Pseudo Programmer

$R,\ S$ : ring. $\phi$ : $R \rightarrow S$ be ring homomorphism. Let $A$ be subring of $R$ and $B$ be ideal of $S$. 1. For any $r \in R$ and any positive integer $n$, $\phi(nr) = n\phi(r)$ and $\phi(r^{n}) = (\phi(r))^{n}$ By operation-preserving property of ring homomorphism, following equations must be satisfied. $\phi(r + \dots + r) = \phi(r) + \dots + \phi(r)$ $\phi(r \cdot r \dots r) = \phi(r) \cdot \phi(r) \dots \phi(r)$ 2. $\phi(A) = \{\phi(a) \mid a \in A\}$ is subring of $S$. Since that $A$ is subring of R, $a - b \in A$ and $ab \in A,\ \forall a,\ b \in A$. Hence, $\phi(a - b) = \phi(a) - \phi(b) \in \phi(A)$ and $\phi(ab) = \phi(a)\phi(b) \in \phi(A),\ \forall a,\ b \in A)$. $\therefore \text{By subring test, } \phi(A) \text{ is subring of }S.$ 3. If $A$ is an ideal and $\phi$ is onto $S$, then $\phi(A)$ is an ideal. $\phi$ is onto $S$ implies that $\phi(r) = s ,\ \exists r \in R,\ \forall s \in S$ or equivalently, $\phi(A) = S$. $A$ is ideal of $R$ impl...

$Q$ : ring of rational number (which is field) $Z$ : ring of integer (which is integral domain, not a field) content of polynomial : greatest common divisor of the coefficients of polynomial (must be nonzero polynomial) Suppose $f(x) \in Z[x]$ and $f(x) = g(x)h(x)$, where $g(x),\ h(x) \in Q[x]$ and $1 \leq deg\ g(x),\ h(x) < deg\ f(x)$. Also, we can assume that $f(x)$ is primitive polynomial. Since that dividing both sides with the content of $f(x)$, say $c_{f(x)}$, derives  $$\frac{f(x)}{c_{f(x)}} = \frac{g(x)h(x)}{c_f(x)}$$ Which makes polynomial in left side primitive, but does not affect degree of $g(x)$ nor $h(x)$. Hence, $f(x)$ can be primitive in any situation. Let $a$ be the LCM(Least Common Divisor) of the denominators of the coefficients of $g(x)$, and $b$ be the LCM of the denominators of the coefficients of $h(x)$. Then, $abf(x) = ag(x) \cdot bh(x)\ (ag(x),\ bh(x) \in Z[x])$. Let $c_{1}$ be the content of $ag(x)$, and $c_{2}$ be the content of $bh(x)$. ...

$F$ : field Reducibility test for degrees 2 and 3 is as follows, $$f(x) \in F[x] \text{ and } deg\ f(x) = 2\text{ or }3$$ $$\implies$$ $$f(x)\text{ is reducible over }F\text{ iff }f(x)\text{ has a zero in }F$$ Suppose that $f(x) = g(x)h(x)$, where $g(x),\ h(x) \in F[x]$. Since that, polynomials of zero degree are the only unit over integral domain , forms containing zero polynomial are not our concern. Hence, $1 \leq g(x),\  h(x) < deg\ f(x)$ Since, in integral domain , deg $f(x) = deg\ g(x) + deg\ h(x)$ and $deg\ f(x)$ is $2$ or $3$, at least one of g(x) and h(x) has degree 1. Say $g(x) = ax + b$. Then, clearly, $-a^{-1}b$ is a zero of $g(x)$ and therefore a zero of $f(x)$. Conversely, suppose that $f(a) = 0, a \in F$. Then by the Factor Theorem, we know that $x - a$ is a factor of $f(x)$ and, therefore $f(x)$ is reducible over $F$.

$\phi$ : $Z_{n} \rightarrow Z_{n}$, $x \mapsto ax$, where $a^{2} = a$ is ring homomorphism. Let $\phi(1) = a$. Then $a = \phi(1) = \phi(1)\phi(1) = a^{2}$. Let $x \in Z_{n}$. Then $\phi(x) = \phi(1 + \dots + 1) = \phi(1)x = ax$. Hence, our original conjecture is true.

$R,\ S$ : ring. $\phi$ : $R \rightarrow S$ be ring homomorphism. Let $Ker(\phi) = \{r \in R \mid \phi(r) = 0 \}$. Since that $\phi(a - b) = \phi(a) - \phi(b) = 0_{R} + 0_{R} = 0_{R}$. Hence, $a - b \in Ker(\phi), \forall a,\ b \in Ker(\phi)$ Also, by as follows, $\phi(rx) = \phi(r) \phi(x) = \phi(r) \cdot 0_{R} = 0_{R},\ \ \forall r \in R,\ \forall x \in Ker(\phi)$ $\phi(xr) = \phi(x) \phi(r) = 0_{R} \cdot \phi(r) = 0_{R},\ \ \forall r \in R,\ \forall x \in Ker(\phi)$ $rx,\ xr \in Ker(\phi),\ \ \forall r \in R,\ \forall x \in Ker(\phi)$ is trivial. $\therefore \text{By ideal test, } Ker(\phi)\text{ is ideal of }R$

$\mathbb Z_{p^{n}}$ : set of integers modulo $p^{n}$ nilpotent : $a^{n} = 0,\ \exists n \in \mathbb Z^{+}$ Note that $a \in \mathbb Z_{n}$ is zero-divisor if and only if $gcd(a, n) = d,\ (d > 1)$. Since that $d \mid p^{n}$, it implies $d = p^{k},\ (k < n)$. $a = ld = lp^{k}$ $a^{n} = (lp^{k})^{n} = (l)^{n}(p^{k})^{n} = (l)^{n}(p^{n})^{k} = 0$.

$F$ : finite field. $char(F)$ : characteristic of $F$ We already know that $(F,\ +)$ is the group under the addition. Since that every field is integral domain,  characteristic of $F$ must be $0$ or some prime $p$. Characteristic of finite field could not be $0$. Since that, by $\textit{Lagrange's Theorem}$, the order of each element of the group divides the order of the group if the order of the group is finite. So, let's consider the case when $char(F)$ is some prime $p$. Then, $0 = 1 \cdot p$ which implies $|1| = p$ on $(F,\ +)$ and $p \mid |F|$. Let's assume that there is some other prime $q$, where $q \mid |F|$. By the $\textit{Cauchy's Theorem}$, if $q \mid |F|$, then $|x| = q, \exists x \in F$. If $char(F) = p$, then $x \cdot p = 0$ and $x \cdot q = 0$ must be satisfied. But, from the fact that $gcd(p,\ q) = 1$, by $Bezout's identity$, there exists some integers a, b which satisfies $ap + bq = 1$. By multiplying $x$ on both sides of equation yiel...

$G$ : undirected simple finite graph, $\bar{G}$ : complementary graph of $G$. self-complementary : $G \cong \bar{G}$ Between complementary graph, $$|E_{G}| + |E_{ \bar{G}}| = \frac{|V_{G}|(|V_{G}| - 1)}{2}$$ should be satisfied. If $G \cong \bar{G}$, then $|E_{G}| = |E_{\bar{G}}|$. Hence, with substitution, $$2|E_{G}| = \frac{|V_{G}|(|V_{G}| - 1)}{2} \\[10pt] 4|E_{G}| = |V_{G}|(|V_{G}| - 1) \\[10pt]$$ If $|V_{G}|$ is even integer, then $|V_{G}| - 1$ is odd integer. Converse is also trivially true. $$\therefore 4 \mid |V_{G}| \ \text{or} \ 4 \mid |V_{G}| - 1$$

$G$ : undirected simple finite graph, $\bar{G}$ : complementary graph of $G$. $C_{n} (n \geq 3)$ : n-cycle graph, which is the graph itself is a cycle. To be $G$ self-complementary, $G \cong \bar{G}$ should be satisfied. Hence, $G$ and $\bar{G}$ must have same number of edges. Which implies that $|E_{G}| = |E_{ \bar{G}}| = \frac{1} {2} \cdot \frac{|V_{G}|(|V_{G}| - 1)}{2}$ (Note that $|V_{G}| = |V_{\bar{G}}|$ by definition of complementary graph). From the fact that $C_{n}$ is cycle graph, $|E_{C_{n}}| = |V_{C_{n}}| = n$ and $|E_{C_{n}}| = |E_{\bar{C_{n}}}|$, $$|E_{C_{n}}|  + |E_{\bar{C_{n}}}| = \frac{|E_{C_{n}}|(|E_{C_{n}}| - 1)}{2} \\[10pt] 2n = n(n-1) \\[10pt] n(n - 5) = 0 \\[10pt] \therefore n = 0, 5$$ Since that $n \geq 3$, the only answer is $5$.

$R$ : ring, $D = \{I_1, \dots , I_n \}$ : arbitrary set of ideals of R. Let's show that $S \subseteq D$, $\bigcap \limits_{I_{i} \in S}{I_{i}}$ is ideal of $R$. Since that D satisfies $a - b \in I_{i}$ ($\forall a, \forall b \in \bigcap \limits_{I_{i} \in S}{I_{i}}$)($\forall I_{i} \in S$) and $ar, ra \in I_{i} \forall I_{i} \in S$. Hence, by ideal test, it is trivial.

$R$ : commutative ring with unity, $I$ : proper ideal of R. Assume that $a \in R$ is unit $\forall a \notin I$. Let's first show that $I$ should be maximal ideal of $R$. We already know that ideal with unit could never be proper ideal.  So we can conclude that $I$ never contains any unit element in $R$. Since that every element outside of $I$ is unit, any ideal of $R$ which properly containing $I$, must have unit of $R$. Which implies $I$ is maximal ideal by definition. Next, we need to prove that $I$ is the unique maximal ideal of $R$. Obviously, any proper ideal of $R$ must be in $I$. Therefore, $I$ is the only maximal ideal.

$R$ : principal ideal domain, $I$ : ideal of $R$. Since that $R$ is the principal ideal domain, every ideal of $R$ is principal ideal. Let such ideal be $H$. Also, let $J =\{a \in R \mid a + I \in H\}$ (Could be interpreted as the set of the representatives of cosets in $R/I$). Let's prove that $J$ is the ideal of $R$. Since that ring must have the additive identity, (which implies that the ring is non-empty set) we can say that $0 + I \in H, 0 \in J$. Also, $a, b \in J, \forall a + I, b + I \in H$ which implies $(a + I) - (b + I) = (a - b) + I \in H$. So, $a - b \in J$. Since that $H$ is the ideal of $R/I$, $ar + I, ra + I \in H, \forall a \in J, \forall r \in R$ which implies $ar, ra \in J$. Therefore, we can conclude that $J$ is the ideal of $R$. Next, let's prove $H = <j + I>, \exists j \in J$. $(\Longleftarrow)$     Obiviously, $<j + I> \subseteq H, \exists j \in J$. $(\Longrightarrow)$     If $a + I \in H$, then $a \in J$. ...