# An irreducible polynomial in $\mathbb Q[x]$ that has all zeros with multiplicity $2$

I'm reading an Abstract Algebra textbook written by Jb-Fraleigh. When learning the Splitting field. I should first give the definition of a perfect field:

Definition: Perfect field A field is perfect if every finite extension is a separable extension.

Now I encounter this problem:

Problem: Give an example of an $$f(x)\in \mathbb Q[x]$$ that has no zeros in $$\mathbb Q$$ but whose zeros in $$\mathbb C$$ are all of multiplicity $$2$$. Explain how this is consistent with Theorem 51.13, which shows that $$\mathbb Q$$ is perfect.

This is Theorem 51.13:

Theorem 51.13 Every field of characteristic zero is perfect.

I notice that in the proof of this Theorem, we first assume $$E$$ is a finite extension of the field of characteristic zero $$F$$, then let $$\alpha\in E$$ and $$f(x)=\operatorname{irr}(\alpha,F)$$ be an irreducible. Then we write $$f(x)=\prod_{i}(x-\alpha_i)^v$$ where $$\alpha_i$$ are distinct zeros and $$v\in \mathbb Z^+$$. In the end we can show that $$v=1$$.

Because of notice this, I wonder in the question is $$f(x)$$ an irreducible or a general (i.e., reducible) polynomial? If it can be reducible then I could give an example $$f(x)=(x^2-2)^2$$

but if what it means is finding an irreducible has all zeros multiplicity $$2$$ when factors in $$\mathbb C[x]$$, I really have no idea. Notice that I knew $$\mathbb C$$ is an infinite extension of $$\mathbb Q$$, so the proof of Theorem 51.13 may not be applied in this case, but I couldn't link everything together in order to give a reasonable proof for this problem.

• You got it right. The only polynomials that fit the Problem description are reducible. More precisely, they have the property that they are divisible by the square of every irreducible factor they have. This is built into the idea of a perfect field. – Jyrki Lahtonen Jul 18 '19 at 10:45
• Also, Theorem 51.13. does mean that $\Bbb{C}$ is also perfect. It does state that every field of characteristic zero is perfect :-) If it bothers you, you can, instead of $\Bbb{C}$ also work with the finite extension of $\Bbb{Q}$ that you get by adjoining all the roots of $f(x)$ (in $\Bbb{C}$) to it. In other words, go to the splitting field of $f$. – Jyrki Lahtonen Jul 18 '19 at 10:49

If $$F$$ is a field of characteristic zero and $$f\in F[x]$$ is irreducible then $$f$$ is separable. The easiest way to see this is define the formal derivative $$f'$$. (which looks exactly like the derivative from calculus but it is formal, there is no convergence here). It is easy to see that $$\alpha$$ is a root of $$f$$ with multiplicity higher than $$1$$ if and only if $$f'(\alpha)=0$$. It follows that $$f$$ is separable if and only if $$\gcd(f,f')=1$$.

Now, if $$f$$ is irreducible then $$\gcd(f,f')\in\{1,f\}$$. Since we are working in a field of characteristic zero we have $$deg(f')=def(f)-1$$ and hence the $$\gcd$$ can't be $$f$$. So it must be $$1$$ and hence $$f$$ is separable. This is actually a way to prove that any algebraic extension of a field of characteristic zero is separable.

So the point is you can't find an irreducible polynomial in $$\mathbb{Q}[x]$$ which has roots with multiplicity $$2$$.

• That is a good point using formal derivative! Thanks for the solution. – kelvin hong 方 Jul 19 '19 at 1:11

There is none; only reducible ones.

If there were an irreducible one, $$f$$, we could adjoin a root, and we'd have a finite extension of $$\Bbb Q$$ (of $$\operatorname {deg}f$$), namely $$\Bbb Q[x]/(f)$$, which wasn't separable.

This violates theorem $$51.13$$, since $$\operatorname {char}\Bbb Q=0$$.