Problem in Jacobson's Basic Algebra (Vol. I)

Solution 1:

Let $E$ be a splitting field of $f(x)=x^p-x-a$ over $F$. Let $b\in E $ be a root of $f$. Then for any $n=1,2,\ldots,(p-1)$, the element $b+n$ is also a root of $F$, because $$f(b+n)=(b+n)^p-(b+n)-a=(b^p-b-a)+(n^p-n)=0+0=0$$ (to be precise, $b+n$ is just shorthand for $b+(n\cdot 1_F)$). Because $f$ is of degree $p$, we've now found all the roots. Clearly, we have that $E=F(b)$.

Let $g(x)\in F[x]$ be the minimal polynomial for $b$ over $F$. Note that $g\mid f$ because $f(b)=0$. Since $E=F(b)$, we have that $[E:F]=\deg(g)$.

Now note that $g(x-n)$ is the minimal polynomial for $b+n$ over $F$; if any monic irreducible polynomial $h(x)\in F[x]$ with $\deg(h)<\deg(g)$ had $h(b+n)=0$, then $h(x+n)$ would have $b$ as a root and still have smaller degree than $g$, contradicting the fact that $g$ is $b$'s minimal polynomial.

Let the factorization of $f$ into monic irreducibles be $f=q_1 q_2\cdots q_r$ (there can't be any repeated factors, because that would lead to repeated roots). Each of the $q_i$ has as a root at least one of the $b+k$, and therefore (being monic and irreducible) is its minimal polynomial. This implies that $\deg(q_i)=\deg(g)$ for all $i$, and therefore $\deg(g)\mid \deg(f)=p$. Thus, either

• $\deg(g)=p$, hence $f=g$ is irreducible, or,

• $\deg(g)=1$, hence $[E:F]=1$, hence $b\in F$.

Solution 2:

Conversely, suppose $f$ is reducible. Then we can write $f=\prod_{i=1}^k f_i$ for some irreducible $f_i$ and some $k\in\mathbb N$. We know that the $f_i$ are distinct factors. Consider any two of the $f_i$; without loss of generality, we consider $f_1,f_2$. Now consider $F[t]/(f_1)$, which contains a root $b$ of $f_1$ which is consequently a root of $f$. Any field which contains a root of $f$ contains all roots of $f$. Thus $F[t]/(f_1)$ is a splitting field of $f$ over $F$, with $[F[t]/(f_1):F]=\deg f_1$. We argue analogously for $f_2$. As splitting fields are isomorphic, we conclude that $\deg(f_1)=\deg(f_2)$.

Hence $f=\prod_{i=1}^k f_i$ where all of the $f_i$ have the same degree, say $\deg(f_i)=n$. Therefore, $p=nk$ and so we have either $n=1$ and $k=p$, or $n=p$ and $k=1$. In the former case, we see that $F$ contains a root of $f$, contradicting our hypothesis. In the latter case, $f$ is irreducible.