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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.

what about the forward implication?

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?

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  • $\begingroup$ The ring $A^X$ is a field (or division ring) iff $A$ is a field (or division ring) and $|X|=1$. $\endgroup$ Jun 24, 2020 at 15:35
  • $\begingroup$ @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? $\endgroup$
    – J.C.VegaO
    Jun 24, 2020 at 15:42

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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?

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  • $\begingroup$ 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? $\endgroup$
    – J.C.VegaO
    Jun 24, 2020 at 15:32
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    $\begingroup$ @J.C.VegaO Maybe it's worth your while to prove this: $A^X\cong \prod_{x\in X} A$. $\endgroup$
    – rschwieb
    Jun 24, 2020 at 15:33
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    $\begingroup$ 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? $\endgroup$
    – J.C.VegaO
    Jun 24, 2020 at 16:03
  • $\begingroup$ @J.C.VegaO, $A$ embeds into $A^X$. $\endgroup$
    – lhf
    Jun 24, 2020 at 16:23
  • $\begingroup$ @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 $A$ $\endgroup$
    – J.C.VegaO
    Jun 24, 2020 at 16:27

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