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Homework 1.pdf

Algebra A

Winter 2024

Homework 1

1. (20 pts) Let \(F\) be a field. Prove the following properties:

   (a) If \(a \cdot b = 0\) for \(a, b \in F\), then \(a = 0\) or \(b = 0\).

   If \(a=0\), then we have proved it.

   If \(a\neq 0\), then there must \(\exists a^{-1}\) such that \(a^{-1}\cdot a\cdot b=a^{-1}\cdot0\Rightarrow1\cdot b=0\Rightarrow b=0\)

   If \(b=0\), then we have proved it

   If \(b\neq 0\), then there must \(\exists b^{-1}\) such that \(b^{-1}\cdot a\cdot b=b^{-1}\cdot0\Rightarrow b^{-1}\cdot b\cdot a=0\Rightarrow1\cdot a=0\Rightarrow a=0\)

   Therefore, \(\forall a,b\in F\), if \(a\cdot b=0\), then \(a=0\) or \(b=0\)

   (b) The additive identity \(0\) is unique.

   Suppose there are two additive identity which is \(0\) and \(\tilde {0}\)

   Then by property of additive identity, \(0+x=x+0=x\) and \(\tilde{0}+x=x+\tilde{0}=x\)

   Then \(0+x=\tilde{0}+x\Rightarrow0+x+\left(-x\right)=\tilde{0}+x+\left(-x\right)\Rightarrow0=\tilde{0}\), which is a contradiction.

   Therefore, the additive identity \(0\) is unique.

   (c) If \((-a) \cdot (-b) = a \cdot b\) for every \(a, b \in F\).

   First, we need to prove \(-1\cdot x=-x\)​

   \(-1\cdot x+x=x\cdot (-1+1)=x\cdot 0=0\)

   Then \(-1\cdot x\) is the additive inverse of \(x\).

   Thus \(-1\cdot x=-x\)

   Then, we can prove it.

   \((-a)(-b)=(-1)\cdot a\cdot(-1)\cdot b=-(-1)\cdot a\cdot b=1\cdot a\cdot b=a\cdot b\), since \(-(-1)=1\) by the additive inverse.

2. (34 pts) Define addition \(\oplus\) and multiplication \(\otimes\) on \(\mathbb{R}\) as follows:

   \(a \oplus b = a + b + 4\)

   \(a\otimes b=2ab\)

   Check whether or not \(\mathbb{R}\) with the addition \(\oplus\) and the multiplication \(\otimes\) satisfies each of the field axioms.

   Let's define \(a,b,c\in R\)​

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   1. Associativity

      \((a\oplus b)\oplus c=\left(a+b+4\right)\oplus c=a+b+4+c+4=a+b+c+8\)​

      \(a\oplus\left(b\oplus c\right)=a\oplus\left(b+c+4\right)=a+b+c+4+4=a+b+c+8\)

      Thus \(\left(a\oplus b\right)\oplus c=a\oplus\left(b\oplus c\right)\)​

      \((a\otimes b)\otimes c=\left(2ab\right)\otimes c=4abc\)

      \(a\otimes\left(b\otimes c\right)=a\otimes\left(2bc\right)=4abc\)

      Thus \(\left(a\otimes b\right)\otimes c=a\otimes\left(b\otimes c\right)\) 2. Commutativity

      \(a\oplus b=a+b+4\)​

      \(b\oplus a=b+a+4=a+b+4\)

      Thus \(a\oplus b=b\oplus a\)

      \(a\otimes b=2ab\)

      \(b\otimes a=2ba=2ab\)

      Thus \(a\otimes b=b\otimes a\) 3. Additive identity and multiplicative identity

      \(a\oplus\left(-4\right)=\left(-4\right)\oplus a=a+\left(-4\right)+4=a\)

      Thus \(-4\) is the additive identity

      \(a\otimes\frac12=\frac12\otimes a=2\cdot\frac12\cdot a=a\)

      Thus \(\frac12\) is the multiplicative identity 4. Additive inverse and multiplicative inverse

      We need to find \(a\oplus b=-4\), then \(b\) is the additive inverse of \(a\)​

      \(a\oplus b=b\oplus a=a+b+4=-4\), then \(b=-8-a\)​

      Thus \(-8-a\) is the additive inverse of \(a\)​

      We need to find \(a\otimes b=\frac12\), then \(b\) is the multiplicative inverse of \(a\)

      \(a\otimes b=b\otimes a=2ab=\frac12\), then \(b=\frac{1}{4a}\)

      Thus \(\frac{1}{4a}\) is the multiplicative inverse of \(a\) 5. Distributivity

      \(a\otimes(b\oplus c)=a\otimes\left(b+c+4\right)=2a\left(b+c+4\right)=2ab+2ac+8a\)​

      \(a\otimes b\oplus a\otimes c=2ab\oplus2ac=2ab+2ac+4\)​

      Since \(a\otimes\left(b\oplus c\right)\neq a\otimes b\oplus a\otimes c\), thus it does not satisfy distributivity

   Therefore, this is not a field in \(\R\) with the addition \(\oplus\) and the multiplication \(\otimes\)

3. (30 pts) Let \(A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}\), \(B = \begin{pmatrix} 5 & 2 \\ 7 & 4 \end{pmatrix}\), and \(C = \begin{pmatrix} 2 & 3 \\ 5 & 8 \end{pmatrix}\). Find a matrix \(D\) such that \(CD - AB = 0\).

   Let \(D=\begin{pmatrix}a & b\\ c & d\end{pmatrix}\), \(CD=AB\)​

   \(CD=\begin{pmatrix}2 & 3\\ 5 & 8\end{pmatrix}\begin{pmatrix}a & b\\ c & d\end{pmatrix}=\begin{pmatrix}2a+3c & 2b+3d\\ 5a+8c & 5b+8d\end{pmatrix}\)​

   \(AB=\begin{pmatrix}2 & -1\\ 3 & 4\end{pmatrix}\begin{pmatrix}5 & 2\\ 7 & 4\end{pmatrix}=\begin{pmatrix}10-7 & 4-4\\ 15+28 & 6+16\end{pmatrix}=\begin{pmatrix}3 & 0\\ 43 & 22\end{pmatrix}\)​

   Then we have \(\begin{cases}2a+3c=3\\ 5a+8c=43\end{cases}\)and \(\begin{cases}2b+3d=0\\ 5b+8d=22\end{cases}\)

   \(a=-105,c=71,b=-66,d=44\)​

   Thus \(D=\begin{pmatrix}-105 & -66\\ 71 & 44\end{pmatrix}\)​

4. (16 pts) Consider the fields \(\mathbb{Z}_7\) and \(\mathbb{Z}_{17}\).

   (a) Find \(3 \cdot 6 + (4^{-1}) + 2 \cdot (-3) + 5\) in \(\mathbb{Z}_7\).

   Since the property of modular arithmetic, we can do the modular at last.

   Since we are in the \(\Z_7\) and \(2\times4\equiv 1\) (mod \(7\)), then the inverse of \(4\) is \(2\)

   Thus \(18+2-6+5=19\) and \(19\equiv 5\) (mod \(7\))

   Therefore, the answer is \(5\)​

   (b) Find \(3 \cdot 6 + (4^{-1}) + 2 \cdot (-3) + 5\) in \(\mathbb{Z}_{17}\).

   Similarly, Since \(4\times 13=17\times 3+1\equiv 1\) (mod \(17\)), then the inverse of \(4\) is \(13\)​

   Thus \(18+13-6+5=30\equiv 13\) (mod \(17\))

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