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Let a ring $R$ be a finite union of fields all having the same unit. I want to prove that $R$ is itself a field.

I wrote $R=\bigcup _{i=0}^{n}F_i$, with $F_0=\{0,1\}$ and $F_i$'s are fields. Since we deal with a finite union, there must exist $j\geq 1$ such that $F_k\subseteq \bigcup _{i=k+1}^nF_i$ for $k<j$, but $F_j\nsubseteq \bigcup _{i=j+1}^nF_i$ . Then I put $B=\bigcup _{i=j+1}^nF_i$. Certainly, we have $R=\bigcup _{i=j}^nF_i$. I tried to show that $R=F_i$, for one of the latter $F_i$'s. Suppose not, so we could choose $b\in B-F_j$ (because $F_j\neq R$) and $a\in F_j-B$. The element $ab\in R$ is either in $F_j$ or in $B$. In the first case, $b=a^{-1}ab\in F_j$, which is a contradiction. In the other case, could we deduce that $a=abb^{-1}\in B$ to reach a contradiction? In fact, $b$ and $b^{-1}$ are both in $B$. But, we are not sure whether $B$ is multiplicatively closed to reach the desired contradiction.

Any help is appreciated!

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karparvar
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    Every element has an inverse. – Elle Najt Jul 19 '17 at 17:17
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    @AreaMan, yes, that's the easier part, but you also have to show that it is commutative. – Marcus M Jul 19 '17 at 17:31
  • I generally assume rings are commutative and I think this is common for a lot of people who do algebraic geometry. Maybe you should mention that $R$ is not necessarily commutative at the start just to clear things up. – Asvin Jul 19 '17 at 17:35
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    Hope this would be helpful: Let $K=\cap F_i$, then $R$ and each $F_i$ are all $K$-vector spaces. If $K$ is infinite, then $R=F_i$ for some $i$. If $R$ is finite, then $R$ is a finite division ring, and hence a field. – chan kifung Jul 19 '17 at 18:09
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    I think @chankifung is on to something. If $K$ is infinite, then a vector space over $K$ cannot be written as a finite union of proper subspaces. So if $K$ is infinite then $R=F_i$, and you are done. – Jyrki Lahtonen Jul 19 '17 at 20:05
  • (cont'd)OTOH if $K$ is finite, then we can proceed as follows. $R$ is also a vector space over $K$, so for the union to play out as in the linked question, we must have enough many fields $F_i$ such that $R/F_i$ is a finite dimensional vector space over $K$. But $R$ is also a vector space over any such $F_i$. This implies that $F_i$ must be finite dimensional over $K$ (otherwise the codimension is also infinite). Consequently $R$ must also be finite dimensional over $K$. Thus $K$ finite implies that $R$ is a finite division ring, and Wedderburn's theorem settles the main claim. – Jyrki Lahtonen Jul 19 '17 at 20:21
  • Probably there are better ways of organizing that train of thought :-/ – Jyrki Lahtonen Jul 19 '17 at 20:22
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    @JyrkiLahtonen, why $F_i$ must be finite dimensional over $K$, what's the problem if $R/K$ is infinite dimensional? – chan kifung Jul 19 '17 at 20:28
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    @chankifung $R/F_i$ has positive dimension as a space over $F_i$ and finite dimension as a space over $K$, so $F_i$ must have finite dimension over $K$. My goal in the case $|K|<\infty$ is to prove that we must also have $|R|<\infty$, so therefore it is essential that we should prove $\dim _KR<\infty$. – Jyrki Lahtonen Jul 19 '17 at 20:41
  • @JyrkiLahtonen, I see, nice solution. – chan kifung Jul 19 '17 at 20:45
  • You probably need to assume the ambient ring is finite too. Otherwise we can take $F_1= $ the real line and $F_2 =$ the Gaussian integers. They share a unit but the union is not a field. But the ambient ring $\mathbb C$ is not finite either – Daron Jul 20 '17 at 12:28

2 Answers2

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Warning: This argument has a gap. I need the following result:

Lemma. Assume that $V$ is a vector space over a finite field $K$ such that it is a finite union of proper subspaces $V_i, i=1,2,\ldots,n$. Then at least for some $i$ we have $\dim_KV/V_i<\infty$.

This is very likely true, and for some reason I thought it would follow from the argument in my linked answer. Unfortunately it doesn't, and I don't see a way to close this hopefully small, but crucial gap. Leaving the rest on for now in case it inspires someone else (sleeeeeepy).

Building upon Chan Kifung's comment.

Let $K=\bigcap_i F_i$. Clearly $K$ is a subfield of all the fields $F_i$ and, in turn all the fields $F_i$ as well as $R$ itself are vector spaces over $K$.

By the results of this thread a vector space can be a finite union of proper subspaces only when the field $K$ is finite and some (actually many) of the subspaces have a finite codimension.

If some $F_i$ is all of $R$ we are done, so we are left with the case $|K|<\infty$ and can infer that there exists at least one subfield $F_i$ such that $\dim_K(R/F_i)$ is finite. But, as $F_i$ was assumed to be a proper subspace of $R$, we have $\dim_{F_i}R>1$ and hence $\dim_{F_i}(R/F_i)\ge1$. On the other hand $$ \dim_K(R/F_i)=\dim_{F_i}(R/F_i)\cdot \dim_K F_i\ge \dim_K F_i, $$ so we can conclude that $\dim_K F_i<\infty$ and hence $F_i$ is a finite field. But by the above reasoning this implies that $\dim_KR=\dim_K F_i+\dim_K(R/F_i)<\infty$.

The conclusion is that $R$ is a finite ring. It was pointed out by many that $R$ is necessarily a division ring. A theorem by Wedderburn states that any finite division ring is commutative, i.e. a field.

Jyrki Lahtonen
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  • Can you explain why $F_i$ is finite codimensional for some $i$. I can't find this in the post https://math.stackexchange.com/questions/60698/if-a-field-f-is-such-that-leftf-rightn-1-why-is-v-a-vector-space-over – chan kifung Jul 19 '17 at 21:19
  • Sorry about that @chankifung. I thought it was part of the argument, but noticed the same weakness myself. It is very likely true, but it is too late an hour here. I need to think about it. – Jyrki Lahtonen Jul 19 '17 at 21:26
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Kaplansky has proved in Canadian Journal of Mathematics, Vol.3 (1951), pp.290-292, that a ring $R$ for each element $x$ of which there exists an integer $n( x)$ with $x^{n(x)}\in Z(R)$, where $Z(R)$ is the center of $R$, is commutative. We prove this for our ring $R$. Indeed, if $n$ is the number of fields in the union, and $a,x\in R$ are arbitrary elements, then at least two of the elements $a,ax,ax^2,\dots ,ax^n$ fall into some field, and hence commute. So, we have $ax^kax^j=ax^jax^k$, for some integers $k>j\geq 0$. Thus, $x^{k-j}a=ax^{k-j}$. Since $0<k-j\leq n$, we may conclude that $$x^{n!}a=ax^{n!}.$$By what is proved by Kaplansky, we infer that the division ring $R$ (whose unit element is the common unit of the fields under union) is commutative, hence a field.

karparvar
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