# Local Rings

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 $p$, $p\mathbb{Z}$ 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. |

**Proof:**Let:

(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 $m_{\mathbf{p}}(V)$. $\blacksquare$