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The vectors of a root-system were originally called "roots" because they are the zeros of a characteristic polynomial that comes from the connection of (crystallographic) root-systems to classifying semisimple Lie algbras. From an answer by José Carlos Santos,

It comes from the roots of the characteristic polynomial of an endomorphism. If $\mathfrak g$ is a complex semisimple Lie algbra, $\mathfrak h$ is a Cartan subalgebra and $\alpha\in\mathfrak{h}^*$, then $\alpha$ is a root if, for every $H\in\mathfrak h$, $\alpha(H)$ is an eigenvalue of the endomorphism of $\mathfrak g$ defined by $X\mapsto[H,X]$.

Is there a way to get at this same polynomial just from the definitions of a (crystallographic) root-system without having to talk about the connection to Lie theory?

To provide some specific motivation, I'm reading through a proof of Gabriel's theorem (classifying quivers of finite representation type) where you define the Tits form $q$ for the ADE Dynkin and Euclidean diagrams, and the set of roots associated to that diagram are the nonzero vectors $x \in \mathbb{Z}^n$ for which $q(x) \leq 1$. I was looking for a way to justify calling these "roots" without getting into Lie theory, and I thought there might be a justification for the word "root" in terms of general abstract root systems.

The vectors of a root-system were originally called "roots" because they are the zeros of a characteristic polynomial that comes from the connection of (crystallographic) root-systems to classifying semisimple Lie algbras. From an answer by José Carlos Santos,

It comes from the roots of the characteristic polynomial of an endomorphism. If $\mathfrak g$ is a complex semisimple Lie algbra, $\mathfrak h$ is a Cartan subalgebra and $\alpha\in\mathfrak{h}^*$, then $\alpha$ is a root if, for every $H\in\mathfrak h$, $\alpha(H)$ is an eigenvalue of the endomorphism of $\mathfrak g$ defined by $X\mapsto[H,X]$.

Is there a way to get at this same polynomial just from the definitions of a (crystallographic) root-system without having to talk about the connection to Lie theory?

To provide some specific motivation, I'm reading through a proof of Gabriel's theorem (classifying quivers of finite representation type) where you define the Tits form $q$ for the ADE Dynkin and Euclidean diagrams, and the set of roots associated to that diagram are the nonzero vectors $x \in \mathbb{Z}^n$ for which $q(x) \leq 1$. Since $q(x) \in \mathbb{Z}$ we could say these are the integer roots of the polynomial $q(x)(q(x)-1)$, but I doubt that this generalizes beyond the root systems you get from the ADE Dynkin diagrams. I was looking for a way to justify calling these "roots" without getting into Lie theory, and I thought there might be a justification for the word "root" in terms of general abstract root systems.

The vectors of a root-system were originally called "roots" because they are the zeros of a characteristic polynomial that comes from the connection of (crystallographic) root-systems to classifying semisimple Lie algbras. From an answer by José Carlos Santos,

It comes from the roots of the characteristic polynomial of an endomorphism. If $\mathfrak g$ is a complex semisimple Lie algbra, $\mathfrak h$ is a Cartan subalgebra and $\alpha\in\mathfrak{h}^*$, then $\alpha$ is a root if, for every $H\in\mathfrak h$, $\alpha(H)$ is an eigenvalue of the endomorphism of $\mathfrak g$ defined by $X\mapsto[H,X]$.

Is there a way to get at this same polynomial just from the definitions of a (crystallographic) root-system without having to talk about the connection to Lie theory?

The vectors of a root-system were originally called "roots" because they are the zeros of a characteristic polynomial that comes from the connection of (crystallographic) root-systems to classifying semisimple Lie algbras. From an answer by José Carlos Santos,

It comes from the roots of the characteristic polynomial of an endomorphism. If $\mathfrak g$ is a complex semisimple Lie algbra, $\mathfrak h$ is a Cartan subalgebra and $\alpha\in\mathfrak{h}^*$, then $\alpha$ is a root if, for every $H\in\mathfrak h$, $\alpha(H)$ is an eigenvalue of the endomorphism of $\mathfrak g$ defined by $X\mapsto[H,X]$.

Is there a way to get at this same polynomial just from the definitions of a (crystallographic) root-system without having to talk about the connection to Lie theory?

To provide some specific motivation, I'm reading through a proof of Gabriel's theorem (classifying quivers of finite representation type) where you define the Tits form $q$ for the ADE Dynkin and Euclidean diagrams, and the set of roots associated to that diagram are the nonzero vectors $x \in \mathbb{Z}^n$ for which $q(x) \leq 1$. I was looking for a way to justify calling these "roots" without getting into Lie theory, and I thought there might be a justification for the word "root" in terms of general abstract root systems.

The vectors of a root-system were originally called "roots" because they are the zeros of a characteristic polynomial that comes from the connection of root-systems to classifying semisimple Lie algbras. From an answer by José Carlos Santos,

It comes from the roots of the characteristic polynomial of an endomorphism. If $\mathfrak g$ is a complex semisimple Lie algbra, $\mathfrak h$ is a Cartan subalgebra and $\alpha\in\mathfrak{h}^*$, then $\alpha$ is a root if, for every $H\in\mathfrak h$, $\alpha(H)$ is an eigenvalue of the endomorphism of $\mathfrak g$ defined by $X\mapsto[H,X]$.

Is there a way to get at this same polynomial just from the definitions of a root-system without having to talk about the connection to Lie theory?

The vectors of a root-system were originally called "roots" because they are the zeros of a characteristic polynomial that comes from the connection of (crystallographic) root-systems to classifying semisimple Lie algbras. From an answer by José Carlos Santos,

It comes from the roots of the characteristic polynomial of an endomorphism. If $\mathfrak g$ is a complex semisimple Lie algbra, $\mathfrak h$ is a Cartan subalgebra and $\alpha\in\mathfrak{h}^*$, then $\alpha$ is a root if, for every $H\in\mathfrak h$, $\alpha(H)$ is an eigenvalue of the endomorphism of $\mathfrak g$ defined by $X\mapsto[H,X]$.

Is there a way to get at this same polynomial just from the definitions of a (crystallographic) root-system without having to talk about the connection to Lie theory?