Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $k$ be a field and $S$ be an infinite set. Assume $|S| \leq |k|$. Why is then every $k$-algebra homomorphism $k^S \to k$ equal to a projection $\mathrm{pr}_s$ for some $s \in S$?

I don't know how to use the cardinality assumption here. The homomorphism corresponds to a special ultrafilter on $S$ and we have to show that it is principal - perhaps this helps?

share|cite|improve this question
up vote 2 down vote accepted

Suppose that $h:k^S\to k$ is a $k$-algebra homomorphism. For $A\subseteq S$ let $\chi_A\in k^S$ be defined by $$\chi_A(s) = \begin{cases}1_k,& s\in A\\ 0_k,& s\in S\setminus A\;.\end{cases}$$ Let $\mathscr{U}=\{A\subseteq S:h(\chi_A)\ne 0_k\}$.

If $A,B\in\mathscr{U}$, $\chi_{A\cap B} = \chi_A \chi_B$, so $h(X_{A\cap B}) = h(\chi_A)h(\chi_B) \ne 0_k$, and hence $A\cap B \in \mathscr{U}$. Now suppose that $A\in \mathscr{U}$ and $A \subseteq B \subseteq S$; then $\chi_A = \chi_A \chi_B$, so $0_k \ne h(\chi_A) = h(\chi_A)h(\chi_B)$, and therefore $h(\chi_B) \ne 0_k$, i.e., $B \in \mathscr{U}$. This shows that $\mathscr{U}$ is a filter on $S$.

Finally, let $A\subseteq S$, and for brevity let $A'=S\setminus A$. $\chi_A \chi_{A'} = 0_{k^S}$, so $h(\chi_A)h(\chi_{A'})=h(0_{k^S})=0_k$, and hence at most one of $A$ and $A'$ belongs to $\mathscr{U}$. On the other hand, $\chi_A+\chi_{A'}=1_{k^S}$, so $h(\chi_A) + h(\chi_{A'}) =h(1_{k^S})=1_k$, and at least one of $A$ and $A'$ belongs to $\mathscr{U}$. Thus, $\mathscr{U}$ is an ultrafilter on $S$. We can go further: since we already know that either $h(\chi_A)=0_k$ or $h(\chi_{A'})=0_k$, we can conclude that for each $A\subseteq S$, $A\in \mathscr{U}$ iff $h(\chi_A)=1_k$.

Let $I=\{\varphi\in k^S:Z(\varphi)\in\mathscr{U}\}$, where $Z(\varphi)=\{s\in S:\varphi(s) = 0_k\}$. It’s easy to check that $I$ is an ideal in $k^S$. $I$ is proper, since $\chi_S \notin I$. Moreover, $I$ is maximal. To see this, suppose that $J\supsetneq I$ is also an ideal, and fix $\varphi \in J\setminus I$. Then $Z(\varphi)\notin \mathscr{U}$, so $S\setminus Z(\varphi) \in \mathscr{U}$, and $\chi_{Z(\varphi)} \in I$ (since $Z(\chi_A) = S\setminus A$ for any $A\subseteq S$). It follows that $\varphi + \chi_{Z(\varphi)} \in I$. But $Z(\varphi + \chi_{Z(\varphi)}) = \varnothing$, so $\varphi + \chi_{Z(\varphi)}$ has a multiplicative inverse, and therefore $1_{k^S} \in J$.

Now let $\varphi \in I$, and let $\psi \in k^S$ be defined by $$\psi(s) = \begin{cases} \varphi(s)^{-1},&s\in S\setminus Z(\varphi)\\ 0_k,&s\in Z(\varphi)\;; \end{cases}\tag{1}$$ then $h(\varphi)h(\psi) = h(\chi_{S\setminus Z(\varphi)}) = 0_k$, so $h(\varphi)=0_k$ or $h(\psi)=0_k$. Say $h(\psi)=0_k$. Then $h(\chi_{Z(\varphi)} + \psi) =$ $h(\chi_{Z(\varphi)}) + h(\psi) = 1_k$, so $h(\varphi) = h((\chi_{Z(\varphi)}+\psi)\varphi) = h(0_{k^S}+\chi_{S\setminus Z(\varphi)}) = 0_k$. Thus, $I \subseteq \ker h$. Conversely, suppose that $h(\varphi) = 0_k$. Define $\psi$ as in $(1)$. Then $h(\chi_{S\setminus Z(\varphi)}) = h(\varphi)h(\psi) = 0_k$, so $Z(\varphi) \in \mathscr{U}$, and $\varphi \in I$. Thus, $I = \ker h$, and $k \cong k^S/I$.

Now suppose that $|S|<|k|$, and let $\varphi \in k^S$ be an injection. Let $\alpha = h(\varphi)$, and let $\psi = \alpha \cdot 1_{k^S}$. Then $h(\psi) = \alpha$, so $\varphi - \psi \in \ker h = I$, and therefore $Z(\varphi - \psi) \in \mathscr{U}$. But since $\varphi$ is injective and $\psi$ is constant, $|Z(\varphi - \psi)|\le 1$. $\varnothing \notin \mathscr{U}$, so $Z(\varphi - \psi) = \{s_0\}$ for some $s_0 \in S$, and $\{s_0\} \in \mathscr{U}$. Thus, $\mathscr{U}$ is principal, and $h(\varphi) = \varphi(s_0)$ for every $\varphi \in k^S$, i.e., $h = \mathrm{pr}_{s_0}$.

share|cite|improve this answer
Thank you! The last paragraph contains the actual proof, the rest is general (and can be proven much faster, think about idempotents). – Martin Brandenburg Oct 18 '11 at 7:24
@Martin: I suspected that it was a bit clumsy; it’s outside my usual haunts, so I was thinking it out as I went along. – Brian M. Scott Oct 18 '11 at 18:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.