Show that localizing a ring at a prime ideal produces a local ring.
Show that \(R\) is a local ring iff for every \(x\in R\), either \(x\) or \(1-x\) is a unit.
Show that if \(R\) is a local ring then \(R\setminus R^\times\) is a proper ideal that is contained in the Jacobson radical \(J(R)\) (and equals it).
Show that if \(R\neq 0\) is a ring in which every non-unit is nilpotent then \(R\) is local.
Show that every prime ideal is primary.
Show that every prime ideal is irreducible.
Solution
Let all rings here be commutative with identity.
If \(\mathfrak p\in\operatorname{Spec}R\), then in \(R_{\mathfrak p}\) every fraction \(a/s\) with \(a\notin\mathfrak p\) is a unit, with inverse \(s/a\). Thus the nonunits are exactly \[ \mathfrak pR_{\mathfrak p}, \] so this is the unique maximal ideal and \(R_{\mathfrak p}\) is local.
Suppose first that \((R,\mathfrak m)\) is local. If both \(x\) and \(1-x\) were nonunits, both would lie in \(\mathfrak m\), forcing \(1\in\mathfrak m\), impossible. Hence one of \(x,1-x\) is a unit.
Conversely, assume that for every \(x\), one of \(x,1-x\) is a unit. Let \(\mathfrak m\) be the set of nonunits. If \(x\in\mathfrak m\) and \(r\in R\), then \(rx\) is a nonunit. If \(x,y\in\mathfrak m\) but \(u=x+y\) were a unit, then \[ 1=u^{-1}x+u^{-1}y. \] Since \(u^{-1}x\) is a nonunit, the hypothesis makes \(u^{-1}y=1-u^{-1}x\) a unit, hence \(y\) a unit, contradiction. Thus \(\mathfrak m\) is an ideal. Every proper ideal consists of nonunits, so every proper ideal is contained in \(\mathfrak m\); hence \(\mathfrak m\) is the unique maximal ideal.
Therefore, in a local ring, \[ R\setminus R^\times=\mathfrak m=J(R). \]
Suppose \(R\ne0\) and every nonunit is nilpotent. A nilpotent element cannot be a unit, so the set of nonunits is exactly the nilradical \(\sqrt{(0)}\), which is an ideal. By part 2, \(R\) is local.
If \(\mathfrak p\) is prime and \(xy\in\mathfrak p\) with \(x\notin\mathfrak p\), then \(y\in\mathfrak p\), so \(\mathfrak p\) is primary (with exponent \(1\)).
If \(\mathfrak p=I\cap J\) and both \(I,J\) strictly contain \(\mathfrak p\), choose \(a\in I\setminus\mathfrak p\) and \(b\in J\setminus\mathfrak p\). Then \(ab\in IJ\subseteq I\cap J=\mathfrak p\), contradicting primality. Hence \(I=\mathfrak p\) or \(J=\mathfrak p\), so every prime ideal is irreducible.