Why is this polynomial separable?

In summary, the conversation discusses the concept of separability in polynomials, where a polynomial is considered separable if all of its roots are different and its irreducible factors are separable. An example is given where the polynomial $f(x)=(x^2-2)^2(x^2+3)$ is shown to be separable because its irreducible factors $x^2-2$ and $x^2+3$ have different roots. The conversation then delves into understanding the concept of reducibility and inseparability through examples and definitions. The proposition that states that all non-constant polynomials in a finite field are separable is discussed and an example of an inseparable polynomial is given. The conversation ends with
  • #1
mathmari
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Hey! :eek:

In my notes there is the following:

Let $F$ be a field. The irresducible $f\in F[x]$ is separable, if all the roots are different.
A non-constant polynomial $f\in K[x]$ is separable, if all the irreducible factors are separable.

Example:
$f(x)=(x^2-2)^2(x^2+3)\in \mathbb{Q}[x]$.
The irreducible factors are:
  • $x^2-2$ $\rightarrow$ Roots : $\pm 2$ different
  • $x^2+3$ $\rightarrow$ Roots : $\pm i\sqrt{3}$ different
So, $f$ is separable. I haven't understood why $f$ is separable. Since we have the factor $(x^2-2)^2$ aren't the roots $\pm 2$ of multiplicity $2$ ? (Wondering)
 
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  • #2
mathmari said:
Hey! :eek:

In my notes there is the following:

Let $F$ be a field. The irresducible $f\in F[x]$ is separable, if all the roots are different.
A non-constant polynomial $f\in K[x]$ is separable, if all the irreducible factors are separable.

Example:
$f(x)=(x^2-2)^2(x^2+3)\in \mathbb{Q}[x]$.
The irreducible factors are:
  • $x^2-2$ $\rightarrow$ Roots : $\pm 2$ different
  • $x^2+3$ $\rightarrow$ Roots : $\pm i\sqrt{3}$ different
So, $f$ is separable. I haven't understood why $f$ is separable. Since we have the factor $(x^2-2)^2$ aren't the roots $\pm 2$ of multiplicity $2$ ? (Wondering)

Hey mathmari!

Aren't the roots $\pm \sqrt 2$? (Wondering)

And isn't $f(x)$ reducible? That would mean that the first statement doesn't apply.
I think we should apply the second statement first. (Thinking)
 
  • #3
I like Serena said:
Aren't the roots $\pm \sqrt 2$? (Wondering)

Ah yes... (Blush)
I like Serena said:
And isn't $f(x)$ reducible? That would mean that the first statement doesn't apply.
I think we should apply the second statement first. (Thinking)

Yes, we apply the second statement first. The irreducible factors $x^2-2$ and $x^2+3$, right? (Wondering)
 
  • #4
So, an inseparable polynomial is for example $f(x)=(x^2-4)^2(x^3-8)$, where the irreducible factors are $x^2-4$ and $x^3-8$, and the roots are $\pm 2$ and $2$ ? (Wondering)
 
  • #5
mathmari said:
Yes, we apply the second statement first. The irreducible factors $x^2-2$ and $x^2+3$, right?
Yes. That's what your example states as well.

mathmari said:
So, an inseparable polynomial is for example $f(x)=(x^2-4)^2(x^3-8)$, where the irreducible factors are $x^2-4$ and $x^3-8$, and the roots are $\pm 2$ and $2$?

Can't we reduce $x^3-8$ to $(x-2)(x^2+2x+4)$? (Wondering)
 
  • #6
I like Serena said:
Can't we reduce $x^3-8$ to $(x-2)(x^2+2x+4)$? (Wondering)

Ah, so it is again separable, right? (Wondering)

Could you give me an example of an inseparable polynomial? (Wondering)
 
  • #7
mathmari said:
Could you give me an example of an inseparable polynomial? (Wondering)

How about $x^3-2$ in $F_3[x]$? (Wondering)
 
  • #8
I like Serena said:
How about $x^3-2$ in $F_3[x]$? (Wondering)

I saw now in my notes the following proposition:
If $F$ is finite, then each non-constant $f\in F[x]$ is separable.

We have that $F_3$ is finite, right? Therefore, $x^3-2$ is separable, isn't it? (Wondering)
 
  • #9
mathmari said:
I saw now in my notes the following proposition:
If $F$ is finite, then each non-constant $f\in F[x]$ is separable.

We have that $F_3$ is finite, right? Therefore, $x^3-2$ is separable, isn't it? (Wondering)

Well... $x^3-2$ is irreducible, but we also have $x^3-2=(x+1)^3$ in $F_3[x]$, so $2$ is a triple root.
If that doesn't mean it's inseparable, I wouldn't know what is.

Do your notes give an example of an inseparable polynomial?
Or do you have a proof for that proposition? (Wondering)
 
  • #10
I like Serena said:
Well... $x^3-2$ is irreducible, but we also have $x^3-2=(x+1)^3$ in $F_3[x]$, so $2$ is a triple root.
If that doesn't mean it's inseparable, I wouldn't know what is.

Do your notes give an example of an inseparable polynomial?
Or do you have a proof for that proposition? (Wondering)

In this book the chapter 3.4 (p.11) is about separability. This proposition is the 3.4.5 and the only example of an inseparable polynomial that I found is the one that is at 3.4.8.

(Thinking)
 
  • #11
mathmari said:
In this book the chapter 3.4 (p.11) is about separability. This proposition is the 3.4.5 and the only example of an inseparable polynomial that I found is the one that is at 3.4.8.

(Thinking)

This book gives example 1.2 with $x^3-2$ in $F_3[x]$ as an inseparable polynomial.
But its definition of inseparable seems to be different from your definition.
Btw, wiki seems to give a different definition as well. (Thinking)

I can see that your book considers $x^3-2$ reducible, since it is reducible over $F_3$. It seems that my book reference considers $x^3-2$ irreducible, but to be honest I'm not sure why. Wiki's definition of irreducible seems to imply that it's reducible.
 

FAQ: Why is this polynomial separable?

Why is it important for a polynomial to be separable?

It is important for a polynomial to be separable because it allows us to easily find the roots or solutions of the polynomial. This makes it easier to solve equations and understand the behavior of the polynomial.

How do you know if a polynomial is separable?

A polynomial is separable if it can be factored into linear or irreducible quadratic factors. This means that each term in the polynomial can be written as a product of factors that cannot be further simplified.

Can a polynomial be separable and not factorable?

No, a polynomial must be factorable in order to be separable. If a polynomial cannot be factored into linear or irreducible quadratic factors, then it is not separable.

What is the difference between a separable and non-separable polynomial?

A separable polynomial can be factored into linear or irreducible quadratic factors, while a non-separable polynomial cannot. This means that a separable polynomial can be easily solved and understood, while a non-separable polynomial may be more difficult to work with.

How does the degree of a polynomial affect its separability?

The degree of a polynomial does not affect its separability. A polynomial of any degree can be separable as long as it can be factored into linear or irreducible quadratic factors. However, higher degree polynomials may be more difficult to factor and analyze, making their separability more challenging to determine.

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