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Charles Pinter A Book of Abstract Algebra chapter 31 problem C5:

Prove:

If $p(x) = x^4 + ax^2 + b$ is irreducible in F[x], then F[x]/$<p(x)>$ is the root field of p(x) over F.

Discussion:

As the hint in the back of the book states, F[x]/$<p(x)>$ = F(c) where c is a root of p(x). The hint then goes on to rewrite p(x) as $(x^2 +a/2)^2 – (a^2/4 – b)$ (actually it has $(x^2 - a/2)^2 – (a^2/4 – b)$, but that’s a typo). The hint then says to let $d_1$ be a root of $x^2 – [(a^2/4) – b]$ and $d_2$ be a root of $x^2 – (a/2 +d_1)$. It says then that F($d_1$, $d_2$) = F(c) and that F($d_1$, $d_2$) contains all the roots of p(x).

$d_1 = {±\sqrt{a^{2}/4-b}}$

$d_2 = {±\sqrt{a/2±\sqrt{a^{2}/4-b}}}$

Note that $d_2$ isn’t necessarily a member of F[x]. I had considered that p(x) may have all its roots in F(c) because it has at least one root in the root field of $x^2 – (a/2 +d_1)$, but because $x^2 – (a/2 +d_1)$ isn't in F[x], that isn't applicable.

The roots of p(x) are ${±\sqrt{a/2±\sqrt{a^{2}/4-b}}}$, the same as $d_2$.

So if I select ${\sqrt{a/2+\sqrt{a^{2}/4-b}}}$ as c, then it’s clear that F($d_1$, $d_2$) = F(c). Since $d_1 = d_2^2 – a/2$, the $d_1$ is extraneous, and F($d_1$, $d_2$) = F($d_2$) = F(c).

But for F(c) to be the root field, it would also have to contain ${\sqrt{a/2-\sqrt{a^{2}/4-b}}}$.

Does there exist some set of $k_0$ to $k_n$ member of F such that $k_0 + k_1c + k_2c^2 + … + k_nc^n = {\sqrt{a/2-\sqrt{a^{2}/4-b}}}$?

The only members of F specified are 0, 1, a and b. The roots of p(x) are specifically excluded. There could be other members, but nothing I can assume.

Letting c = ${\sqrt{a/2+\sqrt{a^{2}/4-b}}}$ and d = ${\sqrt{a/2-\sqrt{a^{2}/4-b}}}$, then F(c, d) would contain all the roots of p(x). In addition, F[x]/$<p(x)>$ is isomorphic to F(c) is isomorphic to F(d), but that’s not the same as F(c, d).

What am I missing here?

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I don't have the book with me; I assume that by root field the author means what is more commonly called a splitting field, i.e. the field extension obtained by adjoining all the roots of $p(x)$.

With this interpretation, the statement is false; $F(c)$ need not be the splitting field. As indicated in this answer, the statement will be false precisely when both $b$ and $b(a^2-4b)$ are not squares in $F$ (e.g. take $F = \mathbb{Q}$, $a=1, b=2$).

To see that this is the case, note that the field $F(c, d)$ contains the element $\sqrt{b} = cd$ and the element $\sqrt{a^2 - 4b} = c^2 - d^2$. Hence it also contains $\sqrt{b(a^2 - 4b)} = \sqrt{b}\sqrt{a^2 - 4b}$, whereby $F' = F(\sqrt{b(a^2-4b)})$ is a subfield of $F(c, d)$. But the polynomial $p(x)$ is still irreducible over $F'$, as $F'$ does not contain the square root of the discriminant $a^2 - 4b$. It follows that $F(c, d)$ has degree at least $4$ over $F'$, whereby $F(c, d)$ has degree at least $8$ over $F$. Thus we cannot have $F(c, d) = F(c)$.

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  • $\begingroup$ Per the Pinter book, given a polynomial a(x) member of F[x], the root field of a(x) over F is “the smallest extension of F containing all the roots of a(x).” The definition of splitting field seems to be equivalent. $\endgroup$ – Alan Ox Jan 12 '20 at 1:37

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