The set of scalar function can be structured to a field.

Question

I ask if it is possible to equip the set of scalar functions $\Bbb F^X$ between a set $X$ and a field $\Bbb F$ with a field structure. Indeed if $\Bbb F$ is a field and if $X$ is a set then for any $f,g\in\Bbb F^X$ I naturally thought to define an addition and multiplication in $\Bbb F^X$ through the following two identities $$ (f+g)(x):=f(x)+g(x)\\ (f\cdot g)(x):=f(x)\cdot g(x) $$ for any $x\in X$ and this seems to me that make the set $\Bbb F^X$ to a field but I'd like to know from otherone if this is effectively true. So could someone help me, please?

Answer 1

No this does not work, because we will always have zero divisors when $|X| > 1$ (which cannot exist in a field).

Indeed, fix distinct elements $x\neq y$ in $X$. Define $f: X \to \mathbb{F}$ to be $1$ in $x$ and $0$ elsewhere. Similarly, define $g: X \to \mathbb{F}$ to be $1$ in $y$ and $0$ elsewhere. Then $fg = 0$ yet $f \neq 0$ and $g \neq 0$. Hence, $\mathbb{F}^X$ is never a field, unless $|X| = 1$ in which case $\mathbb{F}^X\cong \mathbb{F}.$