How is algebraic multiplicity defined for a vector space over $GF(2)$?

Question

I'm reading a book called Linear Algebra by Friedberg, Insel, and Spence. In their chapter on diagonalization, they provide the following definition:

Let $\lambda$ be an eigenvalue of a linear operator or matrix with the characteristic polynomial $f(t)$. The (algebraic) multiplicity of $\lambda$ is the largest positive integer $k$ for which $(t - \lambda)^k$ is a factor of $f(t)$.

This definition makes perfect sense to me when considering a vector space with a scalar field like $\mathbb{R}$ or $\mathbb{C}$. But I find myself running into trouble when considering vector spaces over the field $GF(2)$, a.k.a. $\mathbb{Z}_2$ or $\mathbb{Z}/2\mathbb{Z}$.

Since $0^2 = 0$ and $1^2 = 1$, it seems to me that $GF(2)$ has the property that $x^k = x$ for all $x \in GF(2)$, $k = 1, 2, 3...$. This seems to imply that if $f(t)$ is the characteristic polynomial of a linear operator on a vector space over $GF(2)$, and $(t - \lambda)^k$ is a factor of $f(t)$, then $(t - \lambda)^{k+1}$ may be considered a factor of $f(t)$ as well. But this would imply that there is no largest positive integer $k$ for which $(t - \lambda)^k$ is a factor of $f(t)$, which would in turn make it impossible to define the algebraic multiplicity of $\lambda$.

This seems unlikely to me, because Friedberg et al. define algebraic multiplicity without specifying a field. What am I missing?

Answer 1

In a polynomial ring, the variable $t$ is a formal symbol, not a function.

Thus, for example, in the polynomial ring with coefficients in the field with two elements, the polynomials $t^2$ and $t$ are different (they are the same function, but different polynomials).

When we say a polynomial $f(t)$ divides another polynomial $h(t)$, we mean there exists a polynomial $g(t)$ such that $f(t)g(t) = h(t)$ as polynomials, not as functions.

Answer 2

In addition to @hunter's good answer, it may be worth noting the option of thinking of the minimal polynomial $M(t)$ as factoring over some sufficiently large algebraic extension $GF(2^n)$ of $GF(2)$. While if there's just one linear factor $(t-\lambda)$ that is repeated in $M(t)$, then, yes, using the Euclidean algorithm to compute the gcd of $M$ and $M'$ shows that $\lambda\in GF(2)$ (rather than in a proper extension).

However, for example, for $\lambda$ and $\mu$ in $GF(2^2)$ such that $t^2+t+1=(t-\lambda)(t-\mu)$, and $(t^2+t+1)^k$ divides $M(t)$, that Euclidean-algorithm-gcd argument of course does not prove that $\lambda,\mu\in GF(2)$. So repeated roots of $M(t)$ can lie in proper extensions, if their conjugates also are roots, etc.