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Let $F$ be a field and let $K$ be an associative unital $F-algebra$. If $A$ and $B$ are subsets of $K$, we let $A • B$ be the set of all elements of $K$ of the form $a • b$, with $a ∈ A$ and $b ∈ B$ (in particular, $∅ • B = A • ∅ = ∅$). We know that the set $V$ of all subsets of $K$ is a vector space over $GF(2)$ with $A+B=A\cup B\setminus A\cap B, 0A=∅,1A=A$. Is $V$ a $GF(2)-algebra$?

I think this is false, but I can think of a contradiction.

This is an exercise from the book: The Linear Algebra a Beginning Graduate Student Should Know by Golan.

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  • $\begingroup$ Are you sure there is no condition missing and you stated the problem correctly? What if $K=F$ for example? $\endgroup$ – quid Mar 7 '15 at 17:50
  • $\begingroup$ yes, the problem is exactly like is statement by golan's book, so maybe is a typo $\endgroup$ – Guadalupe Mar 7 '15 at 17:54
  • $\begingroup$ It would be quite strange. What if $v=1_K$? This can never be true. $\endgroup$ – quid Mar 7 '15 at 18:01
  • $\begingroup$ Yes I thought that so, well then I will change the question to a problem were I need such a vector space. $\endgroup$ – Guadalupe Mar 7 '15 at 18:06
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The assumption is not true.

Let $A=\{0,1\}$ then $K=\mathscr{P}(A)$ is an associate and commutative unital $GF(2)-algebra$ with respect to the operation $∩$.

Now $V=\mathscr{P}(K)$ is again a vector space over $GF(2)$ and $\{∅\},\{\{1\}\},\{\{2\}\}\in V$, but $\{∅\}∙(\{\{1\}\}+\{\{2\}\})=\{∅\}∙\{\{1\},\{2\}\}=\{∅,∅\}=\{∅\}$ while $\{∅\}∙\{\{1\}\}+\{∅\}∙\{\{2\}\}=\{∅\}+\{∅\}=∅$ so ${∅}∙({{1}}+{{2}})\neq{∅}∙{{1}}+{∅}∙{{2}}

Therefore $(V,∙)$ is not a $GF(2)-algebra$.

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