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 Algebra Notes
04-06-2015, 07:24 AM
Post: #11
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
2 Field Extensions

2.1 Definition A field extension is a homomorphism $i: F\to L$ between fields. Write $F\hookrightarrow L$

2.2 Propersition If $\phi: F\to R$ is a ring homomorphism and $F$ is a field, then $\phi$ is a monomorphism.
$\qquad$ ($L$'s subfield $\phi(F)$ is isomorphic to $F$. We often write $i$ for $\phi$ and identify $F$ with $\phi(F)\subset L$.)

2.3 Example (1) $i:\mathbb{Q}\to \mathbb{R},\; i:\mathbb{R}\to \mathbb{C},\;i:\mathbb{Q}\to \mathbb{C}.\quad$(2) $i:\mathbb{Q}\to \mathbb{Q}+\sqrt{3}\mathbb{Q}\subset \mathbb{R}$.
$\qquad$(3) The composition of natural homomorphisms $F\to F[t]\to F(t)$ is an extension $F\hookrightarrow F(t)$

2.4 Definition Let $F\hookrightarrow L\supset A$. Then $F(A) = \bigcap \{K: (K\cup A) \subset K\hookrightarrow L\}$ is called the field by
$\qquad$adjoining $A$ to $F$. If $A = \{\alpha_1,\ldots,\alpha_n\}$, wirte $F(\alpha_1,\ldots,\alpha_n)$ for $F(A)$.

2.5 Examples (1) $(A\subset F\hookrightarrow L) \implies (F(A) = F)$
$\qquad$(2) $\mathbb{Q}(2+\sqrt{3}, 2-\sqrt{3}) = \{a +b\sqrt{3}\mid a,b \in\mathbb{Q}\}= \mathbb{Q}(\sqrt{3})\hookrightarrow\mathbb{R}$.
$\qquad$(3) $\mathbb{R}\hookrightarrow \mathbb{C} = \mathbb{R}(i) = \{a + bi \mid a,b \in\mathbb{R}\}.$
$\qquad$(4) $\{f/g \mid f,g\in F[t],\; g\ne 0\} = \bigcap \{K \mid F\cup \{t\}\subset K,\;K\text{ is a field}\}$. So it's fine for
$\qquad\quad\; F(t)$ denoting both the field of rational functions over $F$ and the extend field of $F$ with $\{t\}$.

2.6 Definition The extension $F\hookrightarrow L$ is simple if $L = F(\alpha)$ for some $\alpha\in L$.

Note that $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2} +\sqrt{3})$ hence simple extension may not appears as simple.

2.7 Definition Let $L\hookleftarrow F,\;\alpha\in L$. Then $\alpha$ is algebraic over $F$ if $\exists f\in F[t] - F\, (f(\alpha) = 0)$.
$\qquad$Otherwise it is transcendental over $F$.

2.8 Example (1) $f = t^2 - 4t + 1\in\mathbb{Q}[t],\; f(2+\sqrt{3}) = 0$ hence $2+\sqrt{3}\in\mathbb{R}$ is algebraic over $\mathbb{Q}.$
$\qquad$(2) $e,\;\pi$ are transcendental over $\mathbb{Q}.\quad$ (3) $t\in K(t)$ is transcendental over $K$.

2.9 Theorem If $\alpha\in L\hookleftarrow F$ is algebraic over $F$, then minimal polynomial of $\alpha$ over $F$ exists:
$\qquad\qquad\exists! m\in F[t]\;(m \text{ is monic})\wedge (m(\alpha) = 0)\wedge (\forall g\in F[t]\,(g(\alpha) = 0\implies m\mid g))$

2.10 Example Let $\alpha = \sqrt{2}+\sqrt{3} \in\mathbb{R} -\mathbb{Q}$. Since $\alpha^2 = 5+2\sqrt{6},\; (\alpha -5)^2 = 24$, we have
$\qquad f(\alpha) = 0\;(f = t^4 -10t^2 + 1\in\mathbb{Q}[t])$. By Th.2.9, the minimal polynomial of $\alpha$ over $\mathbb{Q}$ divides $f$.
$\qquad$If $\alpha\in\mathbb{Q}$, then $f$'d have a root over $\mathbb{Z}$ hence $t^4 - t^2 +1$ would have a root over $\mathbb{Z}_3$: not true.
$\qquad$If $f$ has quadratic factor in $\mathbb{Q}[t]$, then $(\sqrt{2}+\sqrt{3})^2 + a(\sqrt{2}+\sqrt{3}) + b = 0$ for some $a,b\in\mathbb{Q}$.
$\qquad$Put $c = b+5$, we get $2\sqrt{6} +a\sqrt{2} = -a\sqrt{3}+c \implies c^2+a^2-24 = 2a(4-c)\sqrt{3}\implies$
$\qquad ((a = 0)\vee (b = 4))\wedge (c^2+a^2-24 = 0)$, impossible either. So $f$ is minimal of $\alpha$ over $\mathbb{Q}.$

2.11 Theorem If $f\in F[t]$ is monic and irreducible, then $f$ is minimal for some $\alpha$ over $F(\hookrightarrow F(\alpha))\underset{\;}{\;}$
Proof. $I = \{g\in F[t]: f\mid g\}$ is an ideal in $F[t]$. If $I \ne g + I\in F[t]/I$ i.e. $g\not\in I$, then $f\nmid g$ hence
$\qquad$the irreducibility of $f$ implies that $1 = af + bg$ for some $a,b\in F[t]$ thus $b + I$ is the inverse
$\qquad$of $g + I$ in $F[t]/I$. Therefore $F[T]/I$ is a field.
$\qquad$The composition of nature inclusion $F\to F[t]$ and the nature quotient map $F[t]\to F[t]/I$ is
$\qquad$a field extension $F\hookrightarrow F[t]/I$. Identifying $F$ with $F + I\subset F[t]/I$ we get $F[t]/I = F(\alpha)$ with
$\qquad \alpha = t + I$, and $f(\alpha) = f(t) + I = 0$. As $f$ is irreducible over $F$, this means that $f$ is minimal
$\qquad$for $\alpha$ over $F$.

2.12 Definition Two extensions $i:F\to L,\; i': F'\to L'$ are isomorphic if there are isomorphisms
$\qquad \phi: L\xrightarrow{\sim} L',\; \psi: F\xrightarrow{\sim} F'$ such that $\phi i = i'\psi$

2.13 Remark Here is a complete description of simple extension $i:F\to F(\alpha).$
$\qquad$Consider the evaluation homomorphism $\phi: F[t]\to F(\alpha)\underset{\,}{.}$ We have two cases
$\qquad\bullet\;\,\alpha$ is transcendental$\qquad$In this case $\phi$ is a monomorphism hence by Proposition 1.3.2,
$\qquad\quad \phi$ extends to a monomorphism $\tilde{\phi}: F(t)\to F(\alpha),\; \tilde{\phi}|_F = \text{id}_F,\,\tilde{\phi}(t) = \alpha$ thus an isomophism.
$\qquad\bullet\;\,\alpha$ is algebraic$\qquad$By Th.2.11, the ideal $I = \ker(\phi) = \{f\in F[t]: m\mid f\}$ ($m$ is the minimal
$\qquad\quad$polynomial of $\alpha$ over $F$), the induced homomorphism $\bar{\phi}: F[t]/I\to F(\alpha)\underset{\,}{\;}$is an isomorphism.
$\qquad$In the algebraic case, every element of $F(\alpha)$ is expressible as $f(\alpha)$ for some $f\in F[t]$.
$\qquad$Of course $f$ can be replaced with its remainder upon division by $m$. Clearly elements of
$\qquad F(\alpha)$ have their 'standard form' in this case.

2.14 Proposition If $\alpha$ is algebraic for the extension $F(\alpha)\hookleftarrow F,\;$with $m$ the minimal polynomial,
$\qquad$then $\quad \forall \gamma\in F(\alpha)\,\exists! f\in F[t]\;((\gamma = f(\alpha))\wedge (\deg(f) <\deg(m)))$

2.15 Theorem If $F(\alpha)\hookleftarrow F\hookrightarrow F(\beta)$ and $\alpha,\beta$ have the same minimal polynomial $m$ over $F$, then
$\qquad$the extensions are isomorphic, there is an isomorphism $\psi:F(\alpha)\to F(\beta)$ such that $\psi(\alpha) = \beta$.
Proof. Let $I = \{f\in F[t]: m|f\}$ then there are isomorphisms $F(\alpha)\overset{\bar{\phi}}{\leftarrow} F[t]/I \overset{\bar{\psi}}{\to} F(\beta)$ such that
$\qquad \bar{\phi}(t+I) = \alpha,\;\bar{\psi}(t+I) = \beta$, and they restrict to the identity on $F$ thus $\bar{\psi}\bar{\phi}^{-1}$ is the desired
$\qquad$isomorphism of extensions.

2.16 Remark Note that isomorphic simple extensions need not share the same minimal polynomial:
$\qquad F(\alpha) = F(\alpha +1)$ is always true. But $\alpha,\;\alpha + 1$ do not in general share minimal polynomials.

Remark 2.17 $f(t),\;f(t -1)$ are different provided that $F$ has characteristic $0$.

2.18 Theorem Let $i:F\xrightarrow{\sim}F'$ be an isomorphism of fields, $F(\alpha)\hookleftarrow F,\; F'(\beta)\hookleftarrow F'$ be simple
$\qquad$extensions with minimal polynomials $m_{\alpha}\in F[t],\;m_{\beta}\in F'[t]$. Suppose that $i(m_{\alpha}) = m_{\beta}$.
$\qquad$Then there are an isomorphism $\hat{i}: F(\alpha) \xrightarrow{\sim} F'(\beta)$ such that $\hat{i}|_F = i$ and $\hat{i}(\alpha) = \beta$.
04-08-2015, 02:58 PM
Post: #12
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
Notes
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04-08-2015, 02:58 PM
Post: #13
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
Notes
..........
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