Exercise 5.3.11

Proof.

(a)

Suppose that $f$ is squarefree.

Let $f=f_1\cdots f_r$ a decomposition of $f$ in irreducible factors in $k[x]$.

If two irreducible factors $f_i,f_j,i\ne j$ in this decomposition are associate (i.e. $f_i\mid f_j,f_j\mid f_i$), then $f_j=\lambda f_i,\lambda \in F^{\text{∗}}$. Then $f_i^2$ divides $f_i{f}_j$ which divides $f$, and $f$ is not squarefree.

Conversely, suppose that $f$ is not squarefree. Then $f$ is divisible by a square factor $g^2$, where $g$ is a nonconstant polynomial. Let $f_1$ an irreducible factor of $g$. The unicity of the decomposition in irreducible factors shows that any decomposition in irreducible factors contains two factors $g_1,g_2$ associate to $f_1$, so $g_1\mid g_2,g_2\mid g_1$.

Conclusion: $f$ is squarefree if and only if $f$ is product of irreducible factors, no two of which are associate.

(b)

Assume that $F$ has characteristic $0$.

Proposition 5.3.7(c) shows that $f$ is separable if and only if $f$ is product of irreducible factors, no two of which are associate.

By part (a), this is equivalent to $f$ is squarefree.

Note: this equivalence remains true in a finite field.

Counterexample in $F={𝔽}_3(t)$: the polynomial $f=x^3-t\in {𝔽}_3(t)[x]$ is irreducible over $F$, so is squarefree, but if $\alpha$ is a root of $f$ in a splitting field $L$, $x^3-t=x^3-{\alpha }^3={(x-\alpha )}^3$ is not separable. This is due to the fact that "squarefree" is a notion which depends on the field: $f$ is squarefree over $F$, not over $L$. □