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Field Theory - Algebraic Extensions - Dummit and Foote Section 13.2 - Example 2 - page 526

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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 [TEX] \mathbb{Q} ( \sqrt{2}, \sqrt{3} )[/TEX] generated over [TEX] \mathbb{Q} [/TEX] by [TEX] \sqrt{2} [/TEX] and [TEX] \sqrt{3} [/TEX].

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

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

Although I may be being pedantic I also have a concern about why exactly [TEX] \sqrt{3} [/TEX] is of degree 2 over [TEX] \mathbb{Q} [/TEX]. I know it is intuitively the case or it seems the case that the minimal polynomial is in this case [TEX] x^2 - 3 [/TEX] 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

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Her Peter. This is related to your previous post http://mathhelpboards.com/linear-ab...ensions-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 [TEX] \mathbb{Q} ( \sqrt{2}, \sqrt{3} )[/TEX] generated over [TEX] \mathbb{Q} [/TEX] by [TEX] \sqrt{2} [/TEX] and [TEX] \sqrt{3} [/TEX].

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

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

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

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Her Peter. This is related to your previous post http://mathhelpboards.com/linear-ab...ensions-dummit-foote-section-13-2-a-6694.html


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 $[\mathbb Q(\sqrt 2, \sqrt 3):\mathbb Q(\sqrt 2)]\leq 2$.
Thanks Caffeinemachine.

So (to repeat your post a bit to emphasize a couple of points) to show that the degree of \(\displaystyle \mathbb Q(\sqrt 2,\sqrt 3) \) as an extension of \(\displaystyle \mathbb Q(\sqrt 2) \) is at most 2, we take \(\displaystyle F = \mathbb Q(\sqrt 2) \) and \(\displaystyle \alpha =\sqrt 3 \).


Then we argue that \(\displaystyle \alpha =\sqrt 3 \) is algebraic over \(\displaystyle F = \mathbb Q(\sqrt 2) \) because it satisfies the polynomial \(\displaystyle x^2 - 3 \).

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

that is \(\displaystyle \ 1 = 1 + 0 \star \sqrt 2\)

and \(\displaystyle 3 = 3 + 0 \star \sqrt 2 \)


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

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

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

and

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

So

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

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

Peter
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Thanks Caffeinemachine.

So (to repeat your post a bit to emphasize a couple of points) to show that the degree of \(\displaystyle \mathbb Q(\sqrt 2,\sqrt 3) \) as an extension of \(\displaystyle \mathbb Q(\sqrt 2) \) is at most 2, we take \(\displaystyle F = \mathbb Q(\sqrt 2) \) and \(\displaystyle \alpha =\sqrt 3 \).


Then we argue that \(\displaystyle \alpha =\sqrt 3 \) is algebraic over \(\displaystyle F = \mathbb Q(\sqrt 2) \) because it satisfies the polynomial \(\displaystyle x^2 - 3 \).

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

that is \(\displaystyle \ 1 = 1 + 0 \star \sqrt 2\)

and \(\displaystyle 3 = 3 + 0 \star \sqrt 2 \)


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

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

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

and

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

So

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

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.