User:J D Bowen/Math735 HW6

7.6.3) If $K\subset R\times S \ $ is an ideal, then for all $(r,s)\in R\times S, \ (i,j)\in K \ $, we have

$(r,s)(i,j)=(ri,sj)\in K \ $.

Define $I=\left\{{i\in R| \exists j:(i,j)\in K }\right\} \ $.

Then for all $r \in R \ $, and all $i\in I \ $, we have $ri\in I \ $, since $(ri,sj)\in K \ $. Hence $I \ $ is an ideal.

A similar argument shows $J=\left\{{j\in S|\exists i:(i,j)\in K }\right\} \ $. Note that $I\times J = \left\{{ (i,j)\in R\times S: (i,j)\in K }\right\}=K \ $.

7.6.7) Note that $n|m \implies \exists k: m=kn \ $.  Suppose $(k,n)=1 \ $.  By the Chinese Remainder Theorem, $\Z_m=\Z_{kn}\cong \Z_k \times \Z_n \ $.

Define this isomophism as $\alpha \ $.

Then define $\hat{\phi}:\Z_k\times\Z_n\to\Z_n \ $ as $(a,b)\mapsto b \ $.

Then define $\phi:\Z_m\to\Z_n \ $ as $x\mapsto\hat{\phi}(\alpha(x)) \ $.

By the corollary noted on the bottom of page 266, we have $\Z_m^\times \cong \Z_k^\times \times \Z_n^\times \ $, and so if $\phi(x)\in \Z_n^\times \ $, then there is a unit $y\in\Z_m^\times \ $ such that $\hat{\phi}(\alpha(y))=x \ $.

Hence, $\phi \ $ is a surjection from $\Z_m^\times \to \Z_n^\times \ $.

8.1.4)  Suppose $(a,b)=1, \ a|bc \ $.  We aim to show $a|c \ $.  Note that the second assumption, and our goal, are equivalent to the statements $(a,bc)=a, \ (a,c)=a \ $, respectively.

Note that $(a,b)=1 \iff 1=ax+by \ $. Multiply this by c and have $c=cax+cby \ $. Then note that a|bc and a|ac, thus a divides $c=cax+cby \ $. Therefore $a|c \ $.

In the general case, we have $a=h\cdot(a,b), \ b=j\cdot(a,b), \ h \not | j \ $. Then $(a,b)=k \implies kh(a,b)=cj(a,b) \implies kh=cj \implies h|cj \implies h|c \ $.