# Thread: Questions regarding field extensions:

1. ## Questions regarding field extensions:

Hi everyone!

So I have a question about field extensions. This theorem is proving that if we have a polynomial field F[x] with an irreducible polynomial p(x) in it, then there is always another field that contains an isomorphic copy of F in which p(x) has a root. They essentially show that the quotient K=F[x]/(p(x)) works as the extension of F in which p(x) has a root. But I'm confused-- isn't a quotient field typically smaller than the field it came from? Because the thing it's "divided by" reduces everything by that, right? If it's smaller, how can it be a field extension? I know I'm missing something stupid, but I'd love it if someone could help me.

Another question: They introduce the concept of the minimal polynomial in F[x] for an element "a" (a member of the extension field of F, K) which is algebraic over F[x], saying that it is a unique monic irreducible polynomial in F[x] which has a as a root. My question is: how do you necessarily find this polynomial? I mean, for sqrt(2) over Q, it's pretty clear that it's x^2-2, but I don't see how you'd know for a more complicated element. For example, one question from the book asks to find the minimal polynomial for (1+i) over Q. I think I found that it is x^4 + 4 (because (i+1) plugged into that is a root, and that polynomial is definitely irreducible in Q), but I don't really know how I got that. I just kind of squared (1+i), then noticed if I squared it again I got -4.

Ahhh, I swear I had more questions that I've forgotten... Anyway, thanks in advance! I really need some things cleared up and my professor isn't very helpful.

2. Originally Posted by declan
Hi everyone!

So I have a question about field extensions. This theorem is proving that if we have a polynomial field F[x] with an irreducible polynomial p(x) in it, then there is always another field that contains an isomorphic copy of F in which p(x) has a root. They essentially show that the quotient K=F[x]/(p(x)) works as the extension of F in which p(x) has a root. But I'm confused-- isn't a quotient field typically smaller than the field it came from? Because the thing it's "divided by" reduces everything by that, right? If it's smaller, how can it be a field extension? I know I'm missing something stupid, but I'd love it if someone could help me.
It is smaller than F[x]. This does not mean that it is smaller than F.

Originally Posted by declan
Another question: They introduce the concept of the minimal polynomial in F[x] for an element "a" (a member of the extension field of F, K) which is algebraic over F[x], saying that it is a unique monic irreducible polynomial in F[x] which has a as a root. My question is: how do you necessarily find this polynomial? I mean, for sqrt(2) over Q, it's pretty clear that it's x^2-2, but I don't see how you'd know for a more complicated element. For example, one question from the book asks to find the minimal polynomial for (1+i) over Q. I think I found that it is x^4 + 4 (because (i+1) plugged into that is a root, and that polynomial is definitely irreducible in Q), but I don't really know how I got that. I just kind of squared (1+i), then noticed if I squared it again I got -4.
$\displaystyle 1+i$ is an element of $\displaystyle \mathbb{Q}[i]$, which, considered as a vector space over $\displaystyle \mathbb{Q}$ has basis 1, i. (This means that $\displaystyle [\mathbb{Q}[i]:\mathbb{Q}]=2$; you should know from your lecture that every element of this extension has a minimal polynomial of degree at most 2.)

Let $\displaystyle a=1+i$. We can compute:
$\displaystyle a^0=1$,
$\displaystyle a^1=1+i$,
$\displaystyle a^2=2i$.
So you have 3 vectors in 2-dimensional space, hence there exists a linear combination which zeroes them. This is a question from linear algebra, you should be able to find coefficients a,b,c such that a(1,0)+b(1,1)+c(0,2)=0.
By solving this you get that all such coefficients are multiples of $\displaystyle (2,-2,1)$, hence the minimal polynomial is $\displaystyle x^2-2x+2$.

A different example: minimal polynomial of $\displaystyle u=\sqrt2+\sqrt3$. This belongs to $\displaystyle [\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]$ and basis for $\displaystyle \mathbb{Q}(\sqrt2,\sqrt3)$ is $\displaystyle 1,\sqrt2,\sqrt3,\sqrt6$. (You might try to prove this from the theory you've already learned.) Now:
$\displaystyle (\sqrt2+\sqrt3)^0=1$
$\displaystyle (\sqrt2+\sqrt3)^1=\sqrt2+\sqrt3$
$\displaystyle (\sqrt2+\sqrt3)^2=5+2\sqrt6$
$\displaystyle (\sqrt2+\sqrt3)^3=11\sqrt2+9\sqrt3$
$\displaystyle (\sqrt2+\sqrt3)^4=49+20\sqrt6$

So you just need to find the coefficients by which you obtain zero as a linear combination of the vectors $\displaystyle (1,0,0,0)$, $\displaystyle (0,1,1,0)$, $\displaystyle (5,0,0,2)$, $\displaystyle (0,11,9,0)$, $\displaystyle (49,0,0,20)$.

You should finally arrive to the minimal polynomial $\displaystyle x^4-10x^2+1$.

3. Originally Posted by kompik
It is smaller than F[x]. This does not mean that it is smaller than F.

Ahhhhh! Of course...thank you so much.

$\displaystyle 1+i$ is an element of $\displaystyle \mathbb{Q}[i]$, which, considered as a vector space over $\displaystyle \mathbb{Q}$ has basis 1, i. (This means that $\displaystyle [\mathbb{Q}[i]:\mathbb{Q}]=2$; you should know from your lecture that every element of this extension has a minimal polynomial of degree at most 2.)

Let $\displaystyle a=1+i$. We can compute:
$\displaystyle a^0=1$,
$\displaystyle a^1=1+i$,
$\displaystyle a^2=2i$.
So you have 3 vectors in 2-dimensional space, hence there exists a linear combination which zeroes them. This is a question from linear algebra, you should be able to find coefficients a,b,c such that a(1,0)+b(1,1)+c(0,2)=0.
By solving this you get that all such coefficients are multiples of $\displaystyle (2,-2,1)$, hence the minimal polynomial is $\displaystyle x^2-2x+2$.
Hmmmm, ok. The way you went around this makes so much sense to me.

A different example: minimal polynomial of $\displaystyle u=\sqrt2+\sqrt3$. This belongs to $\displaystyle [\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]$ and basis for $\displaystyle \mathbb{Q}(\sqrt2,\sqrt3)$ is $\displaystyle 1,\sqrt2,\sqrt3,\sqrt6$. (You might try to prove this from the theory you've already learned.) Now:
$\displaystyle (\sqrt2+\sqrt3)^0=1$
$\displaystyle (\sqrt2+\sqrt3)^1=\sqrt2+\sqrt3$
$\displaystyle (\sqrt2+\sqrt3)^2=5+2\sqrt6$
$\displaystyle (\sqrt2+\sqrt3)^3=11\sqrt2+9\sqrt3$
$\displaystyle (\sqrt2+\sqrt3)^4=49+20\sqrt6$

So you just need to find the coefficients by which you obtain zero as a linear combination of the vectors $\displaystyle (1,0,0,0)$, $\displaystyle (0,1,1,0)$, $\displaystyle (5,0,0,2)$, $\displaystyle (0,11,9,0)$, $\displaystyle (49,0,0,20)$.

You should finally arrive to the minimal polynomial $\displaystyle x^4-10x^2+1$.
Ok, I understand almost all of that. One question though: For those 5 basis vectors, it seems like the last one is a linear combination of the first and the third. I thought they were supposed to be linearly independent?

Thanks!

4. Originally Posted by declan
Ok, I understand almost all of that. One question though: For those 5 basis vectors, it seems like the last one is a linear combination of the first and the third. I thought they were supposed to be linearly independent?

Thanks!
That's exactly the point.
The five vectors are not basis vectors. They have to be linearly dependent - which helps us to find non-zero coefficients (and they give us then the polynomial.)
The basis vectors are, e.g., (1,0,0,0),...,(0,0,0,1).