# Tensor product of $K$-skew fields is nilpotent iff every $D_{i}$ is nilpotent?

an element, x, of a ring, R, is called nilpotent if there exists some positive integer, n, such that $x^n = 0$.

in book Skew Fields by P.K. DRAXL : Definition 3 . A simple ring $A$ with finite $|A:Z(A)|$ is called a "central simple $K-algebra$" ($K := Z(A)$ ). If a central simple $K$-algebra is a skew field then we call it a "$K$-skew field".

now is following statement true ?

let every $D_{i}$ is $K$-skew field then $\otimes _{i=1}^{n} D_{i}$ is nilpotent iff every $D_{i}$ is nilpotent .($\otimes$ : is tensor product over $K$. we say that $D_{i}$ is nilpotent if $D_{i}$ has maximal subfield $F$ and $F/Z(D_i)$ Galois extension and Galois group $Gal(F/Z(D_i))$ be nilpotent group)

I think it is not true but i can't find counterexample.

• What does it mean for $D_i$ to be nilpotent? – Eric Wofsey May 4 '18 at 20:29
• @EricWofsey. I edit my question. – amir bahadory May 11 '18 at 16:14