Let \(B\) be an \(R\)-module. Show that if \(r+r\neq 0\) for all \(r\neq 0 \in R\), then an \(n\)-linear form \(B^n\to R\) is alternating \(\iff\) it is skew-symmetric.
Solution
Let \(f:B^n\to R\) be \(n\)-linear. Here “skew-symmetric” means that interchanging two arguments multiplies the value by \(-1\), and “alternating” means that \(f\) vanishes whenever two arguments are equal.
1. Every alternating \(n\)-linear form is skew-symmetric.
Fix all arguments except positions \(i<j\), and write \(x,y\in B\) in those positions. Alternation gives \[ 0=f(\ldots,x+y,\ldots,x+y,\ldots). \] Expanding by multilinearity, the two terms with equal arguments vanish, leaving \[ f(\ldots,x,\ldots,y,\ldots) +f(\ldots,y,\ldots,x,\ldots)=0. \] Thus interchanging the two arguments changes the sign.
2. Under the hypothesis \(r+r\ne0\) for every \(0\ne r\in R\), everyskew-symmetric \(n\)-linear form is alternating.
Suppose two arguments, say positions \(i<j\), are equal to \(x\). Skew-symmetry under their transposition gives \[ f(\ldots,x,\ldots,x,\ldots) =-f(\ldots,x,\ldots,x,\ldots). \] Hence, with \(r=f(\ldots,x,\ldots,x,\ldots)\), \[ r+r=0. \] By the stated hypothesis this forces \(r=0\). Therefore \(f\) vanishes whenever two arguments coincide, so \(f\) is alternating.
3. Therefore the two notions are equivalent under the stated hypothesis.
Combine 1 and 2.