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Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 15

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading Nicholson: Introduction to Abstract Algebra, Section 6.2 - Algebraic Extensions.

Example 15 on page 282 (see attachment) reads as follows:

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Example 15.

Let [TEX] E = \mathbb{Q} ( \sqrt{2} , \sqrt{5} ) [/TEX].

Find [TEX] [E \ : \ \mathbb{Q} ] [/TEX] , exhibit a [TEX] \mathbb{Q} [/TEX]-basis of E, and show that [TEX] E = \mathbb{Q} ( \sqrt{2} + \sqrt{5} ) [/TEX]. Then find the minimum polynomial of [TEX] \sqrt{2} + \sqrt{5} [/TEX] over [TEX] \mathbb{Q} [/TEX].

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In the solution we read:

Solution: We write [TEX] L = \mathbb{Q} ( \sqrt{2} ) [/TEX] for convenience so that [TEX] E = L(\sqrt{5}) [/TEX] ... ... etc

... ... ... We claim that [TEX] X^2 - 5 [/TEX] is the minimal polynomial of [TEX] \sqrt{5} [/TEX] over L. Because [TEX] \sqrt{5} [/TEX] and [TEX] - \sqrt{5} [/TEX] are the only roots of [TEX] X^2 - 5 [/TEX] in [TEX] \mathbb{R} [/TEX], we merely need to show that [TEX] \sqrt{5} \notin L [/TEX]. ... ... etc




My problem is the following:

How does showing [TEX] \sqrt{5} \notin L [/TEX] imply that [TEX] X^2 - 5 [/TEX] is the minimal polynomial of [TEX] \sqrt{5} [/TEX] over L?


Can someone please help with this issue?

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Note that in the splitting field of $x^2 - 5$, we have:

$x^2 - 5 = (x - \sqrt{5})(x + \sqrt{5})$

so as soon as we have a field that includes 1 and $\sqrt{5},\ x^2 - 5$ splits.

Now any base field we start extending from contains 1, so the only thing keeping $x^2 - 5$ from splitting is the absence of $\sqrt{5}$.

In general, for any field $F$ and any element $a \in F$:

$x^2 - a$ splits in $F$ iff there exists $u \in F$ with $u^2 = a$.

If there is NO such $u$, (and thus no such $-u$), then $x^2 - a$ has no roots in $F$, and thus, being a quadratic over a field, has no LINEAR factors, so it is irreducible (unit factors don't count).

Polynomial rings over a field are UNIQUE FACTORIZATION DOMAINS. This means if a polynomial $p(x)$ splits one way in its splitting field, this is essentially (up to multiplication by constant factors) the ONLY way it splits. A quadratic (over a field) has at most TWO roots (EXACTLY two if you count roots along with their multiplicities). In other words, we ALWAYS have:

$x^2 + bx + c = (x - r_1)(x - r_2)$.

Now suppose $r_1 \not\in F$. Since:

$b = -r_1 - r_2$, we must have $r_2 \not\in F$ as well, or else we have:

$r_1 = -b - r_2 \in F$, a clear contradiction.

If $\text{char}(F) \neq 2$ (which is the case for any extension of $\Bbb Q$) we can use the quadratic formula to obtain:

$r_1 = \dfrac{-b + \sqrt{b^2 - 4c}}{2}$

$r_2 = \dfrac{-b - \sqrt{b^2 - 4c}}{2}$

If, as is the case here, $b = 0$, this becomes:

$r_1 = \sqrt{-c}$

$r_2 = -\sqrt{-c}$

So the only POSSIBLE factors of $x^2 + c$ are $(x - \sqrt{-c}),(x + \sqrt{-c})$.

Again, a LINEAR polynomial $x - u \in F[x] \iff u \in F$. This is the basic idea of what a "square root of $u$" IS, nothing more or less that a root of $x^2 - u$. The basic idea here is to stop thinking of $\sqrt{u}$ as some "element" in some "number field" (which we would have to define, and then show it is indeed a field) and focus simply on polynomials in $F[x]$ (in particular the polynomial $x^2 - u$), which have properties derived from our "starting field" (which presumably, we already know in some detail).

I suggest you consider the alternative to Nicholson's assertion:

Suppose $x^2 - 5$ splits in $L$. Do we then have $\sqrt{5} \in L$?

Another key point: you need to get used to the idea that if:

$p(a) = 0$ and $m(x)$ is the minimal polynomial of $a$, then $m(x)|p(x)$. The division algorithm is all-powerful in this regard:

We know that:

$p(x) = m(x)q(x) + r(x)$ for UNIQUE polynomials $q(x),r(x)$ with deg(r) < deg(m), or $r = 0$.

For $x = a$ this gives:

$0 = p(a) = m(a)q(a) + r(a) = 0q(a) + r(a) = r(a)$.

If deg(r) < deg(m), then $m(x)$ is NOT the minimal polynomial for $a$, so our only viable choice is $r = 0$, that is:

$p(x) = m(x)q(x)$ whence $m(x)|p(x)$.

Moreover, if $m(x)$ is NOT irreducible, say:

$m(x) = h(x)k(x)$, then:

$0 = m(a) = h(a)k(a)$, so (because fields are integral domains) one of $h(a)$ or $k(a)$ must be 0. Since neither $h$ nor $k$ can be a unit, we again have a clear contradiction to the minimality of $m(x)$.

The concept of a minimal polynomial is very powerful, and is essentially the same idea as "irreducibility":

Every minimal polynomial (of some element of some extension field of the base field) is irreducible.

Every irreducible polynomial is a minimal polynomial (of its roots in some extension field).

The whole idea is this: if we want to work with roots of polynomials, we need (sometimes) a bigger field. Otherwise, "some things (polynomials) don't factor (split)".

In this example, we have a polynomial in $L[x]$. This polynomial has degree 2. Possible outcomes:

The polynomial splits into two linear factors, or,
The polynomial is irreducible.

This is because:

2 = 1 + 1

is the ONLY way to get a sum of 2 with positive integers (the degree of a product of polynomials is the sum of the degrees of the factors, with some contortions required for "zero cases", which is why the 0 polynomial is often assigned a degree of negative infinity).

Sums like:

2 = 0 + 2, or
2 = 0 + 1 + 1 + 0

don't count...0-degree polynomials correspond to unit factors (non-zero elements of the field $F$).
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Note that in the splitting field of $x^2 - 5$, we have:

$x^2 - 5 = (x - \sqrt{5})(x + \sqrt{5})$

so as soon as we have a field that includes 1 and $\sqrt{5},\ x^2 - 5$ splits.

Now any base field we start extending from contains 1, so the only thing keeping $x^2 - 5$ from splitting is the absence of $\sqrt{5}$.

In general, for any field $F$ and any element $a \in F$:

$x^2 - a$ splits in $F$ iff there exists $u \in F$ with $u^2 = a$.

If there is NO such $u$, (and thus no such $-u$), then $x^2 - a$ has no roots in $F$, and thus, being a quadratic over a field, has no LINEAR factors, so it is irreducible (unit factors don't count).

Polynomial rings over a field are UNIQUE FACTORIZATION DOMAINS. This means if a polynomial $p(x)$ splits one way in its splitting field, this is essentially (up to multiplication by constant factors) the ONLY way it splits. A quadratic (over a field) has at most TWO roots (EXACTLY two if you count roots along with their multiplicities). In other words, we ALWAYS have:

$x^2 + bx + c = (x - r_1)(x - r_2)$.

Now suppose $r_1 \not\in F$. Since:

$b = -r_1 - r_2$, we must have $r_2 \not\in F$ as well, or else we have:

$r_1 = -b - r_2 \in F$, a clear contradiction.

If $\text{char}(F) \neq 2$ (which is the case for any extension of $\Bbb Q$) we can use the quadratic formula to obtain:

$r_1 = \dfrac{-b + \sqrt{b^2 - 4c}}{2}$

$r_2 = \dfrac{-b - \sqrt{b^2 - 4c}}{2}$

If, as is the case here, $b = 0$, this becomes:

$r_1 = \sqrt{-c}$

$r_2 = -\sqrt{-c}$

So the only POSSIBLE factors of $x^2 + c$ are $(x - \sqrt{-c}),(x + \sqrt{-c})$.

Again, a LINEAR polynomial $x - u \in F[x] \iff u \in F$. This is the basic idea of what a "square root of $u$" IS, nothing more or less that a root of $x^2 - u$. The basic idea here is to stop thinking of $\sqrt{u}$ as some "element" in some "number field" (which we would have to define, and then show it is indeed a field) and focus simply on polynomials in $F[x]$ (in particular the polynomial $x^2 - u$), which have properties derived from our "starting field" (which presumably, we already know in some detail).

I suggest you consider the alternative to Nicholson's assertion:

Suppose $x^2 - 5$ splits in $L$. Do we then have $\sqrt{5} \in L$?

Another key point: you need to get used to the idea that if:

$p(a) = 0$ and $m(x)$ is the minimal polynomial of $a$, then $m(x)|p(x)$. The division algorithm is all-powerful in this regard:

We know that:

$p(x) = m(x)q(x) + r(x)$ for UNIQUE polynomials $q(x),r(x)$ with deg(r) < deg(m), or $r = 0$.

For $x = a$ this gives:

$0 = p(a) = m(a)q(a) + r(a) = 0q(a) + r(a) = r(a)$.

If deg(r) < deg(m), then $m(x)$ is NOT the minimal polynomial for $a$, so our only viable choice is $r = 0$, that is:

$p(x) = m(x)q(x)$ whence $m(x)|p(x)$.

Moreover, if $m(x)$ is NOT irreducible, say:

$m(x) = h(x)k(x)$, then:

$0 = m(a) = h(a)k(a)$, so (because fields are integral domains) one of $h(a)$ or $k(a)$ must be 0. Since neither $h$ nor $k$ can be a unit, we again have a clear contradiction to the minimality of $m(x)$.

The concept of a minimal polynomial is very powerful, and is essentially the same idea as "irreducibility":

Every minimal polynomial (of some element of some extension field of the base field) is irreducible.

Every irreducible polynomial is a minimal polynomial (of its roots in some extension field).

The whole idea is this: if we want to work with roots of polynomials, we need (sometimes) a bigger field. Otherwise, "some things (polynomials) don't factor (split)".

In this example, we have a polynomial in $L[x]$. This polynomial has degree 2. Possible outcomes:

The polynomial splits into two linear factors, or,
The polynomial is irreducible.

This is because:

2 = 1 + 1

is the ONLY way to get a sum of 2 with positive integers (the degree of a product of polynomials is the sum of the degrees of the factors, with some contortions required for "zero cases", which is why the 0 polynomial is often assigned a degree of negative infinity).

Sums like:

2 = 0 + 2, or
2 = 0 + 1 + 1 + 0

don't count...0-degree polynomials correspond to unit factors (non-zero elements of the field $F$).
Thank you for this really helpful post Deveno. I have been reviewing it carefully and trying to follow it all closely

Just two points of carification.

(1) You write:

"\(\displaystyle x^2 - 5 = (x - \sqrt{5})(x + \sqrt{5}) \)

so as soon as we have a field that includes 1 and \(\displaystyle \sqrt{5},\ x^2 - 5 \) splits."
------------------------------------------------------------------------------

I am not sure why you mention the field containing 1 in the context of \(\displaystyle x^2 - 5 \) - of course the field contains 1 but why is this important in this context.

I was expecting you to write: " ... ... so as soon as we have a field that includes \(\displaystyle \sqrt{5},\ x^2 - 5 \) splits.". Why you included 1 in this statement puzzles me ... can you please clarify?

--------------------------------------------------------------------------------

(2) Later in the post you write:

"Moreover, if m(x) is NOT irreducible, say:

\(\displaystyle m(x) = h(x)k(x) \) , then:

\(\displaystyle 0 = m(a) = h(a)k(a) \) , so (because fields are integral domains) one of h(a) or k(a) must be 0. Since neither h nor k can be a unit, we again have a clear contradiction to the minimality of m(x)."

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I do not follow the argument: "Since neither h nor k can be a unit, we again have a clear contradiction to the minimality of m(x)"

Can you be more explicit - why can't they be units and what is the consequence of not being able to be units?

Can you clarify?

Once again, thank you for a most helpful post.

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
The field generated by 1 is the so-called *prime field* which every field extension is an extension OF. This is either $\Bbb Q$ or $\Bbb Z_p$, for some prime $p$.

Field theory uses a fair amount of linear algebra. Linear algebra requires an underlying field. Typically, the base field is a prime field, and the element 1 is the basis element which generates $F$ as a subfield (subspace) of the vector space (field extension) $K$ over $F$.

The DEFINITION of reducible (for polynomials) is that:

$f(x) = g(x)h(x)$ where neither $g$ nor $h$ is a unit.

By extension, the definition of IRREDUCIBLE is that if:

$f(x) = g(x)h(x)$, either $g$ or $h$ is a unit. This is basic ring theory.

In a unique factorization domain (such as the integers, or a polynomial ring over a field), the factorization is unique "only up to units", because we can always add unit pairs like so:

$f(x) = uu^{-1}g(x)h(x) = (ug(x))(u^{-1}h(x)) = \dots$

In the integers, for example, we have the following "distinct" factorizations for 6:

6 = 2*3
6 = (-2)*(-3)
6 = 1*2*(-1)*(-3)

but these are all "essentially the same", we get a unit times a multiple of 2, and a unit times a multiple of 3 (the unit 1 is "invisible").

The point with a minimal polynomial is, if it FACTORED (into non-units) the factors would have lower degree. But the minimal polynomial has BY DEFINITION the lowest degree possible. Therefore, if a minimal polynomial factors at all, one factor must be a unit (and the other has the same degree as the minimal polynomial itself), which is the definition of irreducible.