# Local Rings – The Math

 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\$