Proof: Assume $F$ is finite. Then $|F^\times|=|F^+|-1$, so they cannot be in bijection. Therefore $F$ is infinite: Assume $-1=1$; then $2=0$ implying $a+a=0$ for all $a∈F^+$ so that $u^2=1$ for all $u∈F^\times$, so that there are an infinite number of roots to the equation $x^2-1$, an impossibility. Therefore $1≠-1$ so that $2a=a+a≠0$ for all nonzero $a∈F^+$, so that for all nonidentity $u∈F^\times$ we have $u^2≠1$, a contradiction by $(-1)^2=1$.$~\square$
Tuesday, May 21, 2013
Multiplicative and Additive Independence (9.5.7)
Dummit and Foote Abstract Algebra, section 9.5, exercise 7:
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For $F$ a field, prove $F^\times \not ≅ F^+$.
Proof: Assume $F$ is finite. Then $|F^\times|=|F^+|-1$, so they cannot be in bijection. Therefore $F$ is infinite: Assume $-1=1$; then $2=0$ implying $a+a=0$ for all $a∈F^+$ so that $u^2=1$ for all $u∈F^\times$, so that there are an infinite number of roots to the equation $x^2-1$, an impossibility. Therefore $1≠-1$ so that $2a=a+a≠0$ for all nonzero $a∈F^+$, so that for all nonidentity $u∈F^\times$ we have $u^2≠1$, a contradiction by $(-1)^2=1$.$~\square$
Proof: Assume $F$ is finite. Then $|F^\times|=|F^+|-1$, so they cannot be in bijection. Therefore $F$ is infinite: Assume $-1=1$; then $2=0$ implying $a+a=0$ for all $a∈F^+$ so that $u^2=1$ for all $u∈F^\times$, so that there are an infinite number of roots to the equation $x^2-1$, an impossibility. Therefore $1≠-1$ so that $2a=a+a≠0$ for all nonzero $a∈F^+$, so that for all nonidentity $u∈F^\times$ we have $u^2≠1$, a contradiction by $(-1)^2=1$.$~\square$
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