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We say a polynomial $P \in \mathbb{K}[X]$ is seperable (where $\mathbb{K}$ is a field) if and only if $P$ has only simple roots in the algebraic closure of $K$.

We say an element $x$ is seperable if it's minimal polynomial is separable.

I'm currently searching for non-separable elements (preferable in $\mathbb{Q}, \mathbb{R}, \mathbb{C}$), but cannot seem to find any.

user7802048
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1 Answers1

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You can't find non-separable elements in $\mathbb Q, \mathbb R$ or $ \mathbb C$ because these are all fields of characteristic zero.

You should instead think about infinite fields of characteristic $p$. For example, consider the field extension $\mathbf F_p (t): \mathbf F_p(t^p)$, and consider the element $t \in \mathbf F_p(t)$. Its minimal polynomial over $\mathbf F_p(t^p)$ is $X^p - t^p = (X - t)^p$.

Kenny Wong
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  • Thank you! Could you give an concrete example in let's say $\mathbb{F}_5$. I'm new to the subject and not really familiar with the notation. – user7802048 May 13 '17 at 12:31
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    I'm afraid you can't find examples in finite fields. Your examples must come from fields that are (i) infinite and (ii) characteristic $p$. – Kenny Wong May 13 '17 at 12:32
  • Oh, I think I'm on the right track. Thank you again. I'll accept the answer as soon as possible (Can't upvote for now - don't have enough rep). – user7802048 May 13 '17 at 12:33
  • Also note that it doesn't really make sense to talk about non-separable elements in one given field. You need to specify TWO fields, $L$ and $K$, with $K \subset L$. Then you say that an element $\alpha \in L$ is separable over $K$ iff its minimal polynomial over $K$ has distinct roots in $L$. – Kenny Wong May 13 '17 at 12:34
  • And to explain the notation: $K(t)$ just means "rational functions in $t$ with coefficients in $K$." So $\mathbb F_p(t^p)$ means "rational functions in $t^p$ with coefficients in $\mathbb F_p$. – Kenny Wong May 13 '17 at 12:35