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Let $A$ be a unital commutative $*$-algebra (involutive commutative unital ring + vector space) over the field of the complex numbers. By a projection $p$ in $A$ we mean $p=p^*=p^2$.

As for two projections $p$ and $q$ in $A$ we say $p\leq q$ if $pq=p$. A projection $e$ is called minimal if $p\leq e$ implies that $p=0$ or $p=e$.

True or false: dim$A=n$ if and only if there exist minimal projections $e_1,\cdots,e_n$ with $1_A=e_1+\cdots e_n$ ?!

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I'm not very familiar with $\ast$-algebras, so you might tell me that my example somehow fails to be such an algebra.

Consider $\left\{\begin{bmatrix}a&b\\0&a\end{bmatrix}\middle|\,a,b\in \mathbb C\right\}$ with involution $\begin{bmatrix}a&b\\0&a\end{bmatrix}\mapsto\begin{bmatrix}\bar a&\bar b\\0&\bar a\end{bmatrix} $.

This is a $2$-dimensional commutative local $\mathbb C$ algebra, so it only has one nonzero projection: the identity. This is apparently the only minimal projection.

So the dimension of $A$ and the number of terms in a decomposition of $1$ into minimal projections are not equal for this ring.

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    Thanks for your pay attention. This example does not work. As for the involution, we have that $(ab)^*=b^*a^*$. Therefore the involution that you defined is not correct. – 2017-01-26
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    @AliBagheri I don't understand. It seems to me that $(ab)^\ast=a^\ast b^\ast=b^\ast a^\ast$ in my example. – 2017-01-26
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    Sorry you are completely right. It is a nice example. – 2017-01-26
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    @AliBagheri Great... thank you! – 2017-01-26