HW9_selected_solns

HW9_selected_solns - n are the roots of f x over F Then r...

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College of the Holy Cross, Fall 2009 Math 351 – Homework 9 selected solutions Chapter 20, #20. We will show that F ( c ) F ( ac + b ) and also that F ( ac + b ) F ( c ), which will solve the problem. To show F ( c ) F ( ac + b ), we use the fact that F ( c ) is the smallest extension of F containing the element c (note that c may not be in F , but rather in some extension of F ). In other words, F ( c ) is the smallest field containing both F and c . Now F ( ac + b ) is an extension of F , and so contains F . Also, a,b F , which since F F ( ac + b ) gives us that a,b F ( ac + b ). Because F ( ac + b ) is a field and closed under subtraction and division by non-zero elements, we get that (( ac + b ) - b ) /a F ( ac + b ), and thus c F ( ac + b ). Hence F ( ac + b ) contains both F and c , so F ( c ) F ( ac + b ). To show F ( ac + b ) F ( c ), we use the same kind of argument. Clearly F F ( c ), and a,b F ( c ). Since F ( c ) is a field, it is closed under addition and multiplication, so ac + b F ( c ). Thus F ( ac + b ) F ( c ). Chapter 20, #21. Recall that the splitting field of a polynomial over a field F can be obtained by taking F ( r 1 ,...,r n ), where r 1 ,...,r n are all the roots of the polynomial over F . Here, suppose that r 1 ,...,r
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Unformatted text preview: n are the roots of f ( x ) over F . Then r 1-a,. ..,r n-a are the roots of f ( x + a ) over F . So we need to show that F ( r 1 ,...,r n ) = F ( r 1-a,. ..,r n-a ) . We do this by establishing the inclusions F ( r 1 ,...,r n ) ⊆ F ( r 1-a,. ..,r n-a ) and F ( r 1-a,. ..,r n-a ) ⊆ F ( r 1 ,...,r n ), much like number 20 above. To show F ( r 1 ,...,r n ) ⊆ F ( r 1-a,. ..,r n-a ), we need to show that F ⊆ F ( r 1-a,. ..,r n-a ) (which is obvious) and r 1 ,...,r n ⊆ F ( r 1-a,. ..,r n-a ). For the latter, note that a ∈ F , so a ∈ F ( r 1-a,. ..,r n-a ). Thus by closure of a field under addition, we have r 1-a + a,. ..,r n-a + a ∈ F ( r 1-a,. ..,r n-a ), and thus r 1 ,...,r n ∈ F ( r 1-a,. ..,r n-a ). Therefore F ( r 1 ,...,r n ) ⊆ F ( r 1-a,. ..,r n-a ). A very similar argument shows that F ( r 1-a,. ..,r n-a ) ⊆ F ( r 1 ,...,r n )....
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