# What does the word “ideal” mean in this context?

In Frank W. Warner's book Foundations of Differentiable Manifolds and Lie Groups, it says on page 12

Let $F_m$, a subset of $\bar{F_m}$ (the set of germs at $m$), be the set of germs which vanish at m. Then $F_m$ is an ideal in $\bar{F_m}$, and we let $F_m^k$ denote its kth power. $F_m^k$ is the ideal of $\bar{F_m}$ consisting of all finite linear combinations of k-fold products of elements of $F_m$.

Is an ideal this thing?

What does it mean to take the $k$th power of an ideal?

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Warner has a manifold? :) You should probably mention the name by its correct title. – Mariano Suárez-Alvarez May 8 '12 at 20:11
Ideal is used there in the sense of ring theory. – Mariano Suárez-Alvarez May 8 '12 at 20:12
A few lines before, he says The operations of addition, scalar multiplication, and multiplication of functions induce on F_m tilda the structure of an algebra over $\mathbb R.$ – Will Jagy May 8 '12 at 20:12
Thanks! I guess ring theory is a prerequisite. – sli May 8 '12 at 20:15
See Wikipedia: en.wikipedia.org/wiki/Ideal_(ring_theory) – Dylan Moreland May 8 '12 at 20:24

(this should really be a comment but, embarrassingly, I can't figure out how to make one.)

As others have mentioned, the term ideal comes from ring theory. It's analogous to a normal subgroup - if you know group theory. Do see the wikipedia page. You can take products of two ideals $I$ and $J$ - it's the ideal $IJ$ generate by all products $xy$ for $x \in I$ and $y \in J$. The kth power of an ideal is simply the product with itself k times.

You don't necessarily need a course in ring theory to read Warner, but it's helpful to have a friendly algebra book nearby for the things you haven't seen before. My favourites are Abstract Algebra by Dummit and Foote and Algebra by Artin.

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mebassett: Because you did not have 50 reputation points yet, you could only comment on your own questions and answers. So, no need to have been embarassed; the "add comment" button only appears once you gain 50 points. – Zev Chonoles Jun 9 '12 at 4:01
If you'd like, I can convert this into a comment on the question. However, I think it works equally well as an answer. – Zev Chonoles Jun 9 '12 at 4:04

I'm also starting to read this book. In this definition 1.13, it is written:

The operations of addition, scalar multiplication, and multiplication of functions induce on $\tilde{F}_{m}$ the structure of an algebra over $\mathbb{R}$.

Then you need to know the definition of $\mathbb{R}$-algebra, i.e., a structure of an algebra over $\mathbb{R}$. You'll find a good definition in Algebra, by Birkhoff-Mac Lane. OK, let's continue: since $\tilde{F}_{m}$ is $\mathbb{R}$-algebra, it is in particular a ring. Therefore $F_{m}$ is an ideal of $\tilde{F}_{m}$ with respect to the ring structure that exists on $\tilde{F}_{m}$. The same applies to $F_{m}^{k} \subset \tilde{F}_{m}$; it is also an ideal, but a little more complicated:

$$F_{m}^{k}=\left\lbrace [h]=a_{1}[g_{1}]+\cdots+a_{n}[g_{n}]\;\middle\vert\; {{n\in \mathbb{N}, a_{i}\in \mathbb{R},\text{ and }[g_{i}]=[f_{1}][f_{2}]\cdots[f_{k}],}\atop {\text{where }[f_{j}]\in F_{m}, 1\leq j\leq k,\text{ and }1\leq i\leq n }}\right\rbrace.$$ I hope I have helped; be sure to read the definition of $\mathbb{R}$-algebra, there are two different structures within the same set.

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I've improved the formatting on your post a bit; please check that I haven't changed your intended meaning (if I did, my apologies). – Zev Chonoles Jun 9 '12 at 4:11