# The Unapologetic Mathematician

## Local Rings

Sorry for the break last Friday.

As long as we’re in the neighborhood — so to speak — we may as well define the concept of a “local ring”. This is a commutative ring which contains a unique maximal ideal. Equivalently, it’s one in which the sum of any two noninvertible elements is again noninvertible.

Why are these conditions equivalent? Well, if we have noninvertible elements $r_1$ and $r_2$ with $r_1+r_2$ invertible, then these elements generate principal ideals $(r_1)$ and $(r_2)$. If we add these two ideals, we must get the whole ring, for the sum contains $r_1+r_2$, and so must contain $1$, and thus the whole ring. Thus $(r_1)$ and $(r_2)$ cannot both be contained within the same maximal ideal, and thus we would have to have two distinct maximal ideals.

Conversely, if the sum of any two noninvertible elements is itself noninvertible, then the noninvertible elements form an ideal. And this ideal must be maximal, for if we throw in any other (invertible) element, it would suddenly contain the entire ring.

Why do we care? Well, it turns out that for any manifold $M$ and point $p\in M$ the algebra $\mathcal{O}_p$ of germs of functions at $p$ is a local ring. And in fact this is pretty much the reason for the name “local” ring: it is a ring of functions that’s completely localized to a single point.

To see that this is true, let’s consider which germs are invertible. I say that a germ represented by a function $f:U\to\mathbb{R}$ is invertible if and only if $f(p)\neq0$. Indeed, if $f(p)=0$, then $f$ is certainly not invertible. On the other hand, if $f(p)\neq0$, then continuity tells us that there is some neighborhood $V$ of $p$ where $f(p)\neq0$. Restricting $f$ to this neighborhood if necessary, we have a representative of the germ which never takes the value zero. And thus we can define a function $g(q)=\frac{1}{f(q)}$ for $q\in V$, which represents the multiplicative inverse to the germ of $f$.

With this characterization of the invertible germs in hand, it should be clear that any two noninvertible germs represented by $f_1$ and $f_2$ must have $f_1(p)=f_2(p)=0$. Thus $f_1(p)+f_2(p)=0$, and the germ of $f_1+f_2$ is again noninvertible. Since the sum of any two noninvertible germs is itself noninvertible, the algebra $\mathcal{O}_p$ of germs is local, and its unique maximal ideal $\mathfrak{m}_p$ consists of those functions which vanish at $p$.

Incidentally, we once characterized maximal ideals as those for which the quotient $R/I$ is a field. So which field is it in this case? It’s not hard to see that $\mathcal{O}_p/\mathfrak{m}_p\cong\mathbb{R}$ — any germ is sent to its value at $p$, which is just a real number.

March 28, 2011 - Comment by david holden (@etominusipi) | February 23, 2013 | Reply Comment by isomorphismes | February 23, 2014 | Reply