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Problem
Divisibility Tests could be great examples to put in here. We could do 3, 9, 11, and then invent a few others.
The problems below are still in progress
If $E$ and $E'$ are two splitting fields for $f(x)$ over $F$, prove that they are isomorphic. So Splitting Fields Are Unique Up To Isomorphism
Describe the extension $Q(\pi)$.
Define carefully zeros, factors, divides, multiple zeros.
If $f(x)$ is irreducible over a field of characteristic zero, prove that $f(x)$ has no multiple zeros. So it splits in a very nice way.
Define perfect field (give the definition by just talking about multiple roots)
If $f(x)$ is irreducible over a field of characteristic $p$, must $f(x)$ be perfect? Prove your answer, or give a counter example.
Show that in a perfect field, we must have $F^p=F$, so every element is a $p$th power.
There are several more equivalent ways to define perfect fields. Perhaps we should prove several of them.
Prove that all zeros of an irreducible $f(x)$ over $F$ have the same multiplicity in a splitting field $E$. So when we split a polynomial, it splits in a very nice way.
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theorem}[20.3 - ]
Let $F$ be a field and $p(x)\in F[x]$ be irreducible over $F$. If $a$ is a zero of $p(x)$ in some extension $E$ of $F$, then $F(a)$ is isomorphic to $F[x]/\left<p(x)\right>$. Furthermore, if $\deg p(x) = n$, then every member of $F(a)$ can be uniquely expressed in the form $c_{n-1}a^{n-1} + c_{n-2}a^{n-2} + \cdots + c_{1}a + c_{0}$ where $c_i\in F$. (In other words, the set $\left\{1, a, a^2, \ldots, a^{n-1}\right\}$ is a basis for $F(a)$ over $F$.
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cor}
Let $F$ be a field and let $p(x)\in F[x]$ be irreducible over $F$. If $a$ is a zero of $p(x)$ in some extension $E$ of $F$ and $b$ is a zero of $p(x)$ in some extension $E'$ of $F$, then the fields $F(a)$ and $F(b)$ are isomorphic.
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lemma}
Let $F$ be a field, let $p(x)\in F[x]$ be irreducible over $F$, and let $a$ be a zero of $p(x)$ in some extension of $F$. If $\phi$ is a field isomorphism from $F$ to $F'$ and $b$ is a zero of $\phi(p(x))$ in some extension of $F'$, then there is an isomorphism from $F(a)$ to $F'(b)$ that agrees with $\phi$ on $F$ and carries $a$ to $b$.
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theorem}[20.4 - ]
Let $\phi$ be an isomorphism from a field $F$ to a field $F'$ and let $f(x)\in F[x]$. If $E$ is a splitting field for $f(x)$ over $F$ and $E'$ is a splitting field for $\phi(f(x))$ over $F'$, then there is an isomorphism from $E$ to $E'$ that agrees with $\phi$ on $F$.
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theorem}[20.5 - ]
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theorem}[20.6 - ]
Let $f(x)$ be an irreducible polynomial over a field $F$. If $F$ has characteristic $0$, then $f(x)$ has no multiple zeros. If $F$ has characteristic $p\neq 0$, then $f(x)$ has a multiple zero only if it is of the form $f(x) = g(x^p)$ for some $g(x)\in F[x]$.
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theorem}[20.7 - ]
Every finite field is perfect.
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theorem}[20.8 - ]
If $f(x)$ is an irreducible polynomial over a perfect field $F$, then $f(x)$ has no multiple zeros.
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theorem}[20.9 - ]
Let $f(x)$ be an irreducible polynomial over a field $F$ and let $E$ be a splitting field for $f(x)$ over $F$. Then all the zeros of $f(x)$ in $E$ have the same multiplicity.
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definition}[Types of extenstions]
Let $E$ be an extension field of a field $F$, and let $a\in E$. We call $a$ algebraic over $F$ if $a$ is the zero of some nonzero polynomial in $F[x]$. If $a$ is not algebraic over $F$, it is called transcendental over $F$. An extension $E$ of $F$ is called an algebraic extension of $F$ if every element of $E$ is algebraic over $F$. If $E$ is not an algebraic extension of $F$, it is called a transcendental extension of $F$. An extension of $F$ of the form $F(a)$ is called a simple extension of $F$.
Recall that $F[x]$ is in integral domain, so it has a field of quotients we will call $F(x)$.
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theorem}[21.1 - ]
Let $E$ be an extension field of $F$ and let $a\in E$. If $a$ is transcendental over $F$, then $F(a)\cong F(x)$. If $a$ is algebraic over $F$, then $F(a)\cong F[x]/\left<p(x)\right>$, where $p(x)$ is a polynomial in $F[x]$ of minimum degree such that $p(a)=0$. Moreover, $p(x)$ is irreducible over $F$.
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theorem}[21.2 - ]
If $a$ is algebraic over a field $F$, then there is a unique monic irreducible polynomial $p(x)$ in $F[x]$ such that $p(a)=0$.
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theorem}[21.3 - ]
Let $a$ be algebraic over $F$, and let $p(x)$ be the minimal polynomial for $a$ over $F$. If $f(x)\in F[x]$ and $f(a)=0$, then $p(x)$ divides $f(x)$ in $F[x]$.
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theorem}[21.7 - ]
If $K$ is an algebraic extension of $E$, and $E$ is an algebraic extension of $F$, then $K$ is an algebraic extension of $F$.
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cor}
Let $E$ be an extension of $F$. The set of elements of $E$ that are algebraic over $F$ is a subfield of $E$.
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definition}
We say that a field $F$ is algebraically closed if every polynomial in $F[x]$ splits in $F$. Algebraically closed fields have no non proper algebraic extensions. The fact that $C$ is algebraically closed was first proved by Gauss when he was 22. This is called the fundamental theorem of (classical) algebra.
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theorem}[22.1 -]
For each prime $p$ and each integer $n$, there is, up to isomorphism, a unique field of order $p^n$. We call this field the Galois field of order $p^n$ and denote is $GF(p^n)$.
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theorem}[22.2 -]
As a group under addition, $GF(P^n)\cong Z_p \oplus Z_p\oplus \cdots \oplus Z_p$. As a group under multiplication, $GF(P^n)\backslash \{0\}\cong Z_{p^n-1}$.
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theorem}[22.3 -]
For each divisor $m$ of $n$, $GF(p^n)$ has a unique subfield of order $p^m$. Moreover, these are the only subfields of $GF(P^n)$.
For more problems, see AllProblems