Definition: A Local Ring is a ring $R$ that has a unique maximal ideal. |

*Recall that a proper ideal $I$ of a ring $R$ is a maximal ideal if there exists no other proper ideals of $R$ containing $I$. In general, maximal ideals need not be unique. For example, consider the ring of integers $mathbb{Z}$. The ideals $2 mathbb{Z}$ and $3 mathbb{Z}$ are both maximal ideals. In fact, for prime $p$, $pmathbb{Z}$ is a maximal ideal. So $mathbb{Z}$ is NOT a local ring.*

For example, if $K$ is a field then $K$ is a local ring with unique maximal ideal $(0)$.

The following theorem gives us an alternative criterion for when a ring is a local ring.

Theorem 1: Let $R$ be a ring. Then $R$ is a local ring with maximal ideal $m$ if and only if $m = R setminus R^{times}$ is an ideal in $R$. |

**Proof:**Observe that if $I$ is an ideal of $R$ and if $I$ contains a unit then $I = (R)$.

- $Rightarrow$ Suppose that $R$ is a local ring with maximal ideal $m$. Then from the observation made above, since $m$ is a proper ideal (by definition), $m$ contains no units, and so:

(1)

begin{align} quad m subseteq R setminus R^{times} end{align}

- But $m$ is a maximal ideal, and so $m = R setminus R^{times}$.

- $Leftarrow$ Suppose that $m = R setminus R^{times}$ is an ideal. Then this ideal is maximal since any larger ideal must contain a unit and will not be proper. Furthermore, any other ideal of $R$ must be contained in $m$ and so $m = R setminus R^{times}$ is unique. So $R$ is a local ring with unique maximal ideal $m$. $blacksquare$

Theorem 2: Let $K$ be a field, $V subseteq mathbb{A}^n(K)$ a nonempty affine variety, and $mathbf{p} in V$. Then $O_{mathbf{p}}(V)$ is indeed a local ring. |

(2)

begin{align} quad m_{mathbf{p}}(V) = { f in O_{mathbf{p}}(V) : f(mathbf{p}) = 0 } end{align}

- We will show that $m_{mathbf{p}}$ is a maximal ideal. Let $phi : O_{mathbf{p}}(V) to K$ be the function defined for all $f in O_{mathbf{p}}(V)$ by:

(3)

begin{align} quad phi(f) = f(mathbf{p}) end{align}

- That is, $phi$ is the evaluation mapping. We will show that $phi$ is a ring homomorphism. Let $f, g in O_{mathbf{p}}(V)$. Then:

(4)

begin{align} quad phi(f + g) = (f + g)(mathbf{p}) = f(mathbf{p}) + g(mathbf{p}) = phi(f) + phi(g) end{align}

(5)

begin{align} quad phi(fg) = (fg)(mathbf{p}) = f(mathbf{p})g(mathbf{p}) = phi(f) phi(g) end{align}

- So indeed $phi$ is a ring homomorphism. Now observe that:

(6)

begin{align} quad ker phi = { f in O_{mathbf{p}}(V) : phi(f) = 0 } = { f in O_{mathbf{p}}(V) : f(mathbf{p}) = 0 } = m_{mathbf{p}}(V) end{align}

- Furthermore, $phi$ is surjective, so $phi(O_{mathbf{p}}(V)) = K$. By the first ring isomorphism theorem we have that $O_{mathbf{p}}(V) / ker phi cong phi(O_{mathbf{p}}(V))$, that is:

(7)

begin{align} quad O_{mathbf{p}}(V) / m_{mathbf{p}}(V) cong K end{align}

- But $K$ is a field. Therefore $m_{mathbf{p}}(V)$ is a maximal ideal.

- We now show that $m_{mathbf{p}}(V)$ is a unique maximal ideal. Let $f in O_{mathbf{p}}(V)$ be a unit. Then we must have that $f(mathbf{p}) neq 0$. But then $f not in m_{mathbf{p}}(V)$. Therefore $m_{mathbf{p}}(V)$ contains all of the non-units in $O_{mathbf{p}}(V)$. So every other ideal of $O_{mathbf{p}}(V)$ is contained in $m_{mathbf{p}}(V)$ which shows that $m_{mathbf{p}}(V)$ is the unique maximal ideal of $O_{mathbf{p}}(V)$.

- Hence $O_{mathbf{p}}(V)$ is a local ring with unique maximal ideal $m_{mathbf{p}}(V)$. $blacksquare$