Mg Problem 26 27

FINALS

26) a. Thm: Equality modulo an ideal I of a ring R is an equivalence relation. (In our class, you may always assume (unless stated otherwise) that a ring is commutative and has 1.)

Pf: To show that the relation $a \sim_{I} b \leftrightarrow a - b \in I$ is an equivalence relation, we must show:

1. Reflexivity: $\forall{x} \in R$, $x-x=0$, and $\forall{I} 0 \in I$

2. Symmetry: ($\forall{a,b} \in R(a-b \in I \leftrightarrow b-a \in I)$) Suppose that for some $a,b \in R$, $a-b = c \in I$. Then $b-a = -c$. An ideal with 1 must have the additive identity of 1, or -1. And, since ideals are superclosed, $-c = c(-1) \in I$. The reverse statement is proven similarly.

3. Transitivity: ($\forall{a,b,c} \in R( a\sim_{I}b \wedge b\sim_{I}c \rightarrow a\sim_{I}c)$)

By the equivalence relation, we have $a-b \in I$ and $b-c \in I$. Since ideals are closed under addition, we have $(a-b) + (b-c) \in I$. From this, we simplify to get $a-c \in I$.

QED

b. tThm: Multiplication on the quotient ring R/I is well defined: (r+I)(s+I)=rs+I.

Pf: That multiplication is well-defined may be proven simply by showing that for two pairs of equivalent terms, $\{r, \tilde{r}\}$ and $\{s , \tilde{s}\}$, $(r+I)(s+I) = (\tilde{r} + I)(\tilde{s} + I)$, or $rs + I = \tilde{r}\tilde{s} + I$. If $r \sim \tilde{r}$, then we can say $r = \tilde{r} + a$ for some $a \in I$. Similarly, we have $s = \tilde{s} + b$ for some $b \in I$. Then $(r+I)(s+I) = rs+I$ implies that $(r+I)(s+I) = (\tilde{r}+a)(\tilde{s}+b) + I$. Expanding this, we get $\tilde{r} \tilde{s} + \tilde{r} {b} + \tilde{s} {a} + ab + I$. Since $a,b \in I$ and ideals are superclosed under multiplication, $\tilde{r} {b}, \tilde{s} {a}, ab \in I$. Ideals are closed under addition, so $\tilde{r} {b} + \tilde{s} {a} + ab + I = I$. Therefore, $\tilde{r} \tilde{s} + \tilde{r} {b} + \tilde{s} {a} + ab + I = \tilde{r} \tilde{s} + I$. Hence, $rs + I = \tilde{r} \tilde{s} + I$ and therefore $(r+I)(s+I) = (\tilde{r} + I)(\tilde{s}+I)$, and we conclude that multiplication on the quotient ring $R/I$ is well-defined.

QED

27. a. Thm: Every unit of a ring $R$ has a unique inverse.

Pf: Suppose there exist $u,v,v' \in R$, where $uv = 1$ and $uv' = 1$. In this case, by transitivity, we have $uv = uv'$. We may now multiply by $v$ to get $vuv = vuv'$, implying $1v=1v'$, and finally $v=v'$. Thus, the inverse of the unit is unique.

b. Thm: The units of $\mathbb{Z}/{n\mathbb{Z}}$ are precisely those $[a]$ such that $GCD(a,n) = 1$.

Pf: If $GCD(a,n) = 1$ then there exist $v,w \in \mathbb{Z}$ such that $av + nw = 1$. By the equivalence of cosets, we can say that in the quotient ring $\mathbb{Z}/{n\mathbb{Z}}$ we have $[a][v] + [n][w] = 1$. Since any multiple of $n$ or an equivalent element of $\mathbb{Z}$ modulo $n$ is equivalent to $0$, $[n][w] = [0]$ and therefore $[a][v] = 1$. Therefore, $[a]$ is precisely the units of $R$.

Comments - Andrew Furash
Nice work on 26. You don't know that 1 is in the ideal (in fact 1 is in I iff I=R). You correctly use 'superclosedness' to achieve the result.
Good job on the rest.

page revision: 16, last edited: 16 Apr 2018 17:41