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 $$n|m \implies \exists k: m=kn \ $$.  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.

Since $$(x,y)(x,z)=(x,yz) \ $$, we have $$a=(a,bc)=(a,b)(a,c)=1*(a,c)=(a,c) \ $$, and we know $$(a,c)=a \iff a|c \ $$.

For the general case, we still have $$a=(a,bc) \ $$, and so $$a=(a,bc)=(a,b)(a,c)=k (a,c) \implies (a,c)= a/k \implies \tfrac{a}{k}|c \ $$.