# if $A^X$ is a field, A being a ring, what can be concluded about $A$: field/ unital ring/ division ring?

Let $$A$$ and consider $$A^X$$, with $$X$$ a non-empty a set. The operations are point-wise addition and multiplication. I have proved that $$A^X$$ is a ring, that it is commutative iff A is commutative and that it is unital iff A is unital

Now I am stuck in this part where they ask to prove or disprove: Is $$A^X$$ a field iff A is a field ? Does one of the implication holds? What can be concluded about A?

So far: I know that if A is field, $$A^X$$ will not be a field because I can always define a function that has 0 in the codomain in at least one point and therefore because 0 doesn't have a multiplicative inverse the whole function will be not invertible. Then the reverse and the double implication are false.

If $$A^X$$ is a field, what can I say about $$A$$? is it necessarily a field? or maybe just a division ring or a unital ring?

• The ring $A^X$ is a field (or division ring) iff $A$ is a field (or division ring) and $|X|=1$. Jun 24, 2020 at 15:35
• @Geoffrey Trang For a field the inverse of the function $\phi=\{ (x_0,a)\}$ with $x_0$ the only element of X is $\phi^{-1}=\{ (x_0,a^{-1})\}$ for all $a$ different from zero. Could you help me with the case of a division ring?, in that case the inverse $a^{-1}$ would not unique because since there is no commutativity, there would be left and right inverses? Jun 24, 2020 at 15:42

So far it seems you're not quite coming to the right conclusion.

If $$A$$ is a field and $$|X|=1$$, then $$A^X$$ is certainly a field.

Functions in $$A^X$$ (if that's how you're thinking of it) are simply selecting an element of $$A$$. Therefore there's a function for every element of $$F$$, and addition and multiplication take place exactly as they do in $$F$$. In fact, it's isomorphic to $$F$$.

What you need to ask is

What happens when $$|X|> 1$$?

Then after that,

What happens when $$|X|=1$$, but $$A$$ is not a field?

• Thanks for pointing that out, there was a suggestion about it but I was not sure why is it important to break it into cases according to the cardinality of $X$. If instead of being a field , $A$ were just a division ring, for the case |X|=1, can I in the same manner conclude $A^X$ is a division ring? I am having trouble deciding that , because in a division ring, in general elements would have a right and left inverse right? Jun 24, 2020 at 15:32
• @J.C.VegaO Maybe it's worth your while to prove this: $A^X\cong \prod_{x\in X} A$. Jun 24, 2020 at 15:33
• Ok I have concluded that if $A$ is a field or division ring and $|X| >1$ , $A^X$ is not a field or division ring. and if $|X| =1$, it is. The problem is the inverse implication. If $A^X$ is a field or division ring, is $A$ a field or division ring?Can you help me with that? Jun 24, 2020 at 16:03
• @J.C.VegaO, $A$ embeds into $A^X$.
– lhf
Jun 24, 2020 at 16:23
• @lhf, that means that whatever happens to $A^X$, will happen to $A$? So in this case of $A^X$ is a field , so is $A$ Jun 24, 2020 at 16:27