MHB Field Extensions - Discussion of Dummit and Foote - page 512

[Math Amateur]

I am trying to understand a discussion of field extensions by Dummit and Foote: Abstract Algebra, page 512 ...

The relevant text is as follows:View attachment 4832When I first read the second paragraph, starting "Given any field $$F$$ and any polynomial $$p(x) \in F[x]$$ ... ... ... ", I thought that D&F were trying to construct a situation where $$p(x)$$ did not have a root in $$F$$ ... ... and then go on to construct an extension field $$K = F[x]/p(x)$$ that did indeed have a root of the polynomial $$p(x)$$ ... ...BUT ... ... D&F appear (to me, anyway) to try to ensure that $$p(x)$$ does not have a root in $$F$$ by stipulating or declaring that $$p(x)$$ must be irreducible ... at least I think this is their intention .. ...

BUT ... ... ? ... ... linear polynomials such as $$x - 7$$ are irreducible in $$F[x]$$ when $$F = \mathbb{Q}$$ ... but $$p(x) = x - 7$$ does have a root in $$\mathbb{Q}$$ ... that is no need for an extension field (except the 'extension' $$\mathbb{Q} = \mathbb{Q}$$) ...

So it seems to me I am not fully understanding D&F's discussion ...

Can someone please clarify D&F's discussion for me by explaining what they are really getting at ...

Peter

 

[Deveno]

Lots of polynomials have roots in the underlying field their polynomial coefficients are in. For example, $x^2 - 3x + 2$ does, as it equals $(x - 2)(x - 1)$ (assuming an underlying field of characteristic $0$, so as to avoid ambiguity).

The point being, if we want to find roots of ALL polynomials, it suffices to consider roots of IRREDUCIBLE polynomials, because we can factor polynomials into their irreducible factors (this is, essentially, the statement that $F[x]$ is a UFD. In a UFD, "irreducible" and "prime" mean the same thing- see here).

A polynomial like $x - 7$ is indeed irreducible (over $\Bbb Q$, let's say). All this leads us to conclude is that the only possible root it has is $7$, and yes, $7 \in \Bbb Q$. If a polynomial factors completely, over a field, into linear factors, we say it SPLITS, and the smallest field containing all of its roots is called its SPLITTING FIELD. For a polynomial like $x - 7$, the splitting field is $\Bbb Q$. This is not particularly interesting.

On the other hand, anyone familiar with the quadratic formula knows it entails square roots. So, even if a quadratic polynomial has rational coefficients, it may not have rational roots-because $\Bbb Q$ does not contain all square roots of rational numbers. For example $\sqrt{2} \not\in \Bbb Q$. This is the kind of situation in which we seek to extend $\Bbb Q$.

But how? Imagine we don't yet know about $\Bbb R$ (for square roots of positive rational numbers) or $\Bbb C$ (for square roots of negative rational numbers). How can we, only having knowledge of polynomials and rings, CREATE an extension of $\Bbb Q$ that contains a square root we need as a root of a quadratic? Ideally, we want a "minimal" construction, so we only do the amount of work we HAVE to.

 

[Math Amateur]

Lots of polynomials have roots in the underlying field their polynomial coefficients are in. For example, $x^2 - 3x + 2$ does, as it equals $(x - 2)(x - 1)$ (assuming an underlying field of characteristic $0$, so as to avoid ambiguity).

The point being, if we want to find roots of ALL polynomials, it suffices to consider roots of IRREDUCIBLE polynomials, because we can factor polynomials into their irreducible factors (this is, essentially, the statement that $F[x]$ is a UFD. In a UFD, "irreducible" and "prime" mean the same thing- see here).

A polynomial like $x - 7$ is indeed irreducible (over $\Bbb Q$, let's say). All this leads us to conclude is that the only possible root it has is $7$, and yes, $7 \in \Bbb Q$. If a polynomial factors completely, over a field, into linear factors, we say it SPLITS, and the smallest field containing all of its roots is called its SPLITTING FIELD. For a polynomial like $x - 7$, the splitting field is $\Bbb Q$. This is not particularly interesting.

On the other hand, anyone familiar with the quadratic formula knows it entails square roots. So, even if a quadratic polynomial has rational coefficients, it may not have rational roots-because $\Bbb Q$ does not contain all square roots of rational numbers. For example $\sqrt{2} \not\in \Bbb Q$. This is the kind of situation in which we seek to extend $\Bbb Q$.

But how? Imagine we don't yet know about $\Bbb R$ (for square roots of positive rational numbers) or $\Bbb C$ (for square roots of negative rational numbers). How can we, only having knowledge of polynomials and rings, CREATE an extension of $\Bbb Q$ that contains a square root we need as a root of a quadratic? Ideally, we want a "minimal" construction, so we only do the amount of work we HAVE to.

Thanks for the help, Deveno ...

Will reflect on this more in the morning ... after midnight here in Tasmania ...

Grateful for your help, as usual ...

Peter