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

In summary: Q(\sqrt 2,\sqrt 3)$ is at most 2 and is precisely 2 if and only if $x^2-3$ is irreducible over $\mathbb Q(\sqrt 2)$.
  • #1
Math Amateur
Gold Member
MHB
3,998
48
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]?

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]
 
Physics news on Phys.org
  • #2
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

Peter said:
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 $[\mathbb Q(\sqrt 2, \sqrt 3):\mathbb Q(\sqrt 2)]\leq 2$.
 
  • #3
caffeinemachine said:
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
 
  • #4
Peter said:
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.
 
  • #5


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,
 

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

What are algebraic extensions?

Algebraic extensions are field extensions where every element is algebraic over the base field. In other words, every element in the extension can be expressed as a root of a polynomial with coefficients in the base field.

What is Dummit & Foote Section 13.2?

Dummit & Foote Section 13.2 refers to a specific section in the algebra textbook "Abstract Algebra" by David S. Dummit and Richard M. Foote. This section covers algebraic extensions and their properties.

What is Example 2 on page 526 in Dummit & Foote?

Example 2 on page 526 in Dummit & Foote is a specific example used to illustrate a concept or theorem in the section on algebraic extensions. It may involve solving a polynomial equation or finding the degree of a field extension.

How can I use algebraic extensions in real life?

Algebraic extensions have many applications in fields such as physics, engineering, and computer science. They are used to solve polynomial equations, construct mathematical models, and analyze complex systems.

What are some important theorems related to algebraic extensions?

Some important theorems related to algebraic extensions include the Fundamental Theorem of Galois Theory, the Primitive Element Theorem, and the Artin-Schreier Theorem. These theorems provide important insights and tools for understanding and manipulating algebraic extensions.

Back
Top