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On a real or complex inner product space of finite dimension, a normal operator has a square-free polynomial as its minimal polynomial. To prove this without the spectral theorem, suppose an irreducible polynomial $f$ occurs at least twice in $m_T$. Put $h=m_T/f^2$ and $N=f(T)$. Since polynomials of a normal operator are normal, $N$ is a normal operator. For every $v$,By kernel stabilization for a normal operator, $Nh(T)v=0$. Hence the smaller-degree monic polynomial $m_T/f$ annihilates $T$, contradicting minimality.

Over the complex numbers, all irreducible polynomials are linear, so the primary decomposition theorem for linear maps now gives a basis of eigenvectors. Over the real numbers, irreducible quadratics can remain: a nontrivial planar rotation need not have any real eigenvectors, although its minimal polynomial is square-free.

Put $\gamma =\alpha +\beta$, $E=F(\gamma )$ and $E_0=F({\gamma }^p)$. By the preceding separability criterion using a pth power, it suffices to prove $E=E_0$.

Since $\alpha$ is separable, $\alpha \in F({\alpha }^p)$. But ${\alpha }^p={\gamma }^p-{\beta }^p\in E_0(\beta )$, so $\alpha \in E_0(\beta )$. ConsequentlyLet $h$ and $m$ be the monic minimal polynomials of $\beta$ over $E_0$ and $E$, respectively. Both divide its separable minimal polynomial over $F$, so both are separable. Also $m\mid h$ in $E[T]$.

Every $c\in E=F(\gamma )$ has $c^p\in F({\gamma }^p)=E_0$. Hence $m(T{)}^p$ has coefficients in $E_0$ and vanishes at $\beta$, forcing $h\mid m^p$ in $E_0[T]$. Since $h$ remains a square-free polynomial over $E$, this divisibility implies $h\mid m$ in $E[T]$. Together with $m\mid h$, monicity gives $h=m$. ThereforeThe tower law makes $[E:E_0]=1$. Part (c) now proves that $\gamma =\alpha +\beta$ is separable over $F$. This establishes closure of separable algebraic elements under addition without presupposing that closure as a field theorem.

Suppose some exponent $q_j\ge 2$. DefinePart (iii) makes $N$ a normal operator. For every $v\in V$, put $w=h(T)v$. The minimal polynomial identity givesPart (ii), applied to $N$, forces $Nw=0$. As this holds for every $v$,This contradicts minimality: $m_T/f_j$ is a monic polynomial of smaller degree annihilating $T$. Therefore the minimal polynomial of a normal operator is a square-free polynomial over either of the given fields:This argument does not require the irreducible polynomials to be linear, so it also covers the real case.

Using the assumed square-free polynomial property, factorinto distinct monic polynomials that are irreducible polynomials, with $n_j=\mathrm{deg}{f}_j$. Part (b)(iii) identifies $T={\mathbb{F}}_2[X]/(X^p-1)$ with ${\prod }_j{K}_j$, where $|K_j|=2^{n_j}$.

By the first isomorphism theorem for rings, the image $S$ from part (c)(i) is $T/J$, where $J$ is the kernel of a ring homomorphism. By part (a)(ii), $J={\prod }_j{J}_j$ for ideals $J_j\subseteq K_j$. A field has only the zero ideal and the whole field as ideals, so each $K_j/J_j$ is either $K_j$ or the zero ring. Consequently, for a subset $D\subseteq \{1,\dots ,r\}$,This is the quotient of a finite product of fields principle. The subset is nonempty because $|U(S)|=p>2$, whereas the zero ring has only one unit. By part (a)(i),Each factor is a positive integer. Since $p$ is a prime number, exactly one factor is $p$ and all remaining factors are one. Thusso $p$ is a Mersenne prime. If $n=ab$ with $a,b>1$, then $2^a-1$ would be a proper divisor of $2^{ab}-1$; thus $n$ is itself a prime number, although only the displayed form is needed.

The assumed square-free polynomial property can also be checked directly: in characteristic of a ring two, the formal derivative of $X^p-1$ is $X^{p-1}$ because $p$ is odd. These two polynomials have greatest common divisor one, since $X$ does not divide $X^p-1$, so no repeated irreducible polynomial factor occurs.