# The irreducible polynomial is not separable

• MHB
• mathmari
In summary, we have discussed the properties of $t\in D$ as a prime element and the irreducibility of $x^n-t$ in $K[x]$. We have also shown that the polynomial $x^p-t$ is not separable and has only one root in its algebraic closure $C$.
mathmari
Gold Member
MHB
Hey!

Let $F$ be a field, $D=F[t]$, the polynomial ring of $t$, with coefficients from $F$ and $K=F(t)$ the field of rational functions of $t$.
(a) Show that $t\in D$ is a prime element of $D$.
(b) Show that the polynomial $x^n-t\in K[x]$ is irreducible.
(c) Let $\text{char} F=p$. Show that the polynomial $x^p-t$, even if it is irreducible in $K[t]$, it is not separable, and if $C$ is an algebraic closure of $K$, this polynomial has only one root in $C$. I have done the following:

(a) We assume that $t\in D$ is not prime. Then there are non-constant polynomials $f(t),g(t)\in D$ such that $t=f(t)g(t)$. Then $\deg (t)=\deg (f(t)\cdot g(t)$. Since $D$ is an integral domain, we have that $\deg (t)=\deg (f)+\deg (g)\Rightarrow 1\geq 1+1$, a contradiction.
Is this correct? (Wondering)

(b) From the generalized criterion of Eistenstein we have that the prime $t$ divides the term $t$ but not the coefficient of the highest degree term, so it $x^n-t$ is irreducible in $K[x]$.
Is this correct? (Wondering)

(c) Could you give me a hint how we could show that? Why does this hold? (Wondering)

(c) Let $\text{char} F=p$. Show that the polynomial $x^p-t$, even if it is irreducible in $K[t]$, it is not separable, and if $C$ is an algebraic closure of $K$, this polynomial has only one root in $C$.

Let $c$ be the root of $x^p-t$ in $C$, i.e., $t=c^p$. We have that $x^p-t=x^p-c^p=(x-c)^p$, since $\text{char} F=p$.
Therefore, we have that $c$ is the only root in $C$ of multiplicity $p$, and so $x^p-t$ is not separable.

Is this correct? (Wondering)

## 1. What does it mean for a polynomial to be irreducible?

An irreducible polynomial is one that cannot be factored into smaller polynomials with coefficients from the same field. In other words, it cannot be broken down into simpler components.

## 2. How can you determine if a polynomial is separable or not?

A polynomial is separable if it can be factored into linear factors over its splitting field. This means that all of its roots are distinct and can be written as linear expressions with coefficients from the field. If a polynomial cannot be factored in this way, it is not separable.

## 3. What are the implications of a polynomial being irreducible and not separable?

A polynomial that is both irreducible and not separable has no roots in its splitting field. This means that it cannot be solved algebraically, and therefore has no solutions in terms of familiar numbers like integers or fractions. It may still have solutions in terms of more abstract objects, such as complex numbers or algebraic extensions of the field.

## 4. Can an irreducible polynomial be separable?

Yes, it is possible for a polynomial to be both irreducible and separable. For example, the polynomial x^2 + 1 is irreducible over the real numbers, but it can be factored into (x + i)(x - i) over the complex numbers, making it separable.

## 5. Why is it important to know if a polynomial is irreducible and separable?

Knowing if a polynomial is irreducible and separable can help determine the solvability of equations involving that polynomial. It also has implications in fields such as algebraic geometry and coding theory. Additionally, understanding the properties of irreducible and separable polynomials can help in constructing and manipulating them for various applications.

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