicatively normed rings or to generalize the notion of normed module to only require submultiplicativity?

  • 1
    $\begingroup$ $\|a\|=\sup_p |a|_p$ is a submultiplicative norm on $\Bbb{Q}$ and yes it might be natural to consider modules over it. $\endgroup$
    – reuns
    Mar 6, 2023 at 17:12

1 Answer 1


The right definition is in fact $|xy|\leq|x||y|$, and this is the standard definition to use when talking about normed modules over normed rings. Note, however, that if $x$ is invertible and $|x^{-1}|=|x|^{-1}$, then we also have $|y|=|x^{-1}xy|\leq |x^{-1}||xy|=|x|^{-1}|xy|$ so $|xy|\geq |x||y|$ and thus $|xy|=|x||y|$. In particular, this is always true for the scalar multiplication on a normed vector space, if the norm on the field of scalars is multiplicative (which it is for the usual examples of $\mathbb{R},\mathbb{Q}_p$, etc. So for vector spaces (over a field with a multiplicative norm), submultiplicative norms and multiplicative norms are the same.

  • $\begingroup$ Thank you very much. $\endgroup$ Mar 7, 2023 at 9:24

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