Galois Groups of Polynomials:

Question

I am trying to learn Galois Theory from Dummit and Foote. I just started the exercises in the section regarding Galois Groups of Polynomials and realize I have so many questions that I can only partially answer myself, and am not sure if they are even relevant questions, I think I could use some guidance. The disorganized and stream-of-conscience nature of the discussion below is a reflection of my confusion. Any helpful nudges appreciated.

The first exercises asks:

Show that a cubic with a multiple root has a linear factor. Is the same true for quartics?

I try to make sense of the question-- Suppose $f(x)$ is some cubic polynomial with coefficients in some field $F$. Let $K$ be a splitting field of $f(x)$ over $F$. Then if the factorization of $f(x)$ over $K$ includes a term with power greater than $1$, (i.e.: $f(x)$ is inseparable), then I am trying to show that the factorization of $f(x)$ over $K$ must also include at term with power exactly $1$. That is, that the factorization of $f(x)$ over $K$ must be of the form $(x-\alpha)^2(x-\beta)$.

That does not seem right. Certainly the polynomial $f(x) = (x-\alpha)^3$ is a counter-example? But $(x-\alpha)^3 = x^3 - 3\alpha x^2 + 3\alpha^2 x - \alpha^3$, so actually $K = F$ in this case, is this what Dummit means by ``linear factor''? That question does not even make sense!

Or is it I am trying to show that $f(x)$ is actually reducible over $F$ and includes a linear term in the factorization of $f(x)$ over $F$?

But wait, this whole section is about the Galois groups of separable polynomials? How can I apply anything I have just learned to inseparable polynomials? I find a paragraph in the book:

Note that over $\mathbf{Q}$ or over a finite field (or, more generally, over any perfect field) the splitting field of an arbitrary polynomial f(x) is the same as the splitting field for the product of the irreducible factors of $f(x)$ taken precisely once, which is a separable polynomial.

I try to convince myself that this statement is true. Suppose $f(x)$ is polynomial with coefficients in a field $F$ and $K$ is a splitting field of $f(x)$ over $F$. Suppose $f(x)$ factors as $f(x) = (x-a_1)^{n_1}\ldots (x-a_k)^{n_k}$ over $K$, where $a_1,\ldots, a_k$ are the $k$ distinct roots of $f(x)$. Then $K = F(a_1,\ldots,a_k)$. Now let $g(x) = (x-a_1) \ldots (x-a_k)$. But must the coefficients of $g(x)$ be in $F$? If $s_1, \ldots, s_k$ are the elementary symmetric functions in $a_1, \ldots, a_k$, the best I can say is that the coefficients of $g(x)$ lie in $F(s_1, \ldots, s_k)$. But ok, I can convince myself that the splitting field of $g(x)$ over $F(s_1, \ldots, s_k)$ is $F(s_1, \ldots, s_k, a_1, \ldots, a_k) = F(a_1, \ldots, a_k)$. So the splitting field of $f(x)$ over $F$ is equal to the splitting field of $g(x)$ over $F(s_1, \ldots, s_k)$, but nowhere did I use the fact that $F$ is a perfect field. What did I miss? Further, $F(a_1, \ldots, a_k)$ is Galois over $F(s_1, \ldots, s_k)$, but must $F(a_1, \ldots, a_k)$ be Galois over $F$? Is this where $F$ being perfect is needed? How does it make sense to talk about the Galois group of an inseparable polynomial otherwise?

Answer 1

First question: "That does not seem right. Certainly the polynomial $f(x) = (x-\alpha)^3$ is a counter-example?"

Why that? It has a linear factor $(x-\alpha)$, even three times.

Consider the quartic polynomial $f(x)=(x^2+x+1)^2$. Does it have a linear factor in $\Bbb Q[x]$?

Answer 2

(p. 617) Exercise 1. Show that a cubic with a multiple root has a linear factor. Is the same true for quartics?

The discriminant of a cubic with a repeated factor is zero. Suppose in the discussion following (14.17), the roots $\alpha = \beta$ are repeated. ($\gamma$ may or may not be different from these two roots.) Then $D = [(\alpha - \beta) \dots]^2 = 0$.

You are reminded that the discriminant is zero if and only if its polynomial is not separable: the roots are not all distinct. (p. 610, second to last paragraph). The reminder continues: over a perfect field, this implies the polynomial is reducible. So separability is not what this question can be about.

The following is not the fastest or slickest way to get the solution to the first part of the exercise, but it should give you some idea of how to use the ideas from that section of D&F.

Let's look at a generic (monic) cubic with coefficients in the (not necessarily perfect) field $F$ having a repeated root. (The root $\beta$ may or may not be different from $\alpha$. Whether $\alpha, \beta$ are or are not elements of $F$ is to be discovered.) \begin{align*} p(x) &= (x - \alpha)^2 (x - \beta) \\ &= x^3 + (-2 \alpha - \beta)x^2 + (\alpha^2 + 2\alpha \beta)x - \alpha^2 \beta \text{.} \end{align*} We have that these coefficients are in $F$. In particular, $-2 \alpha - \beta \in F$.

(This paragraph isn't strictly necessary, but it can be convenient to notice characteristic-based shortcuts and obstructions. Here's a shortcut.) If $F$ has characteristic $2$, then this coefficient is $-\beta \in F$, so $\beta \in F$ and $p$ has a linear factor in $F$. We proceed assuming $F$ has characteristic $\neq 2$.

If the Galois group of $p$ (over $F$) is nontrivial, let $\sigma$ be a nontrivial automorphism of the splitting field, $L$, also not characteristic $2$, of $p$ over $F$. If $\sigma$ fixes both $\alpha$ and $\beta$, then $L$ contains a subfield fixed by $\sigma$ and this subfield is the splitting field of $p$, contradicting that $L$ is the splitting field. Therefore, $\sigma$ moves at least one of $\alpha$ and $\beta$, permuting the roots of $p$. The only possible nonidentity permutation of the roots is $\alpha \leftrightarrow \beta$, requiring $\alpha \neq \beta$. Since $\sigma$ fixes $F$, it fixes the coefficients of $p$. This means \begin{align*} p(x) &= \sigma(p(x)) \\ &= (x-\beta)^2(x-\alpha) \\ &= x^3 + (-\alpha -2\beta) x^2 + (2\alpha \beta +\beta^2)x - \alpha \beta^2 \end{align*} and, in particular, $$ -\alpha -2\beta = -2 \alpha - \beta \text{,}$$ from which (working in $L$, by adding $2\alpha + 2\beta$ to both sides), we discover $\alpha = \beta$, contradicting that the Galois group of $p$ (over $F$) is nontrivial.

Therefore, the Galois group of $p$ (over $F$) is trivial, so $p$ is reducible. The only possible partitions of $\deg_x p = 3$ are $2+1$ and $1+1+1$. Either way, $p$ has a linear factor (a factor, $q(x)$, with $\deg_x q(x) = 1$) in $F$. (Notice: this is not the case when $p$ is quartic -- the partitions are $3+1$, $2+2$, $2+1+1$, and $1+1+1+1$ and $2+2$ does not have a linear factor (in $F$). The other answer gives a $2+2$ example.)