Definition 4

Let A be a commutative unital algebra over K. A unital character is a unital morphism of A into K.

When no confusion can result from this, we shall simply say character instead of unital character. The set of unital characters of A is denoted by $\mathsf{X}(A)$. If A is the zero algebra, then $\mathsf{X}(A)$ is empty.

Let A and B be commutative unital algebras over K and $h$ a unital morphism of A into B. The map $\chi \mapsto \chi \circ h$ of $\mathsf{X}(B)$ into $\mathsf{X}(A)$ is denoted by $\mathsf{X}(h)$. If $k$ is a morphism of B into a commutative unital algebra, one has $\mathsf{X}(k\circ h) =\mathsf{X}(h)\circ \mathsf{X}(k)$. The map $\mathsf{X}$(Id$_A$) is the identity map of $\mathsf{X}(A)$.

If $h$ is surjective, $\mathsf{X}(h)$ is a bijection of $\mathsf{X}(B)$ onto the set of characters of A which vanish on the kernel of $h$.

Let $(A_1, e_1), . . . ,(A_n, e_n)$ be commutative unital algebras over K and let A be the unital algebra $A_1\times \cdots \times A_n$, with unit element $(e_1, . . . , e_n)$. For each $i$, identify $A_i$ with an ideal of A and let $\pi_i$ be the canonical mapping of A onto $A_i$. Then $\mathsf{X}(\pi_i)$ is a bijection of $\mathsf{X}(A_i)$ onto the set $\mathsf{X}_i$ of characters of A vanishing on $\prod_{j\not=i}A_j$. The sets $\mathsf{X}_i$ are pairwise disjoint. On the other hand, let $\chi \in \mathsf{X}(A)$. Since $1 =\sum\chi (e_i)$, there exists $i$ such that $\chi (e_i)\not= 0$. For every $j\not=i$ and every $y\in A_j$, one has $\chi (e_i)\chi (y) =\chi (e_iy) =\chi (0) = 0$, hence $\chi (A_j) = 0$. Thus, $\chi$ vanishes on $\prod_{j\not=i}A_j$, so that $\mathsf{X}(A)$ is the union of the $\mathsf{X}_i$.

Let B be the unital algebra $A_1\otimes \cdots \otimes A_n$. Let $h_i$ denote the canonical morphism $A_i\rightarrow B$. Then

is a map from $\mathsf{X}(B)$ into $\mathsf{X}(A_1)\times \cdots \times \mathsf{X}(A_n)$, and

is a map from $\mathsf{X}(A_1)\times \cdots \times \mathsf{X}(A_n)$ into $\mathsf{X}(B)$. One verifies that these maps are reciprocal bijections, by means of which one identifies $\mathsf{X}(B)$ with $\mathsf{X}(A_1)\times \cdots \times \mathsf{X}(A_n)$.

Let A be a commutative unital algebra over K. Let Y be the set of ideals of codimension 1 of A. For every $\chi \in \mathsf{X}(A)$, one has Ker($\chi$ )$\in Y$. The mapping $\chi \mapsto$ Ker($\chi$ ) is a bijection of $\mathsf{X}(A)$ onto Y. In fact, if $I\in Y$, there exists a unique isomorphism of the unital K-algebra $A/I$ onto K and the composite morphism

is the unique character of A with kernel I.