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

#### Peter

##### Well-known member
MHB Site Helper
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]

#### caffeinemachine

##### Well-known member
MHB Math Scholar
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
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
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.