By Huishi Li

Designed for a one-semester direction in arithmetic, this textbook provides a concise and sensible creation to commutative algebra when it comes to basic (normalized) constitution. It indicates how the character of commutative algebra has been utilized by either quantity idea and algebraic geometry. Many labored examples and a couple of challenge (with tricks) are available within the quantity. it's also a handy reference for researchers who use uncomplicated commutative algebra.

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**Sample text**

Let K + 2. 3. 4. 5. equivalently, that a finite field cannot be algebraically closed). , u n } is a finite field, consider the polynomial p ( x ) = ny="=,x- ui) 1 in F [ x ] . ) 6. Let d E Z be square-free. Then every element a E is of the form a = r s d , where r, s E Q. ) = x2 - 2 r x + (r2- s2d). Q(a, 7. Let F = 8). Find a primitive element for F . 8. 14(i). 9. 15. 4. , x , over R. , a,) I > ai E N . , n}. , z], is said to be symmetric if + + + + + For example, 5: 5; xi, ( 2 1 2 2 5 3 ~ ) ( ~ 1 1 ~ 2 ~ 3 a ) ~ .

If S consists of a single element s, then K ( s ) is called a simple extension field of K . 1. Definition Let K be a field, and let f(z)be a polynomial in K [ z ] . , f(z)= a n ( . - ai)in L [ z ] ,and f(x) does not factor completely into linear factors over any proper subfield of L containing K , then L is called a splitting field of f(z). Let K be a field. To see the existence of a splitting field for an arbitrary f(z)E K [ z ] we , start with an irreducible polynomial p ( z ) . Note that the quotient ring + then p(x)h(z) where z is the image of z in L , is a field, for, if p ( z ) ,Y$(z) $(z)g(z) = 1 for some h ( z ) , g ( z )E K [ z ] ,and hence $(z) is invertible in L.

Then nzl(z el + e2 + . . + em = n = degp(s), and each a E L = K ( 8 ) is associated to a monic polynomial in E [ z ]that , is, For convenience, we call fa(z)the total polynomial of a. 2. Proposition Let K K ( 6 ) ,the following hold: L = K ( 6 ) c E be as above. For any a E L = ( 9 f a ( z ) E Wzl. ) E K [ z ]be the minimal polynomial of a over K . Then fa(z)= pa(z)' for some s 2 1. ) of a over K . Proof (i) Since a = ~ ( 2 9 ) = Cyi: A@, where we have i=1 T(X) = CyL; Xixi E K [ z ] , Preliminaries 35 Note that all X i E K .