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Suppose $D$ is a finite dimensional skew field over the field $K$. Futher, take $x \in D\setminus K$ and let $L=K(x)$. My question: is $D\otimes_K L$ a field?

I think not. However I can't seem to give a logical reasoning behind this.

Also, would it help knowing that $L \otimes_K L$ is not a field?

Jyrki Lahtonen
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Eric
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    How is the multiplication defined? Do you mean, 'are these skew fields'? – Berci Nov 20 '12 at 19:42
  • If you know $L \otimes_K L$ is not a field then don't you get a counterexample just taking D=L? – Noah Snyder Nov 20 '12 at 19:55
  • @Berci - Indeed I meant to ask whether $D\otimes_K L$ is a skew field. – Eric Nov 21 '12 at 00:15
  • @NoahSnyder - I thought that you can't just assume $D = L$, right? =) – Eric Nov 21 '12 at 00:16
  • Dear @Eric, Taking $D=L$ will give you a counterexample, but as my answer shows, the answer is "$D\otimes_KL$ is never a field," no matter what $D$ is. Of course I proved this by reducing to the case $D=L$. – Keenan Kidwell Nov 21 '12 at 01:40
  • You should phrase the question more precisely. If you're asking "is the claim $D \otimes_K L$ a skew field true or false" then exhibiting a single counterexample is enough (so you can take D=L). If you're asking "is $D \otimes_K L$ ever a skew field then you have to treat all cases. – Noah Snyder Nov 21 '12 at 09:34

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The ring $L$ is a field (i.e. not a skew field) since $K$ is a field and you're only adjoining one (algebraic) element. Write $L=K[T]/(f(T))$ where $f$ is the minimal polynomial of $x$ over $K$. Note that $f$ has degree $\geq 2$ since $x\notin K$. Then $L\otimes_KL=L[T]/(f(T))$, and since $f(T)$ can be factored as $(T-x)^kg(T)$ in $L[T]$ with $x$ not a root of $g$, i.e., $T-x$ and $g(T)$ relatively prime in $L[x]$, the ring $L[T]/(f(T))$ is isomorphic to $L[T]/(T-x)^k\times L[T]/(g(T))$, which is not a domain, because either $k\geq 2$, in which case you have nilpotents, or $k=1$ and $g(T)\neq 1$, in which case you at least have zero divisors.

In general, if $L/K$ is non-trivial, then $L\otimes_KL$ is not a field. If $L$ is algebraic, pass to a finite subextension and use the argument above. If $L$ is not algebraic, then a theorem of Grothendieck says that the dimension of $L\otimes_KL$ has Krull dimension equal to the transcendence degree of $L$, so not zero (there is almost certainly a much easier way to see this, i.e., reducing to the case $K(T)\otimes_KK(T)$, but I'm not seeing it at the moment).

EDIT: Here's an argument that works in general for the assertion that $L\otimes_KL$ is not a field when $L/K$ is non-trivial. The argument was suggested in the comments by YACP. If $L\neq K$, then there exist distinct elements $\alpha\neq\beta\in L$ which are $K$-linearly independent. We may assume they are contained in a $K$-basis for $L$, in which case $1\otimes\alpha$ and $1\otimes\beta$ are $L$-linearly independent elements of $L\otimes_KL$. In particular, $\beta\otimes\alpha-\alpha\otimes\beta\neq 0$ in $L\otimes_KL$. Now, the $L$-linear multiplication map $x\otimes y\mapsto xy:L\otimes_KL\rightarrow L$ is non-zero ring map (since the target is non-zero). Since $\beta\otimes\alpha-\alpha\otimes\beta$ clearly lies in the kernel, we see that $L\otimes_KL$ has a non-trivial, proper ideal (namely the kernel). Therefore it is not a field.

It follows that $D\otimes_KL$ is not a skew field. If it were, then $L\otimes_KL$, being a commutative subring, would be a domain. But it isn't.

Keenan Kidwell
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    I don't understand if you have troubles with the reduction to the case $K(T)\otimes_KK(T)$ when the extension $K\subset L$ is not algebraic or with the proof that this ring is not a field. In the second case, if $p:K(T)\otimes_KK(T)\to K(T)$ is defined by $p(a\otimes b)=ab$, then $p$ is a surjective ring homomorphism. If $p:K(T)\otimes_KK(T)$ is a field, then $\ker p=0$. But $1\otimes T-T\otimes 1\in\ker p$ and $1\otimes T-T\otimes 1\neq 0$. –  Nov 20 '12 at 21:09
  • Yeah, sure, that works. Thanks. I wasn't talking about the reduction, I was talking about the proof that it's not a field. But it was tangential to my answer. I just added it because it was vaguely related. In fact that same argument works with $K(T)$ replaced by any extension $L\neq K$, the point being that you can choose two $K$-linearly independent elements $\alpha_1,\alpha_2$. Then you can put ${1\otimes\alpha_1,1\otimes\alpha_2}$ inside an $L$-basis for $L\otimes_KL$, and you win by the same argument. – Keenan Kidwell Nov 20 '12 at 21:22
  • @KeenanKidwell - Thank you Sir! – Eric Nov 21 '12 at 00:18
  • Hmm, just curious, in general, given a skew field $D$, and an extension $L$ of $K$, when is $D\otimes_K L$ a skew field? Can this even occur?

    Thanks a lot. I appreciate the responses!

     – Eric Nov 21 '12 at 15:48
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In the following $D$ is a field. Then $D\otimes_KL\cong D[X]/fD[X]$, where $f$ is the minimal polynomial of $x$ over $K$. I see no reason for $f$ to stay irreducible in $D$. So my answer is NO.