# Skew-symmetric implies alternating for $2$ a zero divisor in $R$?

Let $$k\geq 2$$. If $$2\in R^\times$$, then a multilinear function $$f:M^k \to N$$ which is skew-symmetric is alternating.

After the theorem he says

Strictly speaking, the assumption that $$2\in R^\times$$ could be weakened to $$2$$ not being a zero divisor in $$R$$.

Why is this condition not "$$2\notin\text{ann}_R(x)$$ for any $$x\in N$$", i.e., $$2$$ is not a zerodivisor of $$N$$? I thought the key line is $$2f(m,m,m_3,\ldots,m_k)=0\implies f(m,m,m_3,\ldots,m_k)=0$$ (which is true if $$2\in R^\times$$).

However, if $$R=\mathbb{Z}$$ and $$N=\mathbb{Z}/4\mathbb{Z}$$, then if $$f(m,m,m_3,\ldots,m_k)=2$$, $$2f(m,m,m_3,\ldots,m_k)=0\;\not\!\!\!\implies f(m,m,m_3,\ldots,m_k)=0$$, even though $$2$$ is a nonzerodivisor of $$R$$ (but here $$2$$ is a zerodivisor of $$N$$).

But Wikipedia backs up his point, and it's Keith Conrad, so I'm probably missing something obvious or confusing definitions...

• Sorry, I have had tensors in mind lately and that influenced my answer. I'll think about it more. Maybe there is an inaccuracy or missing assumption after all. – rschwieb Feb 22 at 17:55
• Can you point to the Wikipedia article? I am inclined to agree that the Conrad's weakening goes too far (i.e. neglects the possibility of torsion elements in $N$). I'm also puzzled why the definition of 'alternating' is restricted to $k\ge 2$, since it is natural (and useful) to extend it to all $k$. – qman Mar 19 at 22:59