Let \(A\) be a finite-dimensional commutative algebra with identity over \(\mathbb C\), with \[ \dim_{\mathbb C}A\ge2. \] Suppose \[ a^2\ne0 \] for every nonzero \(a\in A\). Show that the equation \[ a^2=a \] has at least three distinct solutions in \(A\).
Solution
1. For every \(x\in A\), the minimal polynomial of \(x\) over \(\CC\) hasno repeated root.
Let \(m_x(t)\) be the minimal polynomial of \(x\). Since \(\CC\) is algebraically closed, \[ m_x(t)=\prod_{i=1}^r(t-\lambda_i)^{e_i} \] for distinct \(\lambda_i\in\CC\) and positive integers \(e_i\).
Suppose some \(e_j\geq2\), and put \[ b(t)\coloneqq\frac{m_x(t)}{t-\lambda_j}. \] Then \(\deg b<\deg m_x\), so minimality of \(m_x\) gives \[ b(x)\neq0. \] On the other hand, \(m_x\) divides \(b^2\): the factor \((t-\lambda_j)^{e_j}\) occurs in \(b^2\) with exponent \(2e_j-2\geq e_j\), and every other factor of \(m_x\) occurs in \(b^2\) with at least its required exponent. Hence \[ b(x)^2=0, \] contrary to the hypothesis. Therefore every \(e_i=1\).
2. There is an element \(x\in A\) whose minimal polynomial has at leasttwo distinct roots.
Since \(\dim_{\CC}A\geq2\), choose \[ x\notin\CC\cdot1. \] If the minimal polynomial of \(x\) had degree \(1\), then \(x=\lambda1\) for some \(\lambda\in\CC\), contrary to the choice of \(x\). Thus \(\deg m_x\geq2\). By step 1, all roots of \(m_x\) are distinct, so there are at least two of them.
3. There is an idempotent \(e\in A\) with\[ e\neq0, \qquad e\neq1. \]
Write \[ m_x(t)=\prod_{i=1}^r(t-\lambda_i), \qquad r\geq2. \] By Lagrange interpolation there is a polynomial \(q(t)\) of degree at most \(r-1\) such that \[ q(\lambda_1)=1, \qquad q(\lambda_i)=0 \quad(2\leq i\leq r). \] The polynomial \(q^2-q\) vanishes at every root of the squarefree polynomial \(m_x\), so \[ m_x\mid(q^2-q). \] Consequently \[ e\coloneqq q(x) \] satisfies \(e^2=e\).
Moreover, \(q\) is nonzero and has degree less than \(\deg m_x\), so \(q(x)\neq0\) by minimality of \(m_x\). Likewise \(q-1\) is nonzero because \(q(\lambda_2)=0\), and \[ \deg(q-1)<\deg m_x, \] so \((q-1)(x)\neq0\). Hence \(e\neq1\).
4. The equation \(a^2=a\) has at least three distinct solutions in \(A\).
The elements \[ \boxed{0,\qquad1,\qquad e} \] are idempotent, and step 3 shows that they are pairwise distinct.
5. Q.E.D.
Step 4 is the required conclusion.