Repeated root in field of char 0

[fishturtle1]

Homework Statement

If F is a field of char 0 and $p(x) \in F[x]$ is irreducible, then $p(x)$ has no repeated roots. Hint: Consider $\gcd(p(x), p'(x))$

Relevant Equations

Definition: Let $F$ be a field. A nonzero polynomial $p(x) \in F[x]$ is irreducible over $F$ if $\deg p(x) \ge 1$ and there is no factorization $p(x) = f(x)g(x)$ in $F[x]$ with $\deg f(x) < \deg p(x)$ and $\deg g(x) < \deg p(x)$.

Definition: Let $R$ be a domain, and let $f(x), g(x) \in R[x]$. The greatest common divisor of $f(x)$ and $g(x)$ is a polynomial $d(x) \in R[x]$ such that
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i) $d(x)$ is a common divisor of $f(x)$ and $g(x)$;
ii) if $c(x)$ is any common divisor of $f(x)$ and $g(x)$ then $c(x) \vert d(x)$;
iii) $d(x)$ is monic.

Definition: The prime field of a field $F$ is the intersection of all the subfields of $F$.

Definition: A field has characteristic $0$ if its prime field is isomorphic to $\mathbb{Q}$; it has characteristic $p$ if its prime field is isomorphic to $\mathbb{Z}_p$.

Definition: If $p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$ then $p'(x) = na_nx^{n-1} + (n-1)a_{n-1}a^{n-2} + \dots + a_1$.

Proof: We will first show $\gcd(p(x), p'(x)) = 1$. Define $d(x) = \gcd(p(x), p'(x))$. Then we can find $q(x) \in F[x]$ such that $p(x) = d(x)q(x)$. But $p(x)$ is irreducible which means $d(x)$ is constant or $q(x)$ is constant. If $q(x)$ is constant, then $\deg d(x) = \deg p(x)$. In particular, $\deg d(x) = \deg p(x) > \deg p'(x)$. But we have also assumed $d(x) \vert p'(x)$. This implies $\deg d(x) \le \deg p'(x)$. We have reached a contradiction. So, we can conclude $d(x)$ is constant. Since $d(x)$ is also monic, we have $d(x) = 1$.

Next, we suppose $p(x)$ has a repeated root, say $a \in F$. Then we can find $k(x) \in F[x]$ such that $p(x) = (x-a)^2k(x)$. Taking the derivative, we have $p'(x) = 2(x-a)k(x) + (x-a)^2k'(x)$. Hence, $(x-a) \vert \gcd(p(x), p'(x))$. This contradicts the fact $\gcd(p(x), p'(x)) = 1$.

We can conclude $p(x)$ has no repeated roots. []My question is, I don't think I used the assumption that $F$ has characteristic $0$. Is there a mistake in the proof?

 

[fresh_42]

Homework Statement:: If F is a field of char 0 and $p(x) \in F[x]$ is irreducible, then $p(x)$ has no repeated roots. Hint: Consider $\gcd(p(x), p'(x))$
Relevant Equations:: Definition: Let $F$ be a field. A nonzero polynomial $p(x) \in F[x]$ is irreducible over $F$ if $\deg p(x) \ge 1$ and there is no factorization $p(x) = f(x)g(x)$ in $F[x]$ with $\deg f(x) < \deg p(x)$ and $\deg g(x) < \deg p(x)$.

Definition: Let $R$ be a domain, and let $f(x), g(x) \in R[x]$. The greatest common divisor of $f(x)$ and $g(x)$ is a polynomial $d(x) \in R[x]$ such that
\\

i) $d(x)$ is a common divisor of $f(x)$ and $g(x)$;
ii) if $c(x)$ is any common divisor of $f(x)$ and $g(x)$ then $c(x) \vert d(x)$;
iii) $d(x)$ is monic.

Definition: The prime field of a field $F$ is the intersection of all the subfields of $F$.

Definition: A field has characteristic $0$ if its prime field is isomorphic to $\mathbb{Q}$; it has characteristic $p$ if its prime field is isomorphic to $\mathbb{Z}_p$.

Definition: If $p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$ then $p'(x) = na_nx^{n-1} + (n-1)a_{n-1}a^{n-2} + \dots + a_1$.

Proof: We will first show $\gcd(p(x), p'(x)) = 1$. Define $d(x) = \gcd(p(x), p'(x))$. Then we can find $q(x) \in F[x]$ such that $p(x) = d(x)q(x)$. But $p(x)$ is irreducible which means $d(x)$ is constant or $q(x)$ is constant. If $q(x)$ is constant, then $\deg d(x) = \deg p(x)$. In particular, $\deg d(x) = \deg p(x) > \deg p'(x)$. But we have also assumed $d(x) \vert p'(x)$. This implies $\deg d(x) \le \deg p'(x)$. We have reached a contradiction. So, we can conclude $d(x)$ is constant. Since $d(x)$ is also monic, we have $d(x) = 1$.

Isn't this automatically the case? If $p(x)$ is irreducible, then $\operatorname{gcd}(p(x),f(x))=1$ for any polynomial $f(x)$, simply because $\operatorname{gcd}\,|\,p.$

Next, we suppose $p(x)$ has a repeated root, say $a \in F$. Then we can find $k(x) \in F[x]$ such that $p(x) = (x-a)^2k(x)$. Taking the derivative, we have $p'(x) = 2(x-a)k(x) + (x-a)^2k'(x)$. Hence, $(x-a) \vert \gcd(p(x), p'(x))$. This contradicts the fact $\gcd(p(x), p'(x)) = 1$.

We can conclude $p(x)$ has no repeated roots. []My question is, I don't think I used the assumption that $F$ has characteristic $0$. Is there a mistake in the proof?

Only a notational. The way you wrote it, you have proven that $p(x)$ is irreducible if and only if $\operatorname{gcd}(p(x),p'(x))=1.$ But we want another conclusion: If $p(x)$ is irreducible with a repeated root $a$, then $(x-a)\,|\,p(x) \wedge (x-a)\,|\,p'(x)$ so it cannot be irreducible.

But even this can be shortened. If $p(x)$ has a root $a$, then $(x-a)\,|\,p(x)$ and $p(x)$ is reducible.

It was valuable to prove what you have proven, but it wasn't necessary for your statement. All you need is the Euclidean algorithm to conclude that $x-a$ divides the polynomial if $a$ is a root. But every polynomial ring over a field (of any characteristic) is a Euclidean ring.

 

[fishturtle1]

That makes sense; I see now there was a simpler proof.

Proof: Assume by contradiction $p(x)$ has a repeated root, say $a \in F$. Since $F[x]$ is a Euclidean ring, we have $p(x) = (x-a)^2 q(x)$ for some $q(x) \in F[x]$. We can rewrite this as $p(x) = (x-a) h(x)$ where $h(x) = (x-a)q(x)$. But $\deg (x-a) < \deg p(x)$ and $\deg h(x) < \deg p(x)$. This shows $p(x)$ is not irreducible. We have reached a contradiction. []

Thank you!

 

[fresh_42]

You should try to prove the following statement:

$p(x)$ has a repeated root $a$ if and only if $(x-a)\,|\,\operatorname{gcd}(p(x),p'(x)).$

You have already shown "$\Longrightarrow$". Can you show the other direction?

 

[fishturtle1]

You should try to prove the following statement:

$p(x)$ has a repeated root $a$ if and only if $(x-a)\,|\,\operatorname{gcd}(p(x),p'(x)).$

You have already shown "$\Longrightarrow$". Can you show the other direction?

Sure, i'll give it a try:

Proof: We want to show if $\gcd(p(x), p'(x)) \neq 1$ then $p(x)$ has a repeated root. We will show the contrapositive. Suppose $p(x)$ has no repeated roots. By Kroneker's theorem, we can find a field $E$ such that $E$ contains $F$ and $p$ splits over $E$. So, we can write $p(x) = \prod_{i=1}^{n} (x-a_i)$ where $a_i \in E$ and $a_i \neq a_j$ whenever $i \neq j$.

Assume by contradiction $\gcd(p(x), p'(x)) \neq 1$. Then $(x-a_i) \vert \gcd(p(x), p'(x))$ for some $a_i$. We can find $q(x)$ such that $p(x) = (x-a_i)q(x)$. Taking the derivative gives

$$p'(x) = 1\cdot q(x) + (x-a_i)q'(x)$$

From the above, we see $(x-a_i)$ does not divide $p'(x)$. We have reached a contradiction. We can conclude $\gcd(p(x), p'(x)) = 1$, which proves the contrapositive. []

 

[fresh_42]

Sure, i'll give it a try:

Proof: We want to show if $\gcd(p(x), p'(x)) \neq 1$ then $p(x)$ has a repeated root. We will show the contrapositive. Suppose $p(x)$ has no repeated roots. By Kroneker's theorem, we can find a field $E$ such that $E$ contains $F$ and $p$ splits over $E$. So, we can write $p(x) = \prod_{i=1}^{n} (x-a_i)$ where $a_i \in E$ and $a_i \neq a_j$ whenever $i \neq j$.

Assume by contradiction $\gcd(p(x), p'(x)) \neq 1$. Then $(x-a_i) \vert \gcd(p(x), p'(x))$ for some $a_i$. We can find $q(x)$ such that $p(x) = (x-a_i)q(x)$. Taking the derivative gives

$$p'(x) = 1\cdot q(x) + (x-a_i)q'(x)$$

From the above, we see $(x-a_i)$ does not divide $p'(x)$. We have reached a contradiction. We can conclude $\gcd(p(x), p'(x)) = 1$, which proves the contrapositive. []

Makes a bit dizzy, doesn't it?

If $(x-a) \,|\,p(x) \wedge (x-a)\,|\,p'(x)$ then $p(x)=...$ and $p'(x)=...$ ...

 

[fishturtle1]

Thanks! Based off the above:

Proof: Suppose $(x-a) \vert \gcd(p(x), p'(x))$. Then we can find $q(x)\in F[x]$ such that $p(x) = (x-a)q(x)$. Taking the derivative of $p(x)$ gives

$$p'(x) = q(x) + (x-a)q'(x)$$

which implies $(x-a) \vert q(x)$. So, we can find $k(x) \in F[x]$ such that

$$p(x) = (x-a)q(x) = (x-a)(x-a)k(x) = (x-a)^2k(x)$$

i.e., $p(x)$ has a repeated root. []

 

[Office_Shredder]

I think I'm missing something from this thread. Is the goal to demonstrate that p(x) doesn't have a repeated root in F, or is the goal to demonstrate that it doesn't have a repeated root in any field extension of F? Because if the root has to be in F, you get an even simpler result, which is that it has no roots unless it's a linear polynomial right. And it seems like that's what's being assumed here, so I assume we're answering the wrong question.

I'm particular, why is $q(x)\in F[x]$? If there's a good reason for it I don't see it in the proof.

 

[fishturtle1]

I think I'm missing something from this thread. Is the goal to demonstrate that p(x) doesn't have a repeated root in F, or is the goal to demonstrate that it doesn't have a repeated root in any field extension of F? Because if the root has to be in F, you get an even simpler result, which is that it has no roots unless it's a linear polynomial right. And it seems like that's what's being assumed here, so I assume we're answering the wrong question.

I'm particular, why is $q(x)\in F[x]$? If there's a good reason for it I don't see it in the proof.

In a previous exercise, the textbook says the following "...Show $f(x)$ has no repeated roots (i.e., $f(x)$ is not a multiple of $(x-a)^2$ for any $a \in F$.) " So, I think (?) we can assume the repeated root belongs to $F$.

And then we have, for $a \in F$, we have $f(a) = 0 \iff f(x) = (x-a)q(x)$ for some $q(x) \in F[x]$.

 

[Office_Shredder]

So f is irreducible, and you just factored it into $k(x)q(x)$ where $k(x)=x-a$. What can you say about $q(x)$, since f is irreducible?

In particular, the hint seems totally unnecessary for this case.

 

[fishturtle1]

In that case, $q(x)$ must be constant. But that means $f(x)$ has degree $1$ which implies $f(x)$ has no repeated roots.

Now that you point it out, it seems (?) likely that the textbook must have meant $f(x)$ has no repeated roots in any field extension. I am not sure.

 

[fresh_42]

$\operatorname{gcd}(f(x),f'(x))$ is used to count multiple roots, so it is in that realm. This is e.g. necessary if you consider sign changes in the values $f(x)$ between roots: $(x-1)^2$ has no change of signs from left to right at $x=1$ whereas $x-1$ does have a change of signs. There are theorems which use this fact.

 

[fishturtle1]

So f is irreducible, and you just factored it into $k(x)q(x)$ where $k(x)=x-a$. What can you say about $q(x)$, since f is irreducible?

In particular, the hint seems totally unnecessary for this case.

I am sooo sorry, I just reread the section and the definition of repeated root was updated:

Definition 1: If $F$ is a field and $f(x) \in F[x]$, then $f(x)$ has repeated roots if there is a field $E$ containing $F$ and a factorization in $E[x]$ of the form $$f(x) = (x-a)^2h(x)$$

-------
I will try again to prove the original problem with the new definition. In the textbook,

Corollary 18: Let $k \subset K$ be fields, and let $f(x), g(x) \in k[x] \subset K[x]$. Then the $\gcd$ of $f$ and $g$ computed in $K[x]$ is the same as the $\gcd$ of $f$ and $g$ computed in $k[x]$.

Exercise 44 (from a couple sections back): Let $f(x) = \prod_{i=1}^{n}(x-a_i) \in F[x]$, where $F$ is a field and $a_i \in F$ for all $i$. Show that $f(x)$ has no repeated roots (i.e., $f(x)$ is not a multiple of $(x-a)^2$ for any $a \in F$) if and only if $\gcd(f(x), f'(x)) = 1$, where $f'(x)$ is the derivative of $f(x)$.We will use the above two statements to prove the OP problem: If $F$ is a field of characteristic $0$ and $p(x) \in F[x]$ is irreducible, then $p(x)$ has no repeated roots. (Hint: Consider $\gcd(p(x), p'(x))$.

Proof: Since $p(x)$ is irreducible over $F$, it's only monic divisors in $F[x]$ are $1$. So, $\gcd(p(x), p'(x)) = 1$ in $F[x]$. By Corollary 18, $\gcd(p(x), p'(x)) = 1$ in any field extension $E$ of $F$. By Exercise 44, this is equivalent to saying $p(x)$ has no repeated roots over any field extension $E$ of $F$. We can conclude $p(x)$ has no repeated roots (Definition 1). []