Give examples of maximal ideals in \(K = R \times R \times R \times \cdots\), the product of countably many copies of R. What about for a product of countably many copies of an arbitrary commutative ring \(R\)?
Solution
Assume throughout that \(R\) is a nonzero commutative ring with identity, and put \[ K=\prod_{i\in\NN}R. \]
1. Every maximal ideal \(\mathfrak m\subset R\) and every coordinate \(j\in\NN\) give a maximal ideal\[ M_{j,\mathfrak m}=\{(a_i)\in K:a_j\in\mathfrak m\}. \]
Let \[ \pi_j:K\to R\to R/\mathfrak m \] be projection to the \(j\)th coordinate followed by the quotient map. This is a surjective ring homomorphism and \[ \ker \pi_j=M_{j,\mathfrak m}. \] Since \(R/\mathfrak m\) is a field, the first isomorphism theorem gives \[ K/M_{j,\mathfrak m}\cong R/\mathfrak m, \] so \(M_{j,\mathfrak m}\) is maximal.
2. More generally, let \(\mathcal U\) be an ultrafilter on \(\NN\) and let \(\mathfrak m\) be a maximal ideal of \(R\). Then\[ M_{\mathcal U,\mathfrak m} = \left\{(a_i)\in K:\{i:a_i\in\mathfrak m\}\in\mathcal U\right\} \]is a maximal ideal of \(K\).
Let \(k=R/\mathfrak m\), a field. Coordinatewise reduction modulo \(\mathfrak m\) gives a surjective homomorphism \[ K\longrightarrow k^{\NN}. \] Compose this with the quotient map to the ultraproduct \[ k^{\NN}\longrightarrow k^{\NN}/\mathcal U. \] The ultraproduct of fields is again a field: if the class of \((x_i)\) is nonzero, then \[ S=\{i:x_i\ne0\}\in\mathcal U, \] and defining \(y_i=x_i^{-1}\) on \(S\) and arbitrarily off \(S\) gives \((x_i)(y_i)=1\) modulo \(\mathcal U\).
The kernel of the composite consists exactly of those \((a_i)\) for which \[ \{i:a_i\bmod\mathfrak m=0\} = \{i:a_i\in\mathfrak m\} \] belongs to \(\mathcal U\). Hence its kernel is \(M_{\mathcal U,\mathfrak m}\), and the quotient is a field. Therefore \(M_{\mathcal U,\mathfrak m}\) is maximal.
3. If \(\mathcal U\) is the principal ultrafilter at \(j\), then \(M_{\mathcal U,\mathfrak m}=M_{j,\mathfrak m}\); if \(\mathcal U\) is nonprincipal, then \(M_{\mathcal U,\mathfrak m}\neq M_{j,\mathfrak m'}\) for every \(j\) and every maximal ideal \(\mathfrak m'\) of \(R\).
If \(\mathcal U\) is principal at \(j\), then \(S\in\mathcal U\iff j\in S\), which gives the first equality. If \(\mathcal U\) is nonprincipal, it contains every cofinite subset of \(\NN\). Let \(e_j\in K\) have \(1\) in coordinate \(j\) and \(0\) elsewhere. Then \(\{i:(e_j)_i\in\mathfrak m\}\supseteq\NN\setminus\{j\}\) lies in \(\mathcal U\), so \(e_j\in M_{\mathcal U,\mathfrak m}\), while \(e_j\notin M_{j,\mathfrak m'}\) because \(1\notin\mathfrak m'\).