V is a variety of commutative semi group satisfying the identity $x^2 = x^3$. I need to prove that: $|F_V(\{x_1\dots,x_n\})|$ = $3^n -1$. Any hints on this ?

$F_V$ is V-free algebra.


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  • $\begingroup$ Judging from the result that you expect, I'm suspecting that you are after the variety of a commutative semigroup ring? Maybe one that satisfies $x_i^2 = x_i^3$ for each indeterminate $x_i$? Please rework this questions, at the moment it makes no sense. $\endgroup$ – Thomas Kahle Apr 15 '14 at 9:38
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    $\begingroup$ @Thomas I suspect you are confusing identities and relations. $\endgroup$ – J.-E. Pin Apr 15 '14 at 10:14
  • $\begingroup$ @J.-E.Pin You are right, I have no idea what the difference between an identity and a relation may be in the context here. $\endgroup$ – Thomas Kahle Apr 15 '14 at 13:50
  • $\begingroup$ Regarding the question on hold. $\endgroup$ – Anurag Sharma Apr 15 '14 at 15:25
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    $\begingroup$ It is phrased in terms of advanced mathematics, but its solution is accessible to students in high school. Have you tried determining the structure of the algebra even for small values of n? $\endgroup$ – The Masked Avenger Apr 15 '14 at 16:02

Let $S_n = F_V(\{x_1, ..., x_n\})$. Since $S_n$ is commutative, its elements can be written in the form $x_1^{r_1} \cdots x_n^{r_n}$ where $r_1 + \dots + r_n > 0$. Morevover, since the semigroup $S_n$ satisfies the identity $x^2 = x^3$, you may assume that each $r_i$ is equal to $0$, $1$ or $2$. It follows that $|S_n| \leqslant 3^n -1$. To prove that this inequality is in fact an equality, it suffices to verify that the set of size $3^n -1$ $$ \{x_1^{r_1} \cdots x_n^{r_n} \mid 0 \leqslant r_i \leqslant 2 \text{ and } r_1 + \dots + r_n > 0\} $$ equipped with the product defined by $$ (x_1^{r_1} \cdots x_n^{r_n})(x_1^{s_1} \cdots x_n^{s_n}) = x_1^{\min\{2, r_1 + s_1\}} \cdots x_n^{\min\{2, r_n + s_n\}} $$ is a semigroup of $V$.


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