(New page: Category:MA453Spring2009Walther From theorem 14.4 we know that R/A is a field if A is maximal. In this case we have R = R[x] and A = <x^2+1>, so we must show that <x^2+1> is maximal....) |
|||
| Line 1: | Line 1: | ||
[[Category:MA453Spring2009Walther]] | [[Category:MA453Spring2009Walther]] | ||
| − | From theorem 14.4 we know that R/A is a field if A is maximal. In this case we have R = R[x] and A = <x^2+1>, so we must show that <x^2+1> is maximal. We can do this by observing x^2+1 = (x+i)(x-i) and neither (x+i) nor (x-i) are contained in R[x]. This means that <x^2+1> is maximal and that R[x]/<x^2+1> is a field. | + | From theorem 14.4 we know that R/A is a field if A is maximal. In this case we have R = R[x] and A = <x^2+1>, so we must show that <x^2+1> is maximal. We can do this by observing x^2+1 = (x+i)(x-i) and neither (x+i) nor (x-i) are contained in R[x]. This means that <x^2+1> is maximal and that R[x]/<x^2+1> is a field.<br> |
--[[User:Jniederh|Jniederh]] 20:17, 25 March 2009 (UTC) | --[[User:Jniederh|Jniederh]] 20:17, 25 March 2009 (UTC) | ||
Latest revision as of 15:18, 25 March 2009
From theorem 14.4 we know that R/A is a field if A is maximal. In this case we have R = R[x] and A = <x^2+1>, so we must show that <x^2+1> is maximal. We can do this by observing x^2+1 = (x+i)(x-i) and neither (x+i) nor (x-i) are contained in R[x]. This means that <x^2+1> is maximal and that R[x]/<x^2+1> is a field.
--Jniederh 20:17, 25 March 2009 (UTC)
