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I want to prove that if $p(x)\in F[x]$, where $F$ is a field, is irreducible and $F$ has characteristic zero, then $p(x)$ has no repeated roots.

I found this argument in a book, but I don't understand something:

If $p(x)$ is irreducible and has a repeated root, then $p'(x)=0$. Now if $F$ has characteristic zero, this implies $p(x)$ is constant and the claim follows.

It is easy to see that $p'(x)=0$ implies $p(x)$ is constant, since $F$ has characteristic zero.

What I don't understand is why if $p(x)$ is irreducible and has a repeated root, then $p'(x)=0$.

If $p(x)$ has a repeated root, then $p(x)=(x-a)^2p_1(x)$ where $a$ is the repeated root in its splitting field. Also, it is $p'(a)=0$. But why it implies $p'(x)=0$?

Thank you.

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Suppose that $p'(x)$ is not identically $0$. If it has a root, then it has degree $\ge 1$.

The $\gcd$ of $p(x)$ and $p'(x)$ over $F$ cannot be $1$. For if it is $1$, there are polynomials $s(x)$ and $t(x)$ in $F[x]$ such that $s(x)p(x)+t(x)p'(x)=1$. But then $p(x)$ and $p'(x)$ cannot have a common root in any field.

So $\gcd(p(x),p'(x))\ne 1$, which contradicts the irreducibility of $p(x)$, since $p'(x)$ has degree less than the degree of $p(x)$.

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  • $\begingroup$ in the last sentence, you're using that if $p(x)$ is irreducible then $(p(x),g(x))=1$ or $p(x)\mid g(x)$ for any g(x). Right? $\endgroup$ – Sandor Jun 5 '15 at 6:11
  • $\begingroup$ If the gcd of $p(x)$ and and a polynomial $g(x)$ of degree $\ge 1$ is something other than $1$, say $d(x)$ of degree $\ge 1$, then $d(x)$ divides $p(x)$ so $p(x)$ is not irreducible. The $p(x)\mid g(x)$ in the above comment is not right. $\endgroup$ – André Nicolas Jun 5 '15 at 6:39
  • $\begingroup$ By is not right, I mean it does not do the job of getting our contradiction. $\endgroup$ – André Nicolas Jun 5 '15 at 6:47
  • $\begingroup$ I see. But $p(x)\mid p'(x)$ isn't a contradiction? Could it even happen? $\endgroup$ – Sandor Jun 5 '15 at 6:53
  • $\begingroup$ An irreducible polynomial doesn't divide every polynomial. If it divides a product then it divides one of the terms. But that will not give us our result. What gives te result is what I wrote. Briefly, if there is a multiple root, and $p'(x)$ is not identically $0$, then $\gcd(p(x),,p'(x))$ has degree $\ge 1$. So there is a polynomial $d(x)$ of degree $\ge 1$ but less than the degree of $p$ (since $p'(x)$ has smaller degree than $p(x)$) such that $d9x)$ divides $p(x)$, contradicting irreducibility. $\endgroup$ – André Nicolas Jun 5 '15 at 7:00

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