Let $f$ be the irreducible polynomial over $K$ and let $L$ be the splitting field ex of $f$. Why should all the roots be simple or none of them can be simple?

I am looking for an example for an extension $K\le L\le M$ such that $M|K$ is normal doesn't mean that $L|K$ and $M|L$ are normal.

  • $\begingroup$ There's a mixup with the order of $K$ and $L$ as well as with $L$ and $M$ in the last sentence. What do you mean by your question? Is your question whether a field extension always has to be separable? $\endgroup$ – k.stm Oct 23 '12 at 19:45
  • $\begingroup$ @K.Stm. : it need not be seperable . $\endgroup$ – Theorem Oct 23 '12 at 19:50
  • $\begingroup$ For your first question (which is entirely separate from the second): if $f$ is irreducible and has a repeated root $\alpha$ in its splitting field, then $\alpha$ is also a root of $f'$, from which it follows that $f' = 0$. But this means $f(x) = g(x^p)$ for some polynomial $g$, so all the roots of $f$ have multiplicity at least $p$ (here $p$ is the characteristic of $K$). $\endgroup$ – Justin Campbell Oct 23 '12 at 19:50
  • $\begingroup$ @JustinCampbell : can you explain me why $f(x)=g(x^p)$, i didn't get the argument . $\endgroup$ – Theorem Oct 23 '12 at 19:55

Picking up where I left off in the comments: if $f$ is irreducible and has a repeated root in its splitting field, then $f' = 0$. But if we write $f(x) = \sum a_nx^n$, then $0 = f'(x) = \sum na_nx^{n-1}$ means that whenever $a_n \neq 0$ we have $n = 0$ in $K$, which is to say $p | n$, where $p$ is the characteristic of $K$. This says precisely that $f(x) = g(x^p)$ for some polynomial $g$.

For your other question: if $M/K$ is normal then $M/L$ is normal automatically, but $K/L$ need not be. For example, take $M = \mathbb{Q}[2^{1/4},i]$, which is Galois (in particular, normal) with group $D_4$, the dihedral group of the square. The subextension $L = \mathbb{Q}[2^{1/4}]$ is not normal: there is an automorphism of $M$ which sends $2^{1/4} \mapsto 2^{1/4}i \notin L$. Alternatively, one can show that $L$ corresponds via Galois theory to a subgroup in $D_4$ generated by a reflection, which is not normal.

  • $\begingroup$ :You have assumed that the field is of character $p$ what if its character is $0$ ? $\endgroup$ – Theorem Oct 23 '12 at 20:31
  • $\begingroup$ If the characteristic is zero then $f' = 0$ implies $f$ is constant. So irreducible polynomials in characteristic zero never have repeated roots: this is strictly a characteristic $p$ phenomenon. $\endgroup$ – Justin Campbell Oct 23 '12 at 23:33

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