10.16 Fields
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Let \(\mathbb{F}\) be a field. Prove the following properties
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\(\forall a,b\in\mathbb{F}\), \(-(a+b)=(-a)+(-b)\)
Remember that the additive inverse is unique, so we just need to check that \((a+b)+[(-a)+(-b)]=0\)
We have that \((a+b)+[(-a)+(-b)]=a+(b+(-a))+(-b)=0\) 2. \(\forall a,b\in\mathbb{F},a,b\neq 0\), \((ab)^{-1}=(b^{-1})(a^{-1})\)
Remember that the multiplicative inverse is unique, so we just need to check that \((ab)(b^{-1})(a^{-1})=1\)
Similarly
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For an arbitrary set \(\mathbb{F}\) of 4 elements find if possible addition and multiplication table such that \(\mathbb{F}\) is a field
Note: \(F_4,F_5....\) must be a field, 有可能有很多个table,实际上也是一个 只是改名字的交换的区别
素数个元素就可以直接写 其中一个table是Z(modular)
合数个元素就需要在素数域上拓展 比如\(F_4\)可以由\(Z_2\)拓展来
涉及到空间的拓展,比如从实数拓展到复数一样的思路
We know that a field must have \(0\) and \(1\), so let's call our set \(\mathbb{F}=\{0,1,a,b\}\)
\(+\) 0 1 a b 0 0 1 a b 1 1 0 b a a a b 0 1 b b a 1 0 \(\times\) 0 1 a b 0 0 0 0 0 1 0 1 a b a 0 a b 1 b 0 b 1 a By commutativity, the tables should be symmetric
By \(a\cdot 0=0\), if we want \(\mathbb{F}\) to be a field, it should satisfy this.
Note that we have 3 non-zero elements: \(1,a,b\)
They must all have additive inverses, so at least one of these elements should be it's own additive inverse
inverse 成对出现 所以有一个会剩余 inverse就是自己本身 uniqueness (because those who aren't come in pairs and uniqueness)
Let's call this element \(c\) (\(c\) is \(1,a,b\)), \(c\) satisfies \(c+c=0\)
Since \(c\neq 0\), \(\exists c^{-1}\in\mathbb{F}\) such that \(c\cdot c^{-1}=1\)
==Trick==
So \(1+1=cc^{-1}+cc^{-1}=(c+c)c^{-1}=0\)
Since \(1+1=0\), then \(a+a=0,b+b=0\)
Then we want to find the value of \(a+1\). We have 4 options
\(a+1=\begin{cases}0\\1\\a\\b\end{cases}\)
If \(a+1=0\), then \(a=1\). Contradiction!
If \(a+1=1\), then \(a=0\). Contradiction!
If \(a+1=a\), then \(0=1\). Contradiction!
If \(a+1=b\), then it is right
Repeating the same process but changing \(a\) for \(b\), we know \(b+1=a\)
We have that \(a+b\) can only be 4 possible options
\(a+b=\begin{cases}0\\ 1\\ a\\ b\end{cases}\)
If \(a+b=0\), then \(a=-b=b\). Contradiction!
If \(a+b=a\), then \(b=0\). Contradiction!
If \(a+b=b\), then \(a=0\). Contradiction!
If \(a+b=1\), then it is right
We have that \(a\cdot a\) can only be 4 possible options
\(a\cdot a=\begin{cases}0\\ 1\\ a\\ b\end{cases}\)
If \(a\cdot a=0\), then \(a=0\). Contradiction!
If \(a\cdot a=a\), then \(a=1\). Contradiction!
If \(a\cdot a=b\), then it is right
If \(a\cdot a=1\), then since \(a\cdot 1=a\), then \(a^3=a\).
Thus \(a=1,-1\)
We have that \(a\cdot b\) can only be 4 possible options
\(a\cdot b=\begin{cases}0\\ 1\\ a\\ b\end{cases}\)
If \(a\cdot b=0\), then \(a=0\vee b=0\). Contradiction!
If \(a\cdot b=a\), then \(b=1\). Contradiction!
If \(a\cdot b=b\), then \(a=1\) Contradiction!
If \(a\cdot b=1\), then it is right
We still have to check associativity and distributivity (exercise)
We have \(0,1,a,b\), more precisely, we have \(0,1,a,a+1\)
Then for associativity
- \((0+1)+a=1+a=0+(1+a)\)
- \((0+1)+a+1=1+a+1=0+(1+a+1)\)
- \((0+a)+a+1=a+a+1=0+(a+a+1)\)
- \((1+a)+a+1=a+1+a+1=1+(a+a+1)\)
- \((0\cdot 1)\cdot a=0\cdot a=0=0\cdot(1\cdot a)\)
- ......
- ......
- .....
For distributivity
- \(0\cdot (1+a)=0\cdot 1+0\cdot a=0\)
- \(1\cdot (a+a+1)=1\cdot a+1\cdot (a+1)=a+a+1\)
- ........
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Consider the field \(\mathbb{Q}(\sqrt2)=\{a+\sqrt2 b:a,b\in \mathbb{Q}\}\)
- Find \((1-\sqrt2)^{-1}\)
- Calculate \((\frac32+\sqrt2)^2\left(1-\sqrt2\right)^{-1}+\left(\frac12+\frac54\sqrt2\right)\)
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Consider the field \(\Z_5\) and \(\mathbb{Z}_{13}\)
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Find \(4\cdot3^{-1}+2\left(-6\right)+3\) in \(\Z_5\)
We have \(2\cdot 3\equiv6\equiv 1\) (mod \(5\)), \(-6\equiv-6+10\equiv4\) (mod \(5\))
Then \(4\cdot 2+2\cdot 4+3=19\equiv 5\) (mod \(5\)) 2. Find \(4\cdot3^{-1}+2\left(-6\right)+3\) in \(\mathbb{Z}_{13}\)
We have \(9\cdot3\equiv27\equiv1\) (mod \(13\)), \(-6\equiv-6+13\equiv 7\) (mod \(13\))
Then \(4\cdot9+2\cdot7+3=53\equiv1\) (mod \(13\))
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Roots of unity
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Find the cube roots of unity
\(z^3=1\Leftrightarrow0=z^3-1=\left(z-1\right)\left(z^2+z+1\right)\)
Easily, \(z=1,z=-\frac12+\frac{\sqrt3}{2}i,z=-\frac12-\frac{\sqrt3}{2}i\) 2. If \(w\) is a cube root of unity, find the value of \(w^{67}\)
\(w^{67}=w^{66}\cdot w=1\cdot w=w\) 3. Find the sixth roots of unity
\(z^6=1\)
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