# A linear map without any minimal polynomial

Given an endomorphism $$T$$ on a finite-dimensional vector space $$V$$ over a field $$\mathbb F$$, the minimal polynomial $$\mu_T$$ of $$T$$ is well defined as the generator (unique up to units in $$\mathbb F$$) of the ideal:$I_T= \{p \in \mathbb F[t]\ ; \ p(T)=0\}.$

For infinite-dimensional vector spaces, the minimal polynomial might not be defined. Let’s provide an example.

We take the real polynomials $$V = \mathbb R [t]$$ as a real vector space and consider the derivative map $$D : P \mapsto P^\prime$$. Let’s prove that $$D$$ doesn’t have any minimal polynomial. By contradiction, suppose that $\mu_D(t) = a_0 + a_1 t + \dots + a_n t^n \text{ with } a_n \neq 0$ is the minimal polynomial of $$D$$, which means that for all $$P \in \mathbb R[t]$$ we have $a_0 + a_1 P^\prime + \dots + a_n P^{(n)} = 0.$ Taking for $$P$$ the polynomial $$t^n$$ we get $a_0 t^n + n a_1 t^{n-1} + \dots + n! a_n = 0,$ which doesn’t make sense as $$n! a_n \neq 0$$, hence $$a_0 t^n + n a_1 t^{n-1} + \dots + n! a_n$$ cannot be the zero polynomial.

We conclude that $$D$$ doesn’t have any minimal polynomial.

# Unique factorization domain that is not a Principal ideal domain

In this article, we provide an example of a unique factorization domain – UFD that is not a principal ideal domain – PID. However, it is known that a PID is a UFD.

We take a field $$F$$, for example $$\mathbb Q$$, $$\mathbb R$$, $$\mathbb F_p$$ (where $$p$$ is a prime) or whatever more exotic.

The polynomial ring $$F[X]$$ is a UFD. This follows from the fact that $$F[X]$$ is a Euclidean domain. It is also known that for a UFD $$R$$, $$R[X]$$ is also a UFD. Therefore the polynomial ring $$F[X_1,X_2]$$ in two variables is a UFD as $$F[X_1,X_2] = F[X_1][X_2]$$. However the ideal $$I=(X_1,X_2)$$ is not principal. Let’s prove it by contradiction.

Suppose that $$(X_1,X_2) = (P)$$ with $$P \in F[X_1,X_2]$$. Then there exist two polynomials $$Q_1,Q_2 \in F[X_1,X_2]$$ such that $$X_1=PQ_1$$ and $$X_2=PQ_2$$. As a polynomial in variable $$X_2$$, the polynomial $$X_1$$ is having degree $$0$$. Therefore, the degree of $$P$$ as a polynomial in variable $$X_2$$ is also equal to $$0$$. By symmetry, we get that the degree of $$P$$ as a polynomial in variable $$X_1$$ is equal to $$0$$ too. Which implies that $$P$$ is an element of the field $$F$$ and consequently that $$(X_1,X_2) = F[X_1,X_2]$$.

But the equality $$(X_1,X_2) = F[X_1,X_2]$$ is absurd. Indeed, the degree of a polynomial $$X_1 T_1 + X_2 T_2$$ cannot be equal to $$0$$ for any $$T_1,T_2 \in F[X_1,X_2]$$. And therefore $$1 \notin F[X_1,X_2]$$.

# Two non similar matrices having same minimal and characteristic polynomials

Consider a square matrix $$A$$ of dimension $$n \ge 1$$ over a field $$\mathbb F$$, i.e. $$A \in \mathcal M_n(\mathbb F)$$. Results discuss below are true for any field $$\mathbb F$$, in particular for $$\mathbb F = \mathbb R$$ or $$\mathbb F = \mathbb C$$.

A polynomial $$P \in \mathbb F[X]$$ is called a vanishing polynomial for $$A$$ if $$P(A) = 0$$. If the matrix $$B$$ is similar to $$B$$ (which means that $$B=Q^{-1} A Q$$ for some invertible matrix $$Q$$), and the polynomial $$P$$ vanishes at $$A$$ then $$P$$ also vanishes at $$B$$. This is easy to prove as we have $$P(B)=P(Q^{-1} A Q)=Q^{-1} P(A) Q$$.

In particular, two similar matrices have the same minimal and characteristic polynomials.

Is the converse true? Are two matrices having the same minimal and characteristic polynomials similar? Continue reading Two non similar matrices having same minimal and characteristic polynomials

# An irreducible integral polynomial reducible over all finite prime fields

A classical way to prove that an integral polynomial $$Q \in \mathbb{Z}[X]$$ is irreducible is to prove that $$Q$$ is irreducible over a finite prime field $$\mathbb{F}_p$$ where $$p$$ is a prime.

This raises the question whether an irreducible integral polynomial is irreducible over at least one finite prime field. The answer is negative and:
$P(X)=X^4+1$ is a counterexample. Continue reading An irreducible integral polynomial reducible over all finite prime fields

# On polynomials having more roots than their degree

Let’s consider a polynomial of degree $$q \ge 1$$ over a field $$K$$. It is well known that the sum of the multiplicities of the roots of $$P$$ is less or equal to $$q$$.

The result remains for polynomials over an integral domain. What is happening for polynomials over a commutative ring? Continue reading On polynomials having more roots than their degree