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Field Theory - Algebraic Extensions - D&F - Section 13.2 - Exercise 7, page 530

Peter

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MHB Site Helper
Jun 22, 2012
2,918
------------------------------------------------------------------------------------------------

7. Prove that [TEX] \mathbb{Q} ( \sqrt{2} + \sqrt{3} ) = \mathbb{Q} ( \sqrt{2} , \sqrt{3} )[/TEX].

Conclude that [TEX] [\mathbb{Q} ( \sqrt{2} + \sqrt{3} ) \ : \ \mathbb{Q} ] = 4 [/TEX].

Find an irreducible polynomial satisfied by [TEX] \sqrt{2} + \sqrt{3} [/TEX]

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I am somewhat overwhelmed by this problem ... can someone advise me on an approach ... and, indeed, get me started?

Peter

[This has also been posted on MHF]
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
First of all it should be clear that $\Bbb Q(\sqrt{2}+\sqrt{3}) \subseteq \Bbb Q(\sqrt{2},\sqrt{3})$, because $\sqrt{2}+\sqrt{3} \in \Bbb Q(\sqrt{2},\sqrt{3})$

So we want to show the other containment:

$Q(\sqrt{2},\sqrt{3}) \subseteq \Bbb Q(\sqrt{2}+\sqrt{3})$

I claim it is sufficient to show that $\sqrt{2} \in \Bbb Q(\sqrt{2}+\sqrt{3})$, because then we can realize $\sqrt{3}$ as:

$(\sqrt{2}+\sqrt{3}) - \sqrt{2}$.

So let's play with $\sqrt{2}+\sqrt{3}$ a bit:

Note that:

$(\sqrt{2}+\sqrt{3})^3 = 2\sqrt{2} + 6\sqrt{3} + 9\sqrt{2} + 3\sqrt{3} = 11\sqrt{2} + 9\sqrt{3}$.

Thus:

$(\sqrt{2}+\sqrt{3})^3 - 9(\sqrt{2}+\sqrt{3}) = 2\sqrt{2}$, that is:

$\sqrt{2} = \frac{1}{2}((\sqrt{2}+\sqrt{3})^3 - 9(\sqrt{2}+\sqrt{3}) \in \Bbb Q(\sqrt{2}+\sqrt{3})$, and we are done with the first part.

Let's let $u = \sqrt{2} + \sqrt{3}$ and look at the first few powers of $u$:

$u^2 = 5 + 2\sqrt{6}$
$u^3 = 11\sqrt{2} + 9\sqrt{3}$
$u^4 = 49 + 20\sqrt{6}$

It appears that:

$u^4 - 10u^2 + 1 = 0$. Conclude that $x^4 - 10x^2 + 1$ is irreducible (why???) over $\Bbb Q$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
First of all it should be clear that $\Bbb Q(\sqrt{2}+\sqrt{3}) \subseteq \Bbb Q(\sqrt{2},\sqrt{3})$, because $\sqrt{2}+\sqrt{3} \in \Bbb Q(\sqrt{2},\sqrt{3})$

So we want to show the other containment:

$Q(\sqrt{2},\sqrt{3}) \subseteq \Bbb Q(\sqrt{2}+\sqrt{3})$

I claim it is sufficient to show that $\sqrt{2} \in \Bbb Q(\sqrt{2}+\sqrt{3})$, because then we can realize $\sqrt{3}$ as:

$(\sqrt{2}+\sqrt{3}) - \sqrt{2}$.

So let's play with $\sqrt{2}+\sqrt{3}$ a bit:

Note that:

$(\sqrt{2}+\sqrt{3})^3 = 2\sqrt{2} + 6\sqrt{3} + 9\sqrt{2} + 3\sqrt{3} = 11\sqrt{2} + 9\sqrt{3}$.

Thus:

$(\sqrt{2}+\sqrt{3})^3 - 9(\sqrt{2}+\sqrt{3}) = 2\sqrt{2}$, that is:

$\sqrt{2} = \frac{1}{2}((\sqrt{2}+\sqrt{3})^3 - 9(\sqrt{2}+\sqrt{3}) \in \Bbb Q(\sqrt{2}+\sqrt{3})$, and we are done with the first part.

Let's let $u = \sqrt{2} + \sqrt{3}$ and look at the first few powers of $u$:

$u^2 = 5 + 2\sqrt{6}$
$u^3 = 11\sqrt{2} + 9\sqrt{3}$
$u^4 = 49 + 20\sqrt{6}$

It appears that:

$u^4 - 10u^2 + 1 = 0$. Conclude that $x^4 - 10x^2 + 1$ is irreducible (why???) over $\Bbb Q$.

Need to show \(\displaystyle p(x) = x^4 - 10x^2 + 1 \) is irreducible.

Consider \(\displaystyle p(x) = x^4 - 10x^2 + 1 = a_4x^4 + a_3x^3 + a_2 x^2 + a_1x + a_0 \)

P(x) is a monic polynomial with \(\displaystyle a_4 = 1, a_3 = 0, a_2 = -10, a_1 = 0 \) and \(\displaystyle a_0 = 1\)

The Rational Roots Theorem (Proposition 11, Dummit and Foote Section 9.4. page 308 states the following:
------------------------------------------------------------------------------------

Proposition 11. Let \(\displaystyle p(x) = a_n x^n + a_{n-1}x^{n-1} + ... ... + a_0 \) be a polynomial of degree n with integer coefficients.

If \(\displaystyle r/s \in \mathbb{Q} \) is in lowest terms (i.e. r and s are relatively prime integers) ans r/s is a root of p(x), then r divides the constant term and s divides the leading coefficient of \(\displaystyle p(x): r \ | \ a_0 \) and \(\displaystyle s \ | \ a_n \).

In particular, if p(x) is a monic polynomial with integer coefficients and \(\displaystyle p(d) \ne 0 \) for all integers d dividing the constant term of p(x), then p(x) has no roots in \(\displaystyle \mathbb{Q} \)

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

In our case P(x) is monic and \(\displaystyle a_0 = 1 \). Integers dividing \(\displaystyle a_0 \) are therefore +1 and - 1 and neither of these give p(x) = 0. Thus we conclude that P(x) is irreducible.

Since the minimal polynomial of the field \(\displaystyle \mathbb{Q}(\sqrt{2} + \sqrt{3}) \) generated over \(\displaystyle \mathbb{Q} \) is of degree 4, then

\(\displaystyle [\mathbb{Q}(\sqrt{2} + \sqrt{3}) \ : \ \mathbb{Q}] = 4 \)

Peter
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Well, yeah...but you did the problem backwards.

By proving: $\Bbb Q(\sqrt{2} + \sqrt{3}) = \Bbb Q(\sqrt{2},\sqrt{3}) = (\Bbb Q(\sqrt{2}))(\sqrt{3})$

we know that $\Bbb Q(\sqrt{2} + \sqrt{3})$ is a degree 2 extension of a degree 2 extension of the rational field. Hence:

$[\Bbb Q(\sqrt{2}+\sqrt{3}:\Bbb Q] = [(\Bbb Q(\sqrt{2}))(\sqrt{3}):\Bbb Q]$

$= [(\Bbb Q(\sqrt{2}))(\sqrt{3}):\Bbb Q(\sqrt{2})]\ast[\Bbb Q(\sqrt{2}):\Bbb Q] = 2 \ast 2 = 4$

Thus $x^4 - 10x^2 + 1$, being a monic polynomial of degree 4 that $\sqrt{2} + \sqrt{3}$ satisfies MUST BE the minimal polynomial of $\sqrt{2} + \sqrt{3}$ and minimal polynomials are irreducible. Always. For if a minimal polynomial FACTORED over $\Bbb Q$, it wouldn't be MINIMAL ($\sqrt{2} + \sqrt{3}$ would have to be a root of one of the factors, which are of lesser degree).

So once we have done the first part of the problem, there is NO NEED to prove irreducibility of the quartic. Once we find a monic quartic that $\sqrt{2} + \sqrt{3}$ satisfies, that's it. We're done.

Watch, it's magic:

We can use the formula

$u^2 = 5 + 2\sqrt{6}$ to also derive this polynomial:

$(u^2 - 5)^2 = 24$

$u^4 - 10u^2 + 25 = 24$

$u^4 - 10u^2 + 1 = 0$

Another way to show $x^4 - 10x^2 + 1$ is irreducible:

It's discriminant $b^2 - 4c$ = $100 - 4 = 96$ is not a perfect square. Thus $u^2 \not\in \Bbb Q$. Hence $u \not\in \Bbb Q$, for if $u$ were rational, $u^2$ would certainly be, since fields are closed under multiplication.

It bears looking into why it suffices to show the discriminant is not a perfect square.

Note if we have $x^2 + bx + c = 0 \in \Bbb Z[x]$, then:

$x = \dfrac{-b \pm \sqrt{b^2 - 4c}}{2}$ in some extension of the field of quotients of the integral domain $\Bbb Z$.

If $x = u$ is some root of this polynomial with $u \in \Bbb Q$, then certainly:

$2u + b \in \Bbb Q$.

Similarly $(2u + b)^2 \in \Bbb Q$. But this is an integer, since it equals $b^2 - 4c$.

Since we are assuming $(2u + b)^2 \in \Bbb Q$, we have an equation:

$\dfrac{p^2}{q^2} = m$, where $p,q,m \in \Bbb Z$ with gcd(p,q) = 1.

We can write this as:

$p^2 = q^2m$, which shows that $q^2 | p^2$, hence $q | p$. Since gcd(p,q) = 1, we must have $q = 1$, which show that $m$ is a perfect square, QED.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Well, yeah...but you did the problem backwards.

By proving: $\Bbb Q(\sqrt{2} + \sqrt{3}) = \Bbb Q(\sqrt{2},\sqrt{3}) = (\Bbb Q(\sqrt{2}))(\sqrt{3})$

we know that $\Bbb Q(\sqrt{2} + \sqrt{3})$ is a degree 2 extension of a degree 2 extension of the rational field. Hence:

$[\Bbb Q(\sqrt{2}+\sqrt{3}:\Bbb Q] = [(\Bbb Q(\sqrt{2}))(\sqrt{3}):\Bbb Q]$

$= [(\Bbb Q(\sqrt{2}))(\sqrt{3}):\Bbb Q(\sqrt{2})]\ast[\Bbb Q(\sqrt{2}):\Bbb Q] = 2 \ast 2 = 4$

Thus $x^4 - 10x^2 + 1$, being a monic polynomial of degree 4 that $\sqrt{2} + \sqrt{3}$ satisfies MUST BE the minimal polynomial of $\sqrt{2} + \sqrt{3}$ and minimal polynomials are irreducible. Always. For if a minimal polynomial FACTORED over $\Bbb Q$, it wouldn't be MINIMAL ($\sqrt{2} + \sqrt{3}$ would have to be a root of one of the factors, which are of lesser degree).

So once we have done the first part of the problem, there is NO NEED to prove irreducibility of the quartic. Once we find a monic quartic that $\sqrt{2} + \sqrt{3}$ satisfies, that's it. We're done.

Watch, it's magic:

We can use the formula

$u^2 = 5 + 2\sqrt{6}$ to also derive this polynomial:

$(u^2 - 5)^2 = 24$

$u^4 - 10u^2 + 25 = 24$

$u^4 - 10u^2 + 1 = 0$

Another way to show $x^4 - 10x^2 + 1$ is irreducible:

It's discriminant $b^2 - 4c$ = $100 - 4 = 96$ is not a perfect square. Thus $u^2 \not\in \Bbb Q$. Hence $u \not\in \Bbb Q$, for if $u$ were rational, $u^2$ would certainly be, since fields are closed under multiplication.

It bears looking into why it suffices to show the discriminant is not a perfect square.

Note if we have $x^2 + bx + c = 0 \in \Bbb Z[x]$, then:

$x = \dfrac{-b \pm \sqrt{b^2 - 4c}}{2}$ in some extension of the field of quotients of the integral domain $\Bbb Z$.

If $x = u$ is some root of this polynomial with $u \in \Bbb Q$, then certainly:

$2u + b \in \Bbb Q$.

Similarly $(2u + b)^2 \in \Bbb Q$. But this is an integer, since it equals $b^2 - 4c$.

Since we are assuming $(2u + b)^2 \in \Bbb Q$, we have an equation:

$\dfrac{p^2}{q^2} = m$, where $p,q,m \in \Bbb Z$ with gcd(p,q) = 1.

We can write this as:

$p^2 = q^2m$, which shows that $q^2 | p^2$, hence $q | p$. Since gcd(p,q) = 1, we must have $q = 1$, which show that $m$ is a perfect square, QED.
Thanks Deveno ... very helpful post.

Just one minor point:

You write:

"By proving: [FONT=MathJax_AMS]Q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_AMS]Q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_AMS]Q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main])[/FONT]

we know that [FONT=MathJax_AMS]Q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main])[/FONT] is a degree 2 extension of a degree 2 extension of the rational field. Hence: ... ... ... "

How (exactly) do we know that [FONT=MathJax_AMS]Q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main])[/FONT] is a degree 2 extension of a degree 2 extension of the rational field.What theorems are we depending on - or are we just taking this result from previous exercises?

Thanks again for your help and guidance!

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Well, one theorem this appeals to is:

$F(a,b) = F(a)(b)$

that is, we can take a field formed from a finite number of adjunctions, and form it "one adjunction at a time".

By definition, the field $F(a,b)$ is the smallest field containing the set $\{a,b\}$ and $F$. Clearly, the field $F(a)(b)$ contains $F$, as it contains $F(a)$, which contains $F$. $F(a)(b)$ also contains $a$, since it contains $F(a)$ and $a \in F(a)$. Finally, $F(a)(b)$ contains $b$, as well, so we clearly have:

$F(a,b) \subseteq F(a)(b)$.

Now $F(a)(b)$ is the smallest field containing $F(a)$, and $b$. But $F(a,b)$ certainly contains $F(a)$, since it contains $F$ and $a$, and $F(a)$ is the smallest field containing $F$ and $a$. Just as clearly, $b \in F(a)(b)$, so we see that:

$F(a)(b) \subseteq F(a,b)$.

Another result this appeals to is:

If $E$ is a finite extension of $F$, and $K$ is a finite extension of $E$, then $K$ is a finite extension of $F$, with:

$[K:F] = [K:E] \ast [E:F]$.

To see this, let $\{u_1,\dots,u_m\}$ be a basis of $E$ over $F$, and let $\{v_1,\dots,v_n\}$ be a basis of $K$ over $E$.

then, for any element $\alpha \in K$, we have:

$\alpha = c_1v_1 + \cdots + c_nv_n$ for some $c_1,\dots,c_n \in E$.

Setting $c_j = b_{1j}u_1 + \cdots + b_{mj}u_m$ for some $b_{ij} \in F$ (for each $j$), which we can do because the $u_i$ form a basis for $E$ over $F$, we get:

$\alpha = b_{11}u_1v_1 + \cdots + b_{m1}u_mv_1 +\cdots + b_{1n}u_1v_n + \cdots + b_{mn}u_mv_n$

which shows that the set $\{u_iv_j: 1 \leq i \leq m, 1 \leq j \leq n\}$ spans $K$ over $F$.

If we set this sum equal to 0, we have (by the linear independence of the $v_j$ over $E$), that in $E$, we have:

$b_{1j}u_1 + \cdots + b_{mj}u_m = 0$ for EACH $j$, and then by the linear independence of the $u_i$ over $F$ we get $b_{ij} = 0$ for each $i$ (and every $j$), which proves linear independence, so we have a basis with $mn$ elements, so:

$[K:F] = mn = nm = [K:E] \ast [E:F]$, since $[K:E] = n$ and $[E:F] = m$.

These results are BASIC (small pun, there) to understanding algebraic extensions of a field (usually we are interested in FINITE algebraic extensions, which are generated by a finite set of elements each of which is algebraic over $F$).

And, why, you might ask, are we interested in such field extensions? So we can solve polynomials, of course! For example, finding a basis of a finite-dimensional vector space over $F$ for which a given $F$-linear transformation is diagonal (which GREATLY simplifies matrix calculations, if it is even possible, which is not always the case), or at least "as diagonal as possible", involves solving a certain polynomial: the characteristic polynomial of that linear transformation. For example, the real matrix:

$\begin{bmatrix}0&-1\\1&0 \end{bmatrix}$

which represents a 90 degree rotation of the plane, has characteristic polynomial $x^2 + 1$, which has no real solutions...so the proper study of such matrices is in the enlarged field of the complex numbers.

It turns out that polynomials shed a great deal of light on the qualitative behavior of linear transformations, but to effectively use this, we need to find the roots of these polynomials, and to do THAT we need to make sure our field is large enough to contain these roots (since irreducible polynomials of degree > 1 exist over some fields).

Moreover, polynomials make great models for more complicated functions, and are very well-behaved. Often we have some "target value" we are concerned about (for a polynomial that measures manufacturing cost this may be the maximum amount we are prepared to spend in said manufacture), which can easily be turned into a question about roots.

As another example, many things in "the real world" are governed by differential equations of order < 3, which in turn can be modelled (approximated) by LINEAR differential equations of order < 3, which in turn involves solving a quadratic to obtain certain parameters. This is the case, for example, with projectile motion (neglecting friction) affected by gravity.

The central theory you are building up to, in your study of field extensions, is a certain correspondence between fields and groups that is rather mind-blowing in and of itself: the Galois correspondence. It turns out that certain "classical" problems, such as solving the general quintic, or squaring the circle, can be shown to be unsolvable using these ideas. It is a bit surprising that groups (as minimally structured as they are) have some insight to offer into the internal structure of fields (which have structure that encompasses most of the math we use up into secondary school...which includes MOST of the math MOST people ever use).
 
Last edited:

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Well, one theorem this appeals to is:

$F(a,b) = F(a)(b)$

that is, we can take a field formed from a finite number of adjunctions, and form it "one adjunction at a time".

By definition, the field $F(a,b)$ is the smallest field containing the set $\{a,b\}$ and $F$. Clearly, the field $F(a)(b)$ contains $F$, as it contains $F(a)$, which contains $F$. $F(a)(b)$ also contains $a$, since it contains $F(a)$ and $a \in F(a)$. Finally, $F(a)(b)$ contains $b$, as well, so we clearly have:

$F(a,b) \subseteq F(a)(b)$.

Now $F(a)(b)$ is the smallest field containing $F(a)$, and $b$. But $F(a,b)$ certainly contains $F(a)$, since it contains $F$ and $a$, and $F(a)$ is the smallest field containing $F$ and $a$. Just as clearly, $b \in F(a)(b)$, so we see that:

$F(a)(b) \subseteq F(a,b)$.

Another result this appeals to is:

If $E$ is a finite extension of $F$, and $K$ is a finite extension of $E$, then $K$ is a finite extension of $F$, with:

$[K:F] = [K:E] \ast [E:F]$.

To see this, let $\{u_1,\dots,u_m\}$ be a basis of $E$ over $F$, and let $\{v_1,\dots,v_n\}$ be a basis of $K$ over $E$.

then, for any element $\alpha \in K$, we have:

$\alpha = c_1v_1 + \cdots + c_nv_n$ for some $c_1,\dots,c_n \in E$.

Setting $c_j = b_{1j}u_1 + \cdots + b_{mj}u_m$ for some $b_{ij} \in F$ (for each $j$), which we can do because the $u_i$ form a basis for $E$ over $F$, we get:

$\alpha = b_{11}u_1v_1 + \cdots + b_{m1}u_mv_1 +\cdots + b_{1n}u_1v_n + \cdots + b_{mn}u_mv_n$

which shows that the set $\{b_{ij}: 1 \leq i \leq m, 1 \leq j \leq n\}$ spans $K$ over $F$.

If we set this sum equal to 0, we have (by the linear independence of the $v_j$ over $E$), that in $E$, we have:

$b_{1j}u_1 + \cdots + b_{mj}u_m = 0$ for EACH $j$, and then by the linear independence of the $u_i$ over $F$ we get $b_{ij} = 0$ for each $i$ (and every $j$), which proves linear independence, so we have a basis with $mn$ elements, so:

$[K:F] = mn = nm = [K:E] \ast [E:F]$, since $[K:E] = n$ and $[E:F] = m$.

These results are BASIC (small pun, there) to understanding algebraic extensions of a field (usually we are interested in FINITE algebraic extensions, which are generated by a finite set of elements each of which is algebraic over $F$).

And, why, you might ask, are we interested in such field extensions? So we can solve polynomials, of course! For example, finding a basis of a finite-dimensional vector space over $F$ for which a given $F$-linear transformation is diagonal (which GREATLY simplifies matrix calculations, if it is even possible, which is not always the case), or at least "as diagonal as possible", involves solving a certain polynomial: the characteristic polynomial of that linear transformation. For example, the real matrix:

$\begin{bmatrix}0&-1\\1&0 \end{bmatrix}$

which represents a 90 degree rotation of the plane, has characteristic polynomial $x^2 + 1$, which has no real solutions...so the proper study of such matrices is in the enlarged field of the complex numbers.

It turns out that polynomials shed a great deal of light on the qualitative behavior of linear transformations, but to effectively use this, we need to find the roots of these polynomials, and to do THAT we need to make sure our field is large enough to contain these roots (since irreducible polynomials of degree > 1 exist over some fields).

Moreover, polynomials make great models for more complicated functions, and are very well-behaved. Often we have some "target value" we are concerned about (for a polynomial that measures manufacturing cost this may be the maximum amount we are prepared to spend in said manufacture), which can easily be turned into a question about roots.

As another example, many things in "the real world" are governed by differential equations of order < 3, which in turn can be modelled (approximated) by LINEAR differential equations of order < 3, which in turn involves solving a quadratic to obtain certain parameters. This is the case, for example, with projectile motion (neglecting friction) affected by gravity.

The central theory you are building up to, in your study of field extensions, is a certain correspondence between fields and groups that is rather mind-blowing in and of itself: the Galois correspondence. It turns out that certain "classical" problems, such as solving the general quintic, or squaring the circle, can be shown to be unsolvable using these ideas. It is a bit surprising that groups (as minimally structured as they are) have some insight to offer into the internal structure of fields (which have structure that encompasses most of the math we use up into secondary school...which includes MOST of the math MOST people ever use).
That helped me very considerably Deveno, thank you

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Well, yeah...but you did the problem backwards.

By proving: $\Bbb Q(\sqrt{2} + \sqrt{3}) = \Bbb Q(\sqrt{2},\sqrt{3}) = (\Bbb Q(\sqrt{2}))(\sqrt{3})$

we know that $\Bbb Q(\sqrt{2} + \sqrt{3})$ is a degree 2 extension of a degree 2 extension of the rational field. Hence:

$[\Bbb Q(\sqrt{2}+\sqrt{3}:\Bbb Q] = [(\Bbb Q(\sqrt{2}))(\sqrt{3}):\Bbb Q]$

$= [(\Bbb Q(\sqrt{2}))(\sqrt{3}):\Bbb Q(\sqrt{2})]\ast[\Bbb Q(\sqrt{2}):\Bbb Q] = 2 \ast 2 = 4$

Thus $x^4 - 10x^2 + 1$, being a monic polynomial of degree 4 that $\sqrt{2} + \sqrt{3}$ satisfies MUST BE the minimal polynomial of $\sqrt{2} + \sqrt{3}$ and minimal polynomials are irreducible. Always. For if a minimal polynomial FACTORED over $\Bbb Q$, it wouldn't be MINIMAL ($\sqrt{2} + \sqrt{3}$ would have to be a root of one of the factors, which are of lesser degree).

So once we have done the first part of the problem, there is NO NEED to prove irreducibility of the quartic. Once we find a monic quartic that $\sqrt{2} + \sqrt{3}$ satisfies, that's it. We're done.

Watch, it's magic:

We can use the formula

$u^2 = 5 + 2\sqrt{6}$ to also derive this polynomial:

$(u^2 - 5)^2 = 24$

$u^4 - 10u^2 + 25 = 24$

$u^4 - 10u^2 + 1 = 0$

Another way to show $x^4 - 10x^2 + 1$ is irreducible:

It's discriminant $b^2 - 4c$ = $100 - 4 = 96$ is not a perfect square. Thus $u^2 \not\in \Bbb Q$. Hence $u \not\in \Bbb Q$, for if $u$ were rational, $u^2$ would certainly be, since fields are closed under multiplication.

It bears looking into why it suffices to show the discriminant is not a perfect square.

Note if we have $x^2 + bx + c = 0 \in \Bbb Z[x]$, then:

$x = \dfrac{-b \pm \sqrt{b^2 - 4c}}{2}$ in some extension of the field of quotients of the integral domain $\Bbb Z$.

If $x = u$ is some root of this polynomial with $u \in \Bbb Q$, then certainly:

$2u + b \in \Bbb Q$.

Similarly $(2u + b)^2 \in \Bbb Q$. But this is an integer, since it equals $b^2 - 4c$.

Since we are assuming $(2u + b)^2 \in \Bbb Q$, we have an equation:

$\dfrac{p^2}{q^2} = m$, where $p,q,m \in \Bbb Z$ with gcd(p,q) = 1.

We can write this as:

$p^2 = q^2m$, which shows that $q^2 | p^2$, hence $q | p$. Since gcd(p,q) = 1, we must have $q = 1$, which show that $m$ is a perfect square, QED.
Deveno,

Just reviewing and reflecting on your post above, I still have an unresolved issue - my apologies if you have actually answered this issue.

You write:

"By proving: \(\displaystyle \Bbb Q(\sqrt{2} + \sqrt{3}) = \Bbb Q(\sqrt{2},\sqrt{3}) = (\Bbb Q(\sqrt{2}))(\sqrt{3}) \)

we know that \(\displaystyle \Bbb Q(\sqrt{2} + \sqrt{3})\) is a degree 2 extension of a degree 2 extension of the rational field. Hence:

\(\displaystyle [\Bbb Q(\sqrt{2}+\sqrt{3}:\Bbb Q] = [(\Bbb Q(\sqrt{2}))(\sqrt{3}):\Bbb Q]\)

= \(\displaystyle [(\Bbb Q(\sqrt{2}))(\sqrt{3}):\Bbb Q(\sqrt{2})]\ast[\Bbb Q(\sqrt{2}):\Bbb Q] = 2 \ast 2 = 4 \)"

My problem/issue is as follows:

How do we know, or on what basis did you determine that

\(\displaystyle [(\Bbb Q(\sqrt{2}))(\sqrt{3}):\Bbb Q(\sqrt{2})] = 2 \)

and

\(\displaystyle [\Bbb Q(\sqrt{2}):\Bbb Q] = 2 \)

Are you, for example, just taking the minimal polynomials in each case to be (obviously?) \(\displaystyle x^2 - 3 \) and \(\displaystyle x^2 - 2 \) and so both degree 2.

Would appreciate some help.

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
It should be obvious that:

$[\Bbb Q(\sqrt{2}):\Bbb Q] = 2$.

On the one hand $\sqrt{2}$ is a root of $x^2 - 2$, so we can express any linear combination:

$c_0 + c_1\sqrt{2} + c_2(\sqrt{2})^2 +\cdots+c_n(\sqrt{2})^n$ as:

$a + b\sqrt{2}: a,b \in \Bbb Q$ which suffices to show that as an extension ring of a field (and thus a vector space over $\Bbb Q$) we have that:

$\{1,\sqrt{2}\}$ spans $\Bbb Q[\sqrt{2}]$.

Now clearly $\Bbb Q[\sqrt{2}] \subseteq \Bbb Q(\sqrt{2})$.

To show that $\Bbb Q[\sqrt{2}] = \Bbb Q(\sqrt{2})$, it suffices to show that $\Bbb Q[\sqrt{2}]$ is already a field.

So consider, given $a + b\sqrt{2} \neq 0 \in \Bbb Q[\sqrt{2}]$, the following element $c+d\sqrt{2} \in \Bbb Q[\sqrt{2}]$, given by:

$c = \dfrac{a}{a^2 - 2b^2}$

$d = \dfrac{-b}{a^2 - 2b^2}$

Of course, we need to be sure that $a^2 - 2b^2 \neq 0$, but this is tantamount to saying:

$\left(\dfrac{a}{b}\right)^2 \neq 2$ if $b \neq 0$, which is clearly true for the rational number $\dfrac{a}{b}$ (since $\sqrt{2} \not\in \Bbb Q$), and when $b = 0$ we have $a \neq 0$ so $a^2 - 2b^2 = a^2 \neq 0$.

It is clear that: $(a+b\sqrt{2})(c + d\sqrt{2}) = 1$, so $\Bbb Q[\sqrt{2}]$ is indeed a field.

Since we have seen that $\{1,\sqrt{2}\}$ spans this field, it has dimension at MOST 2.

On the other hand, since $\sqrt{2} \not\in \Bbb Q$, it has dimension > 1, so the only possible dimension is eaxctly 2.

For further emphasis:

Irreducibility of $x^2 - 2$ over $\Bbb Q \iff \sqrt{2} \not\in \Bbb Q \iff$ linear independence over $\Bbb Q$ of the set $\{1,\sqrt{2}\}$.

It remains to be seen that $[\Bbb Q(\sqrt{2})(\sqrt{3}):\Bbb Q(\sqrt{2})] = 2$. While it should be obvious that in fact $x^2 - 3$ is irreducible over $\Bbb Q$, perhaps it might factor over $\Bbb Q(\sqrt{2})$. But let's think about this:

Suppose $(a + b\sqrt{2})^2 = 3$, for some rational $a,b$. Then:

$a^2 + 2b^2 + 2ab\sqrt{2} = 3 = 3 + 0\sqrt{2}$.

Now $\{1,\sqrt{2}\}$ is a BASIS for $\Bbb Q(\sqrt{2})$ (that is, the $a$ and the $b$ are UNIQUE) so we MUST have:

$a^2 + 2b^2 = 3$
$2ab = 0 \implies ab = 0$.

Since $\Bbb Q$ is a field, and thus an integral domain, either $a = 0$, or $b = 0$.

If $a = 0$, then $2b^2 = 3$, but this is impossible for rational $b$ (why?). If $b = 0$ we have $a^2 = 3$, which is likewise impossible. This means $x^2 - 3$ has no root in $\Bbb Q(\sqrt{2})$, so it is irreducible over this field.

**********

The short answer to your question is: yes, that is exactly what I'm doing...but I want you to see (in elementary terms) that this is valid.

One does need to be careful, though. For example, consider this:

$\Bbb Q(\sqrt{2}) \subseteq \Bbb Q(\omega)$ where $\omega$ is a primitive eighth root of 1:

$\sqrt{2} = \omega + \omega^7$

so $x^2 - 2$ is NOT irreducible over $\Bbb Q(\sqrt{i})$.

(This field example is an interesting one, you should get better acquainted with this field).
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
It should be obvious that:

$[\Bbb Q(\sqrt{2}):\Bbb Q] = 2$.

On the one hand $\sqrt{2}$ is a root of $x^2 - 2$, so we can express any linear combination:

$c_0 + c_1\sqrt{2} + c_2(\sqrt{2})^2 +\cdots+c_n(\sqrt{2})^n$ as:

$a + b\sqrt{2}: a,b \in \Bbb Q$ which suffices to show that as an extension ring of a field (and thus a vector space over $\Bbb Q$) we have that:

$\{1,\sqrt{2}\}$ spans $\Bbb Q[\sqrt{2}]$.

Now clearly $\Bbb Q[\sqrt{2}] \subseteq \Bbb Q(\sqrt{2})$.

To show that $\Bbb Q[\sqrt{2}] = \Bbb Q(\sqrt{2})$, it suffices to show that $\Bbb Q[\sqrt{2}]$ is already a field.

So consider, given $a + b\sqrt{2} \neq 0 \in \Bbb Q[\sqrt{2}]$, the following element $c+d\sqrt{2} \in \Bbb Q[\sqrt{2}]$, given by:

$c = \dfrac{a}{a^2 - 2b^2}$

$d = \dfrac{-b}{a^2 - 2b^2}$

Of course, we need to be sure that $a^2 - 2b^2 \neq 0$, but this is tantamount to saying:

$\left(\dfrac{a}{b}\right)^2 \neq 2$ if $b \neq 0$, which is clearly true for the rational number $\dfrac{a}{b}$ (since $\sqrt{2} \not\in \Bbb Q$), and when $b = 0$ we have $a \neq 0$ so $a^2 - 2b^2 = a^2 \neq 0$.

It is clear that: $(a+b\sqrt{2})(c + d\sqrt{2}) = 1$, so $\Bbb Q[\sqrt{2}]$ is indeed a field.

Since we have seen that $\{1,\sqrt{2}\}$ spans this field, it has dimension at MOST 2.

On the other hand, since $\sqrt{2} \not\in \Bbb Q$, it has dimension > 1, so the only possible dimension is eaxctly 2.

For further emphasis:

Irreducibility of $x^2 - 2$ over $\Bbb Q \iff \sqrt{2} \not\in \Bbb Q \iff$ linear independence over $\Bbb Q$ of the set $\{1,\sqrt{2}\}$.

It remains to be seen that $[\Bbb Q(\sqrt{2})(\sqrt{3}):\Bbb Q(\sqrt{2})] = 2$. While it should be obvious that in fact $x^2 - 3$ is irreducible over $\Bbb Q$, perhaps it might factor over $\Bbb Q(\sqrt{2})$. But let's think about this:

Suppose $(a + b\sqrt{2})^2 = 3$, for some rational $a,b$. Then:

$a^2 + 2b^2 + 2ab\sqrt{2} = 3 = 3 + 0\sqrt{2}$.

Now $\{1,\sqrt{2}\}$ is a BASIS for $\Bbb Q(\sqrt{2})$ (that is, the $a$ and the $b$ are UNIQUE) so we MUST have:

$a^2 + 2b^2 = 3$
$2ab = 0 \implies ab = 0$.

Since $\Bbb Q$ is a field, and thus an integral domain, either $a = 0$, or $b = 0$.

If $a = 0$, then $2b^2 = 3$, but this is impossible for rational $b$ (why?). If $b = 0$ we have $a^2 = 3$, which is likewise impossible. This means $x^2 - 3$ has no root in $\Bbb Q(\sqrt{2})$, so it is irreducible over this field.

**********

The short answer to your question is: yes, that is exactly what I'm doing...but I want you to see (in elementary terms) that this is valid.

One does need to be careful, though. For example, consider this:

$\Bbb Q(\sqrt{2}) \subseteq \Bbb Q(\omega)$ where $\omega$ is a primitive eighth root of 1:

$\sqrt{2} = \omega + \omega^7$

so $x^2 - 2$ is NOT irreducible over $\Bbb Q(\sqrt{i})$.

(This field example is an interesting one, you should get better acquainted with this field).

Thank Deveno, this post I extremely helpful to me.

Working through it carefully now.

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
It should be obvious that:

$[\Bbb Q(\sqrt{2}):\Bbb Q] = 2$.

On the one hand $\sqrt{2}$ is a root of $x^2 - 2$, so we can express any linear combination:

$c_0 + c_1\sqrt{2} + c_2(\sqrt{2})^2 +\cdots+c_n(\sqrt{2})^n$ as:

$a + b\sqrt{2}: a,b \in \Bbb Q$ which suffices to show that as an extension ring of a field (and thus a vector space over $\Bbb Q$) we have that:

$\{1,\sqrt{2}\}$ spans $\Bbb Q[\sqrt{2}]$.

Now clearly $\Bbb Q[\sqrt{2}] \subseteq \Bbb Q(\sqrt{2})$.

To show that $\Bbb Q[\sqrt{2}] = \Bbb Q(\sqrt{2})$, it suffices to show that $\Bbb Q[\sqrt{2}]$ is already a field.

So consider, given $a + b\sqrt{2} \neq 0 \in \Bbb Q[\sqrt{2}]$, the following element $c+d\sqrt{2} \in \Bbb Q[\sqrt{2}]$, given by:

$c = \dfrac{a}{a^2 - 2b^2}$

$d = \dfrac{-b}{a^2 - 2b^2}$

Of course, we need to be sure that $a^2 - 2b^2 \neq 0$, but this is tantamount to saying:

$\left(\dfrac{a}{b}\right)^2 \neq 2$ if $b \neq 0$, which is clearly true for the rational number $\dfrac{a}{b}$ (since $\sqrt{2} \not\in \Bbb Q$), and when $b = 0$ we have $a \neq 0$ so $a^2 - 2b^2 = a^2 \neq 0$.

It is clear that: $(a+b\sqrt{2})(c + d\sqrt{2}) = 1$, so $\Bbb Q[\sqrt{2}]$ is indeed a field.

Since we have seen that $\{1,\sqrt{2}\}$ spans this field, it has dimension at MOST 2.

On the other hand, since $\sqrt{2} \not\in \Bbb Q$, it has dimension > 1, so the only possible dimension is eaxctly 2.

For further emphasis:

Irreducibility of $x^2 - 2$ over $\Bbb Q \iff \sqrt{2} \not\in \Bbb Q \iff$ linear independence over $\Bbb Q$ of the set $\{1,\sqrt{2}\}$.

It remains to be seen that $[\Bbb Q(\sqrt{2})(\sqrt{3}):\Bbb Q(\sqrt{2})] = 2$. While it should be obvious that in fact $x^2 - 3$ is irreducible over $\Bbb Q$, perhaps it might factor over $\Bbb Q(\sqrt{2})$. But let's think about this:

Suppose $(a + b\sqrt{2})^2 = 3$, for some rational $a,b$. Then:

$a^2 + 2b^2 + 2ab\sqrt{2} = 3 = 3 + 0\sqrt{2}$.

Now $\{1,\sqrt{2}\}$ is a BASIS for $\Bbb Q(\sqrt{2})$ (that is, the $a$ and the $b$ are UNIQUE) so we MUST have:

$a^2 + 2b^2 = 3$
$2ab = 0 \implies ab = 0$.

Since $\Bbb Q$ is a field, and thus an integral domain, either $a = 0$, or $b = 0$.

If $a = 0$, then $2b^2 = 3$, but this is impossible for rational $b$ (why?). If $b = 0$ we have $a^2 = 3$, which is likewise impossible. This means $x^2 - 3$ has no root in $\Bbb Q(\sqrt{2})$, so it is irreducible over this field.

**********

The short answer to your question is: yes, that is exactly what I'm doing...but I want you to see (in elementary terms) that this is valid.

One does need to be careful, though. For example, consider this:

$\Bbb Q(\sqrt{2}) \subseteq \Bbb Q(\omega)$ where $\omega$ is a primitive eighth root of 1:

$\sqrt{2} = \omega + \omega^7$

so $x^2 - 2$ is NOT irreducible over $\Bbb Q(\sqrt{i})$.

(This field example is an interesting one, you should get better acquainted with this field).
Just working through your post. The post is very clear and helpful but I wish to be very sure regarding one point - forgive me if I am being obsessive! :(

I am not completely clear about your argument when you say

"Now clearly \(\displaystyle \Bbb Q[\sqrt{2}] \subseteq \Bbb Q(\sqrt{2}) \)"

Now I am not quite sure how you are defining \(\displaystyle \Bbb Q[\sqrt{2}] \) but I suspect it is the following definition:

\(\displaystyle \Bbb Q[\sqrt{2}] = \{a + b\sqrt{2} \ | \ a, b \in \mathbb{Q} \} \)

The definition of \(\displaystyle \mathbb{Q} ( \sqrt{2} ) \) is as follows:

\(\displaystyle \mathbb{Q} ( \sqrt{2} ) \) = the smallest subfield of K (where K is an extension field of \(\displaystyle \mathbb{Q} \) ) containing both \(\displaystyle \sqrt{2} \) and \(\displaystyle \mathbb{Q} \)

Now given these definitions you write

"Now clearly \(\displaystyle \Bbb Q[\sqrt{2}] \subseteq \Bbb Q(\sqrt{2}) \)"

But, ... while it seems likely or intuitive that \(\displaystyle \Bbb Q[\sqrt{2}] \subseteq \Bbb Q(\sqrt{2}) \) as you say, how can we be sure? In other words how do we know that the set of all \(\displaystyle a + b\sqrt{2} \) gives the smallest subfield of K containing both \(\displaystyle \sqrt{2} \) and \(\displaystyle \mathbb{Q} \).

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
No, $\Bbb Q[\sqrt{2}]$ is the RING of all finite $\Bbb Q$-linear combinations of powers of $\sqrt{2}$ ("polynomials in $\sqrt{2}$"), the ring created by formal adjunction of $\sqrt{2}$ to the rationals (the smallest ring containing the rationals and $\sqrt{2}$).

As it turns out, this ring is equal to its field of fractions, since it already contains elements of the form $1/f(\sqrt{2})$ for all non-zero polynomials $f(x)$.

If a field contains $F$,and a set $S$, the closure axioms for a field guarantee it will contain ALL $F$-linear combinations of elements of $S$.

So, in general, for a set $S$ where *every element* of $S$ is algebraic over $F$, we have: $F = F(S)$. It is the algebraic part that actually guarantees we get inverses: if $m(x)$ is the minimal polynomial of $s \in S$, say:

$m(x) = c_0 + c_1x +\cdots+c_{n-1}x^{n-1} + x^n$, then:

$c_0 \neq 0$ (or else $x$ would divide $m(x)$ contradicting its minimality), thus:

$0 = m(s) = c_0 + c_1s +\cdots + c_{n-1}s^{n-1} + s^n$

therefore:

$-c_0 = s(c_1 + \cdots + c_{n-1}s^{n-2} + s^{n-1})$

that is:

$s^{-1} = (-1/c_0)(c_1 + \cdots + c_{n-1}s^{n-2} + s^{n-1}) \in F$.