# Clarification of a statement in Jacobson's book

Jacobson (in Basic Algebra $1$) defines a separable polynomial to be one whose irreducible factors have distinct roots. Then he defines a perfect field to be a field $F$ such that every polynomial in $F[x]$ is separable, and proves that if $F$ is finite or of characteristic $0$, then it is perfect.

At the beginning of the section corresponding to this, he says:

We shall show in this section that if $F$ is of characteristic $0$ or if $F$ is a finite field, then there is no loss in generality in assuming that all the roots are simple.

I don't understand what this is supposed to mean. What is the "generality" in which there is no loss with that assumption?

This section starts at page $229$. The quote is the first sentence on page $230$.

• Wlog. a perfect field $F$ is one where every irreducible polynomial $g \in F[x]$ has no repeated roots. This is equivalent to that any finite extension of $F$ is contained in some Galois extension – reuns Aug 7 '17 at 4:54
• @reuns I don't know any Galois theory (and at that point in the book, the author still hadn't talked about Galois stuff). – Cauchy Aug 7 '17 at 4:55
• Did you read my 1st sentence ? $\mathbb{F}_2(t^2)$ is not perfect because $x^2-t^2 = (x+t)^2$ is irreducible over $\mathbb{F}_2(t^2)$ (its root generates the degree $2$ extension $\mathbb{F}_2(t)/\mathbb{F}_2(t^2)$) – reuns Aug 7 '17 at 4:57
• I read it but I didn't understand it - I thought it was connected to your second sentence. What exactly do you mean in your first sentence? – Cauchy Aug 7 '17 at 4:58
• In a perfect field $F$, if a polynomial $g(x) \in F[x]$ has a double root then it is not irreducible... – reuns Aug 7 '17 at 4:59

Here "without loss of generality" is in the context of describing fields as splitting fields of polynomials. So what it means is that if $E$ is any field extension of $F$ that is the splitting field of some polynomial $f(x)\in F[x]$, then there exists a polynomial $f_0(x)\in F[x]$ such that $E$ is also the splitting field of $f_0(x)$ and all the roots of $f_0(x)$ in $E$ are distinct. (In fact, the discussion shows something stronger: the roots of this $f_0(x)$ are the same as the roots of $f(x)$, just with the repeats removed.)