# 代数|MA20217/MA20219 Algebra 2B代写

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Elementary axiomatic theory of rings. Integral domains, fields, characteristic. Subrings and product of rings. Homomorphisms, ideals and quotient rings. Isomorphism theorems. Fields of fractions.

Use Cramer’s rule to solve the system
\begin{aligned} 2 x_{1}-x_{2} &=1 \ 4 x_{1}+4 x_{2} &=20 . \end{aligned}
Solution. The coefficient matrix and right-hand-side vectors are
$$A=\left[\begin{array}{rr} 2 & -1 \ 4 & 4 \end{array}\right] \text { and } \mathbf{b}=\left[\begin{array}{r} 1 \ 20 \end{array}\right]$$
so that
$$\operatorname{det} A=8-(-4)=12$$
and therefore
$$x_{1}=\frac{\left|\begin{array}{rr} 2 & 1 \ 4 & 20 \end{array}\right|}{\left|\begin{array}{rr} 2 & -1 \ 4 & 4 \end{array}\right|}=\frac{36}{12}=3 \quad \text { and } \quad x_{2}=\frac{\left|\begin{array}{rr} 1 & -1 \ 20 & 4 \end{array}\right|}{\left|\begin{array}{rr} 2 & -1 \ 4 & 4 \end{array}\right|}=\frac{24}{12}=2$$

## MA20217/MA20219 COURSE NOTES ：

Since this holds for all $x$, we conclude that $(f+g)+h=f+(g+h)$, which is the associative law for addition of vectors.

Next, if 0 denotes the constant function with value 0 , then for any $f \in V$ we have that for all $0 \leq x \leq 1$,
$$(f+0)(x)=f(x)+0=f(x) .$$
(We don’t write the zero element of this vector space in boldface because it’s customary not to write functions in bold.) Since this is true for all $x$ we have that $f+0=f$, which establishes the additive identity law. Also, we define $(-f)(x)=-(f(x))$ so that for all $0 \leq x \leq 1$,
$$(f+(-f))(x)=f(x)-f(x)=0,$$

# 代数 Algebra II MATH2581-WE01

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\begin{aligned}
\varphi_{L}(a) \vee \varphi_{L}(b)=\left(a^{\star \uparrow} \vee b^{\star \dagger}\right) \cap D(L) &=\left(a^{\star} \wedge b^{\star}\right)^{\dagger} \cap D(L) \
&=(a \vee b)^{\star \uparrow} \cap D(L) \
&=\varphi_{L}(a \vee b)
\end{aligned}

\begin{aligned} \varphi_{L}(a) \cap \varphi_{L}(b)=a^{\star \uparrow} \cap b^{\star \uparrow} \cap D(L) &=\left(a^{\star} \vee b^{\star}\right)^{\dagger} \cap D(L) \ &=(a \wedge b)^{\star \uparrow} \cap D(L) \ &=\varphi_{L}(a \wedge b) . \end{aligned}
Since also $\varphi_{L}(0)={1}$ and $\varphi_{L}(1)=D(L)$ it follows that $\varphi_{L}$ is a $(0,1)$-lattice morphism.
Now, by the distributivity, we have $x=x^{\star \star} \wedge\left(x \vee x^{\star}\right)$. It follows that
$$(\forall x \in L) \quad x^{\dagger}=x^{\star \star \uparrow} \vee\left(x^{\dagger} \cap D(L)\right) \text {. }$$
In fact, if $t \in x^{\dagger}$ then $t=t \vee x=\left(t \vee x^{\star \star}\right) \wedge\left(t \vee x \vee x^{\star}\right)$ where $x \vee x^{\star} \in D(L)$; and conversely if $t \in x^{\star \star \uparrow} \vee\left(x^{\dagger} \cap D(L)\right)$ then $t \geqslant y \wedge z$ where $y \geqslant x^{\star \star} \geqslant x$ and $z \geqslant x$ and thus $t \geqslant x$.

## MATH2581-WE01COURSE NOTES ：

Proof $\Rightarrow$ : If $L$ is a Heyting algebra follow from the above whereas (3) is immediate from the fact that $x:(x \wedge y)=1$.
$\Leftarrow$ : Conversely, if the identites hold then $x \wedge(y: x)=x \wedge y \leqslant y$; and if $x \wedge z \leqslant y$ then
$$z \wedge(y: x)=z \wedge[(z \wedge y):(z \wedge x)]=z \wedge[(y \wedge z):(x \wedge y \wedge z)]=z$$
and so $z \leqslant y: x$. Thus $(L ;:)$ is a Heyting algebra.

# 代数 Algebra MAT00010C

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first box. Since the third box is formed from $x_{1}\left(\begin{array}{l}1 \ 3\end{array}\right)+x_{2}\left(\begin{array}{l}2 \ 1\end{array}\right)=\left(\begin{array}{l}6 \ 8\end{array}\right)$ and $\left(\begin{array}{l}2 \ 1\end{array}\right)$, and the determinant is unchanged by adding $x_{2}$ times the second column to the first column, the size of the third box equals that of the second. We have this.
$$\left|\begin{array}{ll} 6 & 2 \ 8 & 1 \end{array}\right|=\left|\begin{array}{ll} x_{1} \cdot 1 & 2 \ x_{1} \cdot 3 & 1 \end{array}\right|=x_{1} \cdot\left|\begin{array}{ll} 1 & 2 \ 3 & 1 \end{array}\right|$$

Solving gives the value of one of the variables.
$$x_{1}=\frac{\left|\begin{array}{ll} 6 & 2 \ 8 & 1 \end{array}\right|}{\left|\begin{array}{ll} 1 & 2 \ 3 & 1 \end{array}\right|}=\frac{-10}{-5}=2$$
The theorem that generalizes this example, Cramer’s Rule, is: if $|A| \neq 0$ then the system $A \vec{x}=\vec{b}$ has the unique solution $x_{i}=\left|B_{i}\right| /|A|$ where the matrix $B_{i}$ is formed from $A$ by replacing column $i$ with the vector $\vec{b}$. Exercise 3 asks for a proof.
For instance, to solve this system for $x_{2}$
$$\left(\begin{array}{ccc} 1 & 0 & 4 \ 2 & 1 & -1 \ 1 & 0 & 1 \end{array}\right)\left(\begin{array}{l} x_{1} \ x_{2} \ x_{3} \end{array}\right)=\left(\begin{array}{c} 2 \ 1 \ -1 \end{array}\right)$$

$$(a+b i)+(c+d i)=(a+c)+(b+d) i$$
\begin{aligned} (a+b i)(c+d i) &=a c+a d i+b c i+b d(-1) \ &=(a c-b d)+(a d+b c) i \end{aligned}