1.3.6 Algebraic extensions
Let \(K/F\) be a field extension and let \(\alpha \in K\). Then we say \(\alpha \) is algebraic over \(F\) if there exists a polynomial \(f(x) \in F[x]\) such that \(f(\alpha )=0\). Otherwise we say \(\alpha \) is transcendental.
\(\sqrt{2}\) is algebraic over \(\mathbb {Q}\).
\(\sqrt{-1}\) is algebraic over \(\mathbb {R}\).
Prove that if \(\alpha \in K\) is algebraic over \(F\), then it is also algebraic over any field extension of \(F\).
Let \(\alpha \) be algebraic over \(F\). Then there is a unique monic irreducible polynomial \(m_{\alpha ,F}(x)\) in \(F[x]\) which has \(\alpha \) as a root. Moreover, \(m_{\alpha ,F}(x)\) divides any other polynomial in \(F[x]\) with \(\alpha \) as a root. This polynomial is called the minimal polynomial of \(\alpha \) over \(F\).
Lets take \(m_{\alpha ,F}(x)\) to be a polynomial of minimal degree having \(\alpha \) as a root. Moreover, since we are working over a field, we can multiply by some element in \(F\) to make sure its monic. This is our candidate for the minimal polynomial. We need to show its unique and irreducible.
If \(m_{\alpha ,F}(x)\) were reducible, then we would have
so \(f(\alpha )g(\alpha )=0\). Therefore, since we are in a field, this means either \(f(\alpha )=0\) or \(g(\alpha )=0\). But \(m_{\alpha ,F}(x)\) has minimal degree, so one of \(f\) or \(g\) must have degree \(0\) (i.e. a constant.) thus \(m_{\alpha ,F}(x)\) is irreducible.
So we just need to check its unique. Note that if \(f(x)\) is any polynomial with \(\alpha \) as a root, then by polynomial long division we have
with \(\deg (r(x)){\lt} \deg (m_{\alpha ,F}(x))\). But then evaluating this at \(\alpha \) we would have \(r(\alpha )=0\), but which can’t happen as it has smaller degree than \(m_{\alpha ,F}(x)\), so \(r(x)=0\). Therefore, \(m_{\alpha ,F}(x)\) divides any polynomial with \(\alpha \) as a root. From this it follows that if we had two minimal polynomials \(m_1(x),m_2(x)\) then \(m_1(x)|m_2(x)\) and \(m_2(x)|m_1(x)\) so \(m_1(x)=a m_2(x)\) for some \(a \in F\), but now being monic comes to the rescue to say that \(a=1\), giving uniqueness and finishing the proof.
Let \(L/F\) be a field extension and let \(\alpha \) be algebraic over \(F\), then prove that \(m_{\alpha ,L}(x)\) divides \(m_{\alpha ,F}(x)\) in \(L[x]\).
Now, from Theorem 1.3.4, Proposition 1.3.2 and Proposition 1.3.10 we have the following:
Let \(K/F\) be a field extension and \(\alpha \in K\) algebraic over \(F\). Then
and moreover \([F(\alpha ):F]=\deg (m_{\alpha ,F}(x))\).
Let \(K/F\) be a field extension. Then we say \(K\) is algebraic over \(F\) if every element of \(K\) is algebraic over \(F\).
\(\mathbb {Q}(\sqrt{2})\) is algebraic over \(\mathbb {Q}\).
\(\mathbb {C}\) is algebraic over \(\mathbb {R}\).
Let \(K/F\) be a field extension. Then \(\alpha \in K\) is algebriac over \(F\) if and only if there is a finite extension of \(F\) (inside \(K\)) containing \(\alpha \).
If \(\alpha \) is algebraic over \(F\), then we know that \(F(\alpha )\) is a field isomorphic to \(F[x]/(m_{\alpha ,K}(x))\). Moreover, \([F(\alpha ):F]=\deg (m_{\alpha ,F}(x))\) by Corollary 1.3.12 and by definition of \(F(\alpha )\) is is the the smallest such field, so \(F(\alpha )\) is contained in \(K\).
Conversely, if \(\alpha \) is in a finite extension \(L/F\) of degree \(n\). Then consider \(1,\alpha ,\alpha ^2,\dots ,\alpha ^n\). These \(n+1\) elements must be linearly dependent over \(F\), so there exist \(b_i\) such that
which proves that \(\alpha \) is algebraic over \(F\) since its a root of \(b_0+b_1x+\cdots +b_n x^n\).
If \(K/F\) is a finite extension, then is it algebraic.
Is the converse of this statement true? In other words, if \(K/F\) is an algebraic extension, then does it have to be finite?
We say a field extension \(K/F\) is finitely generated, if there are finitely many elements \(\alpha _1,\dots ,\alpha _n\) such that \(K=F(\alpha _1,\dots ,\alpha _n)\).
Now, here are a couple of facts you might remember from Galois theory
Let \(K/F\) be a field extension, \(\alpha _i\) be elements of \(K\) which are algebraic over \(F\) and let \(F_i=F(\alpha _1,\dots ,\alpha _i)\).
\(F_n=F_{n-1}(\alpha _n)\) and in general \(F_i=F_{i-1}(\alpha _i)\).
\([F_n:F]=[F_n:F_{n-1}][F_{n-1}:F_{n-2}]\dots [F_1:F]\) (note \(F_0=F\))
Proposition 1.3.19 and Corollary 1.3.16 gives us:
A field extension \(K/F\) is finite if and only if \(K\) is generated by a finite number of algebraic elements over \(F\).
Let \(\alpha ,\beta \in K\) be non-zero and algebraic over \(F\). Then \(\alpha \pm \beta \), \(\alpha \beta \), \(\alpha /\beta \), \(\alpha ^{-1}\), \(\beta ^{-1}\),etc are also algebraic over \(F\).