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Inspired by this question we ask: Is there a name for each of the following properties about fields? what are some examples other than $\mathbb{Q}$?

  1. A field $K$ with the property that $K\otimes_{\mathbb{Z}} K$ is a field.

  2. A field $K$ is isomorphic to $K\otimes_{\mathbb{Z}} K$ (as field, or equivalently as rings)

The tensor products are taken over $\mathbb{Z}$ (the tensor product of $\mathbb{Z}$-modules, i.e., Abelian groups).

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    $\begingroup$ The rings $R$ such that the map $R\otimes_\mathbb{Z} R\to R$ are called solid rings and are completely classified. $\endgroup$
    – Denis Nardin
    Commented Feb 14, 2016 at 21:02
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    $\begingroup$ @DenisNardin Did you mean such that the multiplication map is an isomorphism? $\endgroup$
    – Todd Trimble
    Commented Feb 14, 2016 at 21:04
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    $\begingroup$ @ToddTrimble Yes, at least that is what I interpreted from the question. If I recall correctly the only solid fields are finite fields $\mathbb{F}_p$ and $\mathbb{Q}$. $\endgroup$
    – Denis Nardin
    Commented Feb 14, 2016 at 21:08
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    $\begingroup$ Actually now that I think about it if $R\otimes R$ and $R$ are fields the map $R\otimes R\to R$ needs to be an isomorphism, since it is a surjection and all maps of fields are monomorphisms so I think they exhaust all examples. $\endgroup$
    – Denis Nardin
    Commented Feb 14, 2016 at 21:12
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    $\begingroup$ @ToddTrimble let me rephrase Denis's comment. TFAE: 1) $K\otimes_{\mathbb Z}K$ is a field. 2) $K\otimes_{\mathbb Z}K\cong K$. Proof (hint): The multiplication map is always a surjective ring homomorphism. $\endgroup$
    – Fernando Muro
    Commented Feb 14, 2016 at 21:30

2 Answers 2

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I already wrote this in the comments but I think this might be worth of an answer. I think we can classify all fields $K$ such that $K\otimes K$ is a field.

Claim If $K$ is a field such that $K\otimes_\mathbb{Z}K$ is a field then the multiplication map $K\otimes_\mathbb{Z} K\to K$ is an isomorphism

In fact the multiplication map is always a surjection because $x\otimes 1$ gets sent to $x$. Moreover all maps of fields are injections, so the map is a bijection. But a bijective map of rings is an isomorphism.

Now the rings $R$ such that the multiplication map $R\otimes_\mathbb{Z}R\to R$ is an isomorphim have been classified by Bousfield and Kan in their paper The core of a ring. For our purposes we need lemmas 3.6 and 3.7 in said paper.

Now there are two possibilities for our field $K$: either it is of characteristic $p$ or it is of characteristic 0. If it is of characteristic 0 it cannot have any torsion subgroup. Hence from lemma 3.7 it needs to be a localization of $\mathbb{Z}$. Since it is a field it must be $\mathbb{Q}$.

If the characteristic of $K$ is $p$ it coincides with its own torsion subgroup. Hence from lemma 3.6 it is of the form $\mathbb{Z}/n$ for some $n$. Since $K$ is a field it must be $n=p$.

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    $\begingroup$ Once you've made the nice observation that $K\otimes_{\mathbb{Z}}K$ is a field iff the multiplication map is an isomorphism, I think it's a bit of overkill to look at the general classification of solid rings, as $K\otimes_{\mathbb{Z}}K$ is the same as $K\otimes_kK$, where $k$ is the prime subfield, and the multiplication map $K\otimes_kK\to K$ is never an isomorphism unless $k=K$, as if $x\in K\setminus k$ then $x\otimes1 - 1\otimes x$ is in the kernel of the multiplication map. $\endgroup$
    – Jeremy Rickard
    Commented Feb 14, 2016 at 21:47
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    $\begingroup$ @JeremyRickard I'm sure that the classification of solid rings is an overkill but I think you need to say something to explain why $x\otimes 1 = 1\otimes x$ implies $x\in k$ (something like choosing a basis of $K/k$ containing 1 etc.) $\endgroup$
    – Denis Nardin
    Commented Feb 14, 2016 at 21:55
  • $\begingroup$ @DenisNardin Thank you very much for your very interesting answer and helpful link "The core of a ring". $\endgroup$
    – Ali Taghavi
    Commented Feb 15, 2016 at 13:44
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Here is a self-contained argument. First, as Jeremy Rickard observes, $K \otimes K \cong K \otimes_k K$, where $k$ is the prime subfield of $K$ (so $\mathbb{Q}$ if $K$ has characteristic zero and $\mathbb{F}_p$ if $K$ has characteristic $p$). If $K \otimes_k K$ is a field, then as Denis Nardin observes, the multiplication map

$$K \otimes_k K \xrightarrow{m} K$$

must be injective, and since it's surjective it must be an isomorphism. But $K$ has dimension $1$ as a $K$-vector space, while $K \otimes_k K$ has dimension $\dim_k K$. It follows that $\dim_k K = 1$, hence that $K = k$.

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  • $\begingroup$ Thank you very much for your very interesting self contained argument. $\endgroup$
    – Ali Taghavi
    Commented Feb 15, 2016 at 13:43
  • $\begingroup$ It seems that your proof depends on the axiom of choice (or, at least, some weaker form of it). $\endgroup$
    – Todd Leason
    Commented Mar 18, 2020 at 20:55

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