Algebraic Extensions: Dummit & Foote Section 13.2, Example 2 pg 526 - Help!

[Math Amateur]

I am reading Dummit and Foote on algebraic extensions. I am having some issues understanding Example 2 on page 526 - see attachment.

Example 2 on page 526 reads as follows:

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(2) Consider the field $$\mathbb{Q} ( \sqrt{2}, \sqrt{3} )$$ generated over $$\mathbb{Q}$$ by $$\sqrt{2}$$ and $$\sqrt{3}$$.

Since $$\sqrt{3}$$ is of degree 2 over $$\mathbb{Q}$$ the degree of the extension $$\mathbb{Q} ( \sqrt{2}, \sqrt{3} )$$ is at most 2 and is precisely 2 if and only if $$x^2 - 3$$ is irreducible over $$\mathbb{Q} ( \sqrt{2} )$$. ... ... etc etc

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My question is: why exactly does it follow that the degree of the extension $$\mathbb{Q} ( \sqrt{2}, \sqrt{3} )$$ is at most 2 and is precisely 2 if and only if $$x^2 - 3$$ is irreducible over $$\mathbb{Q} ( \sqrt{2} )$$?

Although I may be being pedantic I also have a concern about why exactly $$\sqrt{3}$$ is of degree 2 over $$\mathbb{Q}$$. I know it is intuitively the case or it seems the case that the minimal polynomial is in this case $$x^2 - 3$$ but how do we demonstrate this for sure - or is it obvious? (I may be overthinking this??)

Can someone help with the above issues/problems?

Peter

[Note: This has also been posted on MHF]

 

[caffeinemachine]

Her Peter. This is related to your previous post http://mathhelpboards.com/linear-abstract-algebra-14/field-theory-algebrais-extensions-dummit-foote-section-13-2-a-6694.html

I am reading Dummit and Foote on algebraic extensions. I am having some issues understanding Example 2 on page 526 - see attachment.

Example 2 on page 526 reads as follows:

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

(2) Consider the field $$\mathbb{Q} ( \sqrt{2}, \sqrt{3} )$$ generated over $$\mathbb{Q}$$ by $$\sqrt{2}$$ and $$\sqrt{3}$$.

Since $$\sqrt{3}$$ is of degree 2 over $$\mathbb{Q}$$ the degree of the extension $$\mathbb{Q} ( \sqrt{2}, \sqrt{3} )$$ is at most 2 and is precisely 2 if and only if $$x^2 - 3$$ is irreducible over $$\mathbb{Q} ( \sqrt{2} )$$. ... ... etc etc

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

My question is: why exactly does it follow that the degree of the extension $$\mathbb{Q} ( \sqrt{2}, \sqrt{3} )$$ is at most 2 and is precisely 2 if and only if $$x^2 - 3$$ is irreducible over $$\mathbb{Q} ( \sqrt{2} )$$?

One thing you should always make sure is that whenever you talk about degree of the extension field $K$, you should also mention the field over which you are viewing $K$ as an extension. Here you say that the degree if the extension $\mathbb Q(\sqrt 2,\sqrt 3)$ is is at most two... You didn't mention over which field are you viewing $\mathbb Q(\sqrt 2,\sqrt 3)$ as an extension field.

What we can show is that the degree of $\mathbb Q(\sqrt 2,\sqrt 3)$ as an extension of $\mathbb Q(\sqrt 2)$ is at most $2$. Say $F=\mathbb Q(\sqrt 2)$ and $\alpha =\sqrt 3$. It is easy to see that $\alpha$ is algebraic over $F$ simply because $\alpha$ satisfies $x^2-3$ which is a polynomial over $F$. This also shows that $\deg m_\alpha$, where $m_\alpha$ is the minimal polynomial of $\alpha$ over $F$, is at most $2$. Thus $F(\alpha)\cong F[x]/\langle m_\alpha(x)\rangle$ is a vector space over $F$ with dimension $\deg m_\alpha$. This is same as saying that $[F(\alpha):F]=\deg m_\alpha$. But $\deg m_\alpha\leq 2$ and hence $[F(\alpha):F]\leq 2$. This is same as $[\mathbb Q(\sqrt 2, \sqrt 3):\mathbb Q(\sqrt 2)]\leq 2$.

 

[Math Amateur]

Her Peter. This is related to your previous post http://mathhelpboards.com/linear-abstract-algebra-14/field-theory-algebrais-extensions-dummit-foote-section-13-2-a-6694.htmlOne thing you should always make sure is that whenever you talk about degree of the extension field $K$, you should also mention the field over which you are viewing $K$ as an extension. Here you say that the degree if the extension $\mathbb Q(\sqrt 2,\sqrt 3)$ is is at most two... You didn't mention over which field are you viewing $\mathbb Q(\sqrt 2,\sqrt 3)$ as an extension field.

What we can show is that the degree of $\mathbb Q(\sqrt 2,\sqrt 3)$ as an extension of $\mathbb Q(\sqrt 2)$ is at most $2$. Say $F=\mathbb Q(\sqrt 2)$ and $\alpha =\sqrt 3$. It is easy to see that $\alpha$ is algebraic over $F$ simply because $\alpha$ satisfies $x^2-3$ which is a polynomial over $F$. This also shows that $\deg m_\alpha$, where $m_\alpha$ is the minimal polynomial of $\alpha$ over $F$, is at most $2$. Thus $F(\alpha)\cong F[x]/\langle m_\alpha(x)\rangle$ is a vector space over $F$ with dimension $\deg m_\alpha$. This is same as saying that $[F(\alpha):F]=\deg m_\alpha$. But $\deg m_\alpha\leq 2$ and hence $[F(\alpha):F]\leq 2$. This is same as $[\mathbb Q(\sqrt 2, \sqrt 3):\mathbb Q(\sqrt 2)]\leq 2$.

Thanks Caffeinemachine.

So (to repeat your post a bit to emphasize a couple of points) to show that the degree of $$ \mathbb Q(\sqrt 2,\sqrt 3) $$ as an extension of $$ \mathbb Q(\sqrt 2) $$ is at most 2, we take $$ F = \mathbb Q(\sqrt 2) $$ and $$ \alpha =\sqrt 3 $$.Then we argue that $$ \alpha =\sqrt 3 $$ is algebraic over $$ F = \mathbb Q(\sqrt 2) $$ because it satisfies the polynomial $$ x^2 - 3 $$.

$$ x^2 - 3 $$ is a polynomial over $$ F = \mathbb Q(\sqrt 2) $$ because its coefficients (i.e. 1 and 3) are both in $$ \mathbb Q(\sqrt 2) $$

that is $$ \ 1 = 1 + 0 \star \sqrt 2$$

and $$ 3 = 3 + 0 \star \sqrt 2 $$

Then, given the above we have $$ [ \mathbb Q(\sqrt 2,\sqrt 3) \ : \ \mathbb Q(\sqrt 2)] \le 2 $$ ... ... ... ... ... (1)

Now, similarly we can show that $$ [ \mathbb Q(\sqrt 2) \ : \ \mathbb Q] \le 2 $$ ... ... ... ... ... (2)

and the equality is actually the case in both (1) and (2) above because the polynomials concerned are irreducible (is that correct?)

So we have the following

$$ [ \mathbb Q(\sqrt 2,\sqrt 3) \ : \ \mathbb Q(\sqrt 2)] = 2 $$

and

$$ [ \mathbb Q(\sqrt 2) \ : \ \mathbb Q] = 2 $$

So

$$ [\mathbb Q(\sqrt 2,\sqrt 3) \ : \ \mathbb Q] = [ \mathbb Q(\sqrt 2,\sqrt 3) \ : \ \mathbb Q(\sqrt 2)] \ [ \mathbb Q(\sqrt 2) \ : \ \mathbb Q] = 4 $$

Can someone confirm that the above reasoning is correct ... or not?

Peter

 

[caffeinemachine]

Thanks Caffeinemachine.

So (to repeat your post a bit to emphasize a couple of points) to show that the degree of $$ \mathbb Q(\sqrt 2,\sqrt 3) $$ as an extension of $$ \mathbb Q(\sqrt 2) $$ is at most 2, we take $$ F = \mathbb Q(\sqrt 2) $$ and $$ \alpha =\sqrt 3 $$.Then we argue that $$ \alpha =\sqrt 3 $$ is algebraic over $$ F = \mathbb Q(\sqrt 2) $$ because it satisfies the polynomial $$ x^2 - 3 $$.

$$ x^2 - 3 $$ is a polynomial over $$ F = \mathbb Q(\sqrt 2) $$ because its coefficients (i.e. 1 and 3) are both in $$ \mathbb Q(\sqrt 2) $$

that is $$ \ 1 = 1 + 0 \star \sqrt 2$$

and $$ 3 = 3 + 0 \star \sqrt 2 $$

Then, given the above we have $$ [ \mathbb Q(\sqrt 2,\sqrt 3) \ : \ \mathbb Q(\sqrt 2)] \le 2 $$ ... ... ... ... ... (1)

Now, similarly we can show that $$ [ \mathbb Q(\sqrt 2) \ : \ \mathbb Q] \le 2 $$ ... ... ... ... ... (2)

and the equality is actually the case in both (1) and (2) above because the polynomials concerned are irreducible (is that correct?)

So we have the following

$$ [ \mathbb Q(\sqrt 2,\sqrt 3) \ : \ \mathbb Q(\sqrt 2)] = 2 $$

and

$$ [ \mathbb Q(\sqrt 2) \ : \ \mathbb Q] = 2 $$

So

$$ [\mathbb Q(\sqrt 2,\sqrt 3) \ : \ \mathbb Q] = [ \mathbb Q(\sqrt 2,\sqrt 3) \ : \ \mathbb Q(\sqrt 2)] \ [ \mathbb Q(\sqrt 2) \ : \ \mathbb Q] = 4 $$

Can someone confirm that the above reasoning is correct ... or not?

Peter

Yes. This is correct. Only thing, I think you need to fill in some details as to why the polynomials in question were irreducibles. Otherwise it's fantastic.