Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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 0:15
@NoahSnyder - I thought that you can't just assume $D = L$, right? =) –  Eric Nov 21 '12 at 0:16
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 1:40

2 Answers 2

up vote 4 down vote accepted

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.

share|improve this answer
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 $1\otimes T-T\otimes 1\neq 0$. –  user26857 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 0:18
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

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.