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What does the notation $(F_p^\times)^2$ mean, where $F=\mathbb{Z}/p\mathbb{Z}$?

I was taking this (in two previous questions) to mean the cross product of $F=\mathbb{Z}/p\mathbb{Z}$, but I am beginning to realize that this might be what I am confusing here.


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You write superscript $x$. Perhaps it was a superscript $\times$ instead? – msh210 Dec 14 '11 at 0:10
first take the multiplicative group of units. then square them all. e.g. $p=5, (\mathbb{Z}/5\mathbb{Z})^{\times}=\{1,2,3,4\}, ((\mathbb{Z}/5\mathbb{Z})^{\times})^2=\{1,4\}$ – yoyo Dec 14 '11 at 0:12
Thanks, that helps a lot! – Jason Smith Dec 14 '11 at 0:21
up vote 3 down vote accepted

If $A$ is a ring, then $A^*$ or $A^\times$ denotes the set of units of $A$. Under the multiplication of the ring, this is a group, e.g. $\mathbf Z^* = \{\pm 1\}$. In a field $k$ all non-zero elements have inverses, so $k^* = k - \{0\}$.

Then $k^{*n}$ is the image of the map $k^* \to k^*$ given by $x \mapsto x^n$. It's somewhat confusing, since you could also imagine this standing for an $n$-fold direct product of copies of the group $k^*$, or for the set of all $x_1 \cdots x_n$ with $x_i \in k^*$.

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