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 [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]

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 [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]?

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 \(\displaystyle \mathbb Q(\sqrt 2,\sqrt 3) \) 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]

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.

 

[blue_raver22]

Dear Peter,

Thank you for reaching out for help with Example 2 on page 526 of Dummit and Foote's algebraic extensions section. I can understand your confusion and concerns regarding the degree of the extension and the minimality of the polynomial.

Firstly, let's address why the degree of the extension is at most 2 and is precisely 2 if and only if x^2 - 3 is irreducible over \mathbb{Q}(\sqrt{2}). This follows from the definition of degree of an extension. The degree of an extension is the degree of the minimal polynomial of the element over the base field. In this case, we have the field \mathbb{Q}(\sqrt{2}, \sqrt{3}) and the element \sqrt{3}. Now, we know that \sqrt{3} is a root of the polynomial x^2 - 3. This polynomial is in \mathbb{Q}(\sqrt{2}) and it is irreducible over \mathbb{Q}(\sqrt{2}) (since it has no roots in \mathbb{Q}(\sqrt{2})). Therefore, the minimal polynomial of \sqrt{3} over \mathbb{Q}(\sqrt{2}) is x^2 - 3. Hence, 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}).

Regarding your concern about why \sqrt{3} is of degree 2 over \mathbb{Q}, it is indeed obvious. We know that \sqrt{3} is a root of the polynomial x^2 - 3 over \mathbb{Q}. This polynomial is irreducible over \mathbb{Q} (since it has no roots in \mathbb{Q}). Therefore, the minimal polynomial of \sqrt{3} over \mathbb{Q} is x^2 - 3, which has degree 2. Hence, \sqrt{3} is of degree 2 over \mathbb{Q}.

I hope this helps clarify your doubts and concerns. If you have any further questions, please do not hesitate to ask.

Best,