# Is the field extension $K(t):K$ normal? Is separable?

The field extension $K(t):K$ where $K$ is any field. I don't know how to apply definition of normal and separable in a transcendental case.

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As far as I know, normal and separable are not defined for transcendental extensions. –  Qiaochu Yuan Sep 30 '12 at 18:30

Normal extensions are only defined for algebraic extensions, so that it doesn't make sense to ask whether the extension $K(t)/K$ is normal.

For "separable" the situation is more interesting.
The modern point of view, systematically used in algebraic geometry and developed in Bourbaki's Algebra, Chapter V for example, is that a field extension $L/K$ is separable if for every field extension $K'/K$ the ring $L\otimes _K K'$ is reduced i.e. its only nilpotent element is zero.
In other words, separable extensions are universally reduced extensions (one also says geometrically reduced extensions).
It is then not very difficult to prove that any purely transcental extension $K(...,T_i,...)/K$ in any set (even infinite) of indeterminates is separable, and this answers your question: yes $K(t)/K$ is a separable extension.

However for algebraic extensions $K \subset L$ there is a classical definition of separable extension: the condition is that the minimal polynomial $Irr_K(a,X)$ over $K$ of any element $a\in L$ be separable, i.e. that it have no multiple rooot in an algebraic closure of $\bar K$ of $K$.
Reassuringly, as I am sure you expected, it turns out that for algebraic extensions both notions of "separable" coincide.

Finally let me illustrate these definitions on a simple example.
Let $K=\mathbb F_p(x)$ the rational function field over $\mathbb F_p$ and consider the algebraic extension $K=\mathbb F_p(x)\subset L=K(\xi)$, where $\xi$ is the unique zero of $X^p-x$ in $\bar K$.
The extension $L/K$ is not classically separable because $X^p-x=Irr_K(\xi,X)$ has $\xi$ as a root of multiplicity $p$.
On the other hand, we have $$L\otimes_K \bar K=\frac{K[X]}{(X^p-x)} \otimes_K \bar K= \frac{\bar K[X]}{(X^p-x)}= \frac{\bar K[X]}{(X-\xi)^p}$$ a ring where the class of $X-\xi$ is nilpotent but not zero, which proves that $L/K$ is not universally reduced, hence not separable in the modern sense either.

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