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Question: $L \supset K$, $K$ has characteristic $3$, $[L:K]=3$. Find an example where $L$ is
(a) separable
(b) non-separable.


What I know: $L$ is a finite field extension of $K$. So, $K$ is its subfield and $L$ contains finitely many elements.

$K$ has characteristic $3$. So, any $a, b \in K, (a+b)^3=a^3+b^3$. An example of a field of characteristic $3$ is $\mathbb{F_3}$

$[L:K]=3$. This implies that the dimension of the vector space $L$ over $K$ is equal to $3$

Now, for $L$ to be separable over $K$, if any element of $L$ is separable over $K$ (i.e. $\forall l \in L$ the minimal polynomial over $K$ has no multiple roots in its slitting field.)


That is what I know but I am having trouble conjuring up examples where $L$ is and isn't separable. Would taking $K=F_3$ be a good start?

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    $\begingroup$ Be careful, "finite extension" does not mean that $L$ is a finite field. In fact, theorem : every algebraic extension of a finite field is separable. $\endgroup$ – Captain Lama Mar 29 '16 at 14:39
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Firstly, a question to you.

Does a field extension of prime degree have nontrivial subextensions? I.e., if $L|K$ is a finite field extension of prime degree $[L:K]=p$ a prime number, is there a subfield $F$ of $L$ containing $K$ such that $F$ is neither $L$ or $K$?

Secondly, have you learnt what a perfect field is? A perfect field (of characteristic $p$) is one such that every element $\alpha\in K$ has a $p$-th root.

This is in fact a characterisation of perfect fields. The definition of a perfect field is: one whose algebraic extensions are all separable. This is always the case in characteristic zero, and in characteristic $p>0$ you need to prove the following:

CLAIM: Let $f(x)\in K[x]$ be an irreducible polynomial. If $f(x)$ has multiple roots, then $f'(x)\equiv 0$, and so $f(x)=g(x^p)$ for some polynomial $g(t)\in K[t].$

It is easy to see that $\mathbb{F}_q$ (in your case, $\mathbb{F}_p=\mathbb{Z}/p\mathbb{Z}$ is a perfect field. Now, every field of characteristic $p$ contains $\mathbb{F}_p$, so what does it tell you? You must choose a transcendental extension of $\mathbb{F}_p$ if you wish to find an example of a field with a finite inseparable extension.

EXAMPLE: Let $k=\mathbb{F}_p$, and let $L=k(T)$ be the field of rational functions over $k$. Consider $K$ to be $k(T^p)$, and call $\alpha=T.$ The bigger field $L$ is an extension of $K$ by adding the element $\alpha$ such that $\alpha^p\in K.$

This is an example of a purely inseparable field extension, which is monogenous as $L=K[\alpha]=K(\alpha)$, but at the same time $\alpha^p=T^p\in K$ means that the only root of the irreducible polynomial $(X-T)^p=X^p-T^p\in K[X]$ is $\alpha$.

In one variable, we have Lüroth's theorem, which characterises all subextensions $K$ of $L=k(T)$ such that $L|K$ is finite. The example given is but one of the many one could provide, and perhaps the simplest.

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  • $\begingroup$ Thank you for your response. I have not come across perfect fields before and am a bit confused by this. Would $K$ be $\mathbb{F_q}$ in this ok? $\endgroup$ – thinker Mar 29 '16 at 21:01
  • $\begingroup$ No, since it can only have separable finite extensions. The reason is the claim above: $x^q=x$ for any $x\in\mathbb{F}_q$, so $x=y^p$ for some $y$ indeed. $\endgroup$ – Theon Alexander Mar 29 '16 at 21:02
  • $\begingroup$ Do you know what a transcendental extension is? I mention it at the end of my answer. $\endgroup$ – Theon Alexander Mar 29 '16 at 21:03
  • $\begingroup$ A non-algebraic extension? So L/K is transcendental if NOT every element of L is a root of some non-zero polynomial with coefficients in K $\endgroup$ – thinker Mar 29 '16 at 21:37
  • $\begingroup$ No, it is that $K$ that must be taken as your example of a non-perfect field, since, given the above, a finite field won't do. $\endgroup$ – Theon Alexander Mar 29 '16 at 21:38

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