Short questions relating to field extensions

[Kiwi1]

my book gives a number of short questions, one of which seems trivially simple. Is my solution correct or am I missing the point?

8. If p(x) is irreducible and has degree 2, prove that F[x]/<p(x)> contains both roots of p(x).
Soln.
Let the two roots be a and b.

F(a) is isomorphic to F[x]/<p(x)> which is isomorphic to F(b). This is proven in the text.

Therefore; F(a) is isomorphic to F(b). So F(a) and F(b) contain the same elements but F(a) contains a and F(b) contains b so F[x]/<p(x)> contains both a and b.

This argument would work for any number of roots of a polynomial p(x) or any order.

 

[caffeinemachine]

Let the two roots be a and b.

I feel that you do not understand the statement you make here. What is meant by saying that let $a$ and $b$ be the two roots of $p(x)$? Where are these two roots contained?

F(a) is isomorphic to F[x]/<p(x)> which is isomorphic to F(b). This is proven in the text. Therefore; F(a) is isomorphic to F(b).

This is correct. But i suggest you examine the proof given in book again once you have answered the question the I just posed.

So F(a) and F(b) contain the same elements

This is not true.
$F(a)$ and $F(b)$ may be two different sets. To understand this you first need to be clear on the first question I asked.

In summary, the argument you gave is not correct. I am willing to discuss this further once you respond.

 

[Deveno]

Your problem begins:

Let...(relevant information here). Prove that...(statement to be proved).

You must exhibit two elements of $F[x]/\langle p(x)\rangle$:

$f(x) + \langle p(x)\rangle$ and $g(x) + \langle p(x)\rangle$ that are roots of $p$, or otherwise show that two such roots *must* exist in $F[x]/\langle p(x)\rangle$.

I'll give you a hint: it suffices to use $f,g$ of degree 1 (why?).

It's also a good idea to state some facts about $F$, is it an arbitrary field, or one of characteristic $0$, or a finite field? Another slight technicality, is $p$ irreducible over $F$? The irreducibility of a polynomial depends crucially on the field its coefficients lie in, for example:

$x^2 - 2$ is irreducible over $\Bbb Q$, but is not irreducible over $\Bbb R$.

You say: "let $a$ and $b$ be roots..." my question to you is, how do you know such roots exist? Furthermore, how do you know they are not possibly the SAME? Does this even make a difference? You should think on these things for a bit, because to stumble at this stage of the game is to build a house of cards atop shifting sands.

 

[Kiwi1]

Thanks Guys. Cafinemachine's post has been rattling around my brain all night and I don't have time to absorb Deveno's post before I take the family on an 8 hour drive followed by a bachelor party. I will get back to it.

p(x) is irreducible in the field F so I extend F by simply appending the root a.

F(a) is the smallest field that contains both a and all elements of F. It is the set of elements \[\{a_0+a_1a+a_2a^2+...+a_na^n:a_i \in F\}\]

This clearly contains a and I must show it contains the other root. Now:

Since I am in a field I can assume p(x) is monic without loss of generality so let \(p(x)=x^2+bx+c\) then if I choose a to be the smaller root:

\(a=\frac{-b-\sqrt{b^2-4c}}{2} \in F(a)\), \(b,c \in F\)

\[\therefore -(2a+b)=\sqrt{b^2-4c} \in F(a)\]

\[\therefore \frac{-(2a+b)-b}{2}=\frac{-b+\sqrt{b^2-4c}}{2} \in F(a)\]

But this is the other root that I was looking for. So both roots are in F(a).

Is this correct?

 

[Deveno]

Thanks Guys. Cafinemachine's post has been rattling around my brain all night and I don't have time to absorb Deveno's post before I take the family on an 8 hour drive followed by a bachelor party. I will get back to it.

p(x) is irreducible in the field F so I extend F by simply appending the root a.

F(a) is the smallest field that contains both a and all elements of F. It is the set of elements \[\{a_0+a_1a+a_2a^2+...+a_na^n:a_i \in F\}\]

How do you know that such a root even exists? And more to the point, how do you know it is an element of $F[x]/\langle p(x)\rangle$, which is what you are supposed to prove?

This clearly contains a and I must show it contains the other root.

Typically one can only assume such a field $F(a)$ if one can show that there is an *algebraic closure* of $F$. That is *much* more involved than what this problem is trying to do, and I suggest you delay such assumptions until you can physically produce such a root, $a$.

Now:

Since I am in a field I can assume p(x) is monic without loss of generality so let \(p(x)=x^2+bx+c\) then if I choose a to be the smaller root:

\(a=\frac{-b-\sqrt{b^2-4c}}{2} \in F(a)\), \(b,c \in F\)

\[\therefore -(2a+b)=\sqrt{b^2-4c} \in F(a)\]

\[\therefore \frac{-(2a+b)-b}{2}=\frac{-b+\sqrt{b^2-4c}}{2} \in F(a)\]

But this is the other root that I was looking for. So both roots are in F(a).

Is this correct?

It would be a *lot* easier to write:

$p(x) = x^2 + bx + c = (x - r_1)(x - r_2)$, and note that if some extension field $E$ of $F$ contains $r_1$, then it contains:

$r_2 = -r_1 - b$.

This still does not show how you arrived at the root $r_1$. It is a bit premature to just "adjoin" it (explicitly defining it as an element of $F[x]/\langle p(x)\rangle$ is something very basic in arriving at a proof that every polynomial HAS a root in some extension field, which you need for the very *existence* of $F(a)$), you are missing a crucial step. Re-read my last post.

 

[caffeinemachine]

Thanks Guys. Cafinemachine's post has been rattling around my brain all night and I don't have time to absorb Deveno's post before I take the family on an 8 hour drive followed by a bachelor party. I will get back to it.

p(x) is irreducible in the field F so I extend F by simply appending the root a.

F(a) is the smallest field that contains both a and all elements of F. It is the set of elements \[\{a_0+a_1a+a_2a^2+...+a_na^n:a_i \in F\}\]

This clearly contains a and I must show it contains the other root. Now:

Since I am in a field I can assume p(x) is monic without loss of generality so let \(p(x)=x^2+bx+c\) then if I choose a to be the smaller root:

\(a=\frac{-b-\sqrt{b^2-4c}}{2} \in F(a)\), \(b,c \in F\)

\[\therefore -(2a+b)=\sqrt{b^2-4c} \in F(a)\]

\[\therefore \frac{-(2a+b)-b}{2}=\frac{-b+\sqrt{b^2-4c}}{2} \in F(a)\]

But this is the other root that I was looking for. So both roots are in F(a).

Is this correct?

Hello Kiwi,

My response to this post would essentially be a rephrasing of Deveno's response.

Basically, you should first be clear on as to why any root of $p(x)$ exists. An equation like $x^2+1=0$ has no root in reals. So given an arbitrary irreducible polynomial, there is no guarantee a priori that one can find a root of it in some extension field.

 

[Kiwi1]

Every polynomial has n roots in the complex numbers so if F is a subfield of the complex numbers then surely there must be an extension sitting somewhere between F and C that contains the root?

My text also makes the following statent and then offers a proof that I cannot fault:

'Basic theorem of field extensions: let F be a field and a(x) a non constant polynomial in F[x]. There exists an extension field E of F and an element c in E such that c is a root of a(x).'

I interpreted the question as being entirely about the second root.

 

[Deveno]

Every polynomial has n roots in the complex numbers so if F is a subfield of the complex numbers then surely there must be an extension sitting somewhere between F and C that contains the root?

My text also makes the following statent and then offers a proof that I cannot fault:

'Basic theorem of field extensions: let F be a field and a(x) a non constant polynomial in F[x]. There exists an extension field E of F and an element c in E such that c is a root of a(x).'

I interpreted the question as being entirely about the second root.

Not all fields are subfields of the complex numbers. For example, any finite field is not. The only polynomials which have $n$ complex roots are degree $n$ polynomials whose coefficients lie in some subfield of the complex numbers. This is not "every" polynomial.

You are not asked in your problem to prove $p(x)$ has two roots, or that if a field contains one root, it contains them both. You are asked to show a SPECIFIC field contains both roots, namely:

$F[x]/\langle p(x)\rangle$.

In other words, you are being asked to explicitly display a polynomial of the form:

$c_0 + c_1x$

whose coset in $F[x]/\langle p(x)\rangle$, $c_0 + c_1x + \langle p(x)\rangle$ is a solution to $p$.

Once you have done this, a second root is easy to find.

I repeat for emphasis, that without further information about the field $F$, you cannot blithely assume that $F$ is a subfield of the complex numbers, and invoke the Fundamental Theorem of Algebra to produce a root $a$.

To underscore the difficulty involved, suppose that $F = \Bbb Z/2\Bbb Z$ (this is a field with just two elements), and that $p(x) = x^2 + x + 1$ (this is an irreducible quadratic polynomial in that field). One cannot apply the "quadratic formula" to this polynomial, since division by $2$ (which is equal to $0$) is not allowed. Furthermore, one cannot even speak of the "greater" of two roots, since there is no possible order on this (rather small) field consistent with the field operations.

Nevertheless, it is possible to find two roots in $F[x]/\langle x^2 + x + 1\rangle$, which is a field with four elements.

 

[Kiwi1]

OK I think I have finally solved it. Thanks so much for your help.

Let F be any field.
p(x) can be made monic without loss of generality.
let \(p(x) = x^2+bx+c\)
Let \(J=<p(x)>\)

Now:
\(x+J \in F[x]/<p(x)>\)
\(p(x+J)=(x+J)^2+(b+J)(x+J)+(c+J)\)
\(\therefore p(x+J)=x^2+bx+c+J=p(x)+J\)
but \(p(x) \in J\) so:
\[p(x+J)=J\]
and x+J is one of the required roots of p(x) in F[x]/J.

Now let \(\overline x\) be any member of F[x]/J which is necessarily of the form:

\(\overline x = a_i+a_jx+J\) where \(a_i,a_j \in F\)

then \(p(\overline x)=a \overline x^2+b \overline x + c +J=(\overline x -x)(\overline x -r_2)+J=\overline x^2+(-r_2-x)\overline x+xr_2+J\)

therefore \(-r_2-x=b\)
\[\therefore r_2=-b-x\]

Check:
\[p(-b-x+J)=(-b-x)^2+b(-b-x)+c+J=b^2+2bx+x^2-b^2-bx+c+J=p(x)+J=J\]

The required roots are x+J and -b-x+J.

Some examples:
The roots of p(x) in \(\mathbb{Z}_5/<x^2+2x+4>\) are x and -2-x

The roots of \(x^2-2\) are equal in \(F(\sqrt{2})\) but they are not in F[x]/J. Instead they are x and -x.

The roots of \(x^2+1\) are complex conjugate in F(i) but they are not in F[x]/J. Instead they are x and -x

The roots of p(x) in \(\mathbb{Z}_2/<x^2+x+1>\) are x and x+1.

I speculate that it might not be possible to get repeated roots in F[x]/<p(x)> when p(x) has degree 2.

 

[Deveno]

OK I think I have finally solved it. Thanks so much for your help.

Let F be any field.
p(x) can be made monic without loss of generality.
let \(p(x) = x^2+bx+c\)
Let \(J=<p(x)>\)

Now:
\(x+J \in F[x]/<p(x)>\)
\(p(x+J)=(x+J)^2+(b+J)(x+J)+(c+J)\)
\(\therefore p(x+J)=x^2+bx+c+J=p(x)+J\)
but \(p(x) \in J\) so:
\[p(x+J)=J\]
and x+J is one of the required roots of p(x) in F[x]/J.

Now let \(\overline x\) be any member of F[x]/J which is necessarily of the form:

\(\overline x = a_i+a_jx+J\) where \(a_i,a_j \in F\)

then \(p(\overline x)=a \overline x^2+b \overline x + c +J=(\overline x -x)(\overline x -r_2)+J=\overline x^2+(-r_2-x)\overline x+xr_2+J\)

therefore \(-r_2-x=b\)
\[\therefore r_2=-b-x\]

Check:
\[p(-b-x+J)=(-b-x)^2+b(-b-x)+c+J=b^2+2bx+x^2-b^2-bx+c+J=p(x)+J=J\]

The required roots are x+J and -b-x+J.

Some examples:
The roots of p(x) in \(\mathbb{Z}_5/<x^2+2x+4>\) are x and -2-x

The roots of \(x^2-2\) are equal in \(F(\sqrt{2})\) but they are not in F[x]/J. Instead they are x and -x.

The roots of \(x^2+1\) are complex conjugate in F(i) but they are not in F[x]/J. Instead they are x and -x

The roots of p(x) in \(\mathbb{Z}_2/<x^2+x+1>\) are x and x+1.

I speculate that it might not be possible to get repeated roots in F[x]/<p(x)> when p(x) has degree 2.

Yes! The coset of $x$ is the root we were looking for, after which finding the second root is pretty easy.

You would be mistaken, however, to speculate that an irreducible quadratic cannot have repeated roots. In a field of characteristic $0$, this is indeed true, but there are many kinds of fields (some of which may seem unusual to you).

As proof of my bold statement above, I offer the following example:

Let $F= \Bbb Z_2(x) = \left\{\dfrac{p(x)}{q(x)}: p,q \in \Bbb Z_2[x], q \neq 0\right\}$ (this is the field of fractions of $\Bbb Z_2[x]$, or "rational functions in $x$ with coefficients in $\Bbb Z_2$").

Consider the following polynomial in the indeterminate $t$:

$t^2 - x$.

It can be shown that this is irreducible over $F$ (try proving this!), but given a root $a = \sqrt{x}$ in some extension of $F$, we have:

$t^2 - x = (t - a)^2$, that is, the quadratic has only one root (this is because $F$ has characteristic $2$).