1. ## Irreducible, inseparable polynomials

Let K be a field of characteristic p which is not a perfect field (K neq K^{p})

Prove there exist irreducible inseparable polynomials over K. Conclude that there exist inseparable finite extensions of K.

I am a little confused, because I thought that fields of characteristic p were isomorphic to the finite field with p elements. In this finite field, every element is a p'th power. This is what I thought the definition of a perfect field was.

2. Originally Posted by robeuler
Let K be a field of characteristic p which is not a perfect field (K neq K^{p})

Prove there exist irreducible inseparable polynomials over K. Conclude that there exist inseparable finite extensions of K.
Since $K$ is not perfect it means $K^p \not = K$ and so there exists $a\in K$ so that $a\not \in K^p$. Define the polynomial $f(x) = x^p - a$. Let $L$ be an extension field having a zero $\alpha$ of $f(x)$. Notice that if $\alpha \in L$ is a zero of this polynomial then $\alpha^p - a = 0 \implies a = \alpha^p$, but this means $x^p - \alpha^p = (x-\alpha)^p$ therefore we have shown that $f(x)$ splits over any extension field that has a zero of $f(x)$. This means either $f(x)$ is irreducible or it factors into linear factors (this is a result about polynomials of prime degree, if you do not know I can prove it for you) over $F$. It cannot factor into linear factors since it has no zeros in $F$ and so $f(x)$ is irreducible. Thus, $L/F$ is an inseperable extension. In fact, what we have constructed is a purely inseperable extension, this is like the worst case scenario of seperability.

3. Originally Posted by ThePerfectHacker
Since $K$ is not perfect it means $K^p \not = K$ and so there exists $a\in K$ so that $a\not \in K^p$. Define the polynomial $f(x) = x^p - a$. Let $L$ be an extension field having a zero $\alpha$ of $f(x)$. Notice that if $\alpha \in L$ is a zero of this polynomial then $\alpha^p - a = 0 \implies a = \alpha^p$, but this means $x^p - \alpha^p = (x-\alpha)^p$ therefore we have shown that $f(x)$ splits over any extension field that has a zero of $f(x)$. This means either $f(x)$ is irreducible or it factors into linear factors (this is a result about polynomials of prime degree, if you do not know I can prove it for you) over $F$. It cannot factor into linear factors since it has no zeros in $F$ and so $f(x)$ is irreducible. Thus, $L/F$ is an inseperable extension. In fact, what we have constructed is a purely inseperable extension, this is like the worst case scenario of seperability.
thank you. I know the result about irreducible/linear factors.

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