Let \(F\) be a field and let \(X\) be a finite set. Let \[ R(X,F)=\{f:X\to F\} \] with pointwise addition and multiplication. Determine all maximal ideals of \(R(X,F)\).
Solution
Assume first that \(X\neq\varnothing\). For \(x\in X\), define \[ M_x \coloneqq \{f\in R(X,F):f(x)=0\}. \]
1. Each \(M_x\) is a maximal ideal.
Evaluation at \(x\) is a surjective ring homomorphism \[ \operatorname{ev}_x:R(X,F)\longrightarrow F, \qquad f\longmapsto f(x). \] Its kernel is \(M_x\). Hence \[ R(X,F)/M_x\cong F, \] which is a field, so \(M_x\) is maximal.
2. For each \(x\in X\), let \(e_x\) be the characteristic function of\(\{x\}\):\[ e_x(y) = \begin{cases} 1,&y=x,\\ 0,&y\neq x. \end{cases} \]Then\[ e_x^2=e_x, \qquad e_xe_y=0\quad(x\neq y), \qquad \sum_{x\in X}e_x=1. \]
All three identities hold pointwise. The last sum is finite because \(X\) is finite.
3. Let \(M\) be any maximal ideal of \(R(X,F)\). There is a unique\(x_0\in X\) such that\[ e_{x_0}\notin M, \]and for every \(x\neq x_0\),\[ e_x\in M. \]
The quotient \[ K\coloneqq R(X,F)/M \] is a field. Let \(\bar e_x\) denote the image of \(e_x\) in \(K\). By step 2, each \(\bar e_x\) is idempotent. A field has only the idempotents \(0\) and \(1\), so \[ \bar e_x\in\{0,1\}. \] Also \[ \sum_{x\in X}\bar e_x=1, \] so at least one \(\bar e_x\) equals \(1\). Two distinct ones cannot both equal \(1\), because step 2 would then give \[ 1 = \bar e_x\bar e_y = 0. \] Thus exactly one, say \(\bar e_{x_0}\), equals \(1\), and all the others are \(0\). This is equivalent to the stated membership assertions.
4. Every maximal ideal \(M\) equals \(M_{x_0}\) for the point \(x_0\)from step
3.
Every \(f\in R(X,F)\) has the finite decomposition \[ f = \sum_{x\in X}f(x)e_x. \] Modulo \(M\), step 3 gives \[ \bar f = f(x_0)\bar e_{x_0} = f(x_0). \] The map from \(F\) to the quotient induced by constant functions is injective, because its kernel is an ideal of the field \(F\) and does not contain \(1\). Hence \[ \bar f=0 \quad\Longleftrightarrow\quad f(x_0)=0. \] Thus \[ f\in M \quad\Longleftrightarrow\quad f\in M_{x_0}, \] so \(M=M_{x_0}\).
5. If \(X\neq\varnothing\), the maximal ideals are exactly\[ \boxed{ M_x=\{f:X\to F:f(x)=0\}, \qquad x\in X. } \]
Step 1 shows that all displayed ideals are maximal, and step 4 shows that there are no others.
6. If \(X=\varnothing\), then \(R(X,F)\) is the zero ring and has nomaximal ideals.
There is exactly one function from the empty set to \(F\), so the additive and multiplicative identities coincide. The zero ring has no proper ideals, hence no maximal ideals.
7. Q.E.D.
Steps <1>5–<1>6 cover every finite set \(X\).