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Using Algebraic Geometry in Computing_HW1_Elimination Ideals.pdf

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CIS6930 Homework 1 Problem 1. Rings Recall that a commutative ring R is a set with two associative and commutative operations, addition and multiplication, (+, ·) satisfying the followings 12 : 1. Associativity: for a, b ∈ R, a · (b · c) = (a · b) · c and a + (b + c) = (a + b) + c. Commutativity: a + b = b + a and a · b = b · a. 2. Additive Identity: there is an element 0 ∈ R such that a + 0 = 0 + a = a. 3. Additive Inverse: for each a ∈ R, there exists an element (denoted as) −a such tha
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  CIS6930 Homework 1 Problem 1. Rings Recall that a commutative ring  R  is a set with two associative and commutative operations, addition andmultiplication,  (+ ,  · )  satisfying the followings 12 :1. Associativity: for  a,b  ∈  R ,  a · ( b · c ) = ( a · b ) · c  and  a  + ( b  +  c ) = ( a  +  b ) +  c . Commutativity: a  +  b  =  b  +  a  and  a · b  =  b · a .2. Additive Identity: there is an element  0  ∈  R  such that  a  + 0 = 0 +  a  =  a .3. Additive Inverse: for each  a  ∈  R , there exists an element (denoted as) − a  such that  a +( − a ) = 0 .4. Distributive Law:  a · ( b  +  c ) = ( a · b ) + ( a · c ) . A.  Show that additive identity is unique in  R . B.  A multiplicative identity is an element  1  ∈  R  such that  1 · a  =  a  =  a · 1  for all  a  ∈  R . Show that if multiplicative identity exists, then it is also unique. C.  Let  R  be a commutative ring. Show that the set  R [ x ]  of polynomials with coefficients in  R  is also acommutative ring. In particular, if   R  has a multiplicative identity, so does  R [ x ] . Note that elements in R [ x ]  are polynomials a n x n +  a n − 1 x n − 1 + ··· +  a 1 x  +  a 0 with  a i  ∈  R , 0  ≤  i  ≤  n . Problem 2. Ideals Recall that an ideal  I   ⊂  R  is a subring of   R  such that for any  a  ∈  I   and  b  ∈  R ,  ab  ∈  I  . Nowconsider the polynomial ring  R  =  C [ x 1 , ···  ,x n ] . Let  f  x , ···  ,f  s  be polynomials in  C [ x 1 , ···  ,x n ] , let < f  1 , ···  ,f  s  >  denote the set of polynomials < f  1 , ···  ,f  s  > =  {  p 1 f  1  + ··· +  p s f  s  :  p i  ∈ C [ x 1 , ···  ,x n ] , for i  = 1 , ···  ,s } . A.  Show that  < f  1 , ···  ,f  s  >  is an ideal in  R . B.  Show that  < f  1 , ···  ,f  s  >  is the smallest ideal in  C [ x 1 , ···  ,x n ]  containing  f  1 , ···  ,f  s  in the sensethat if   J   is any idea containing  f  1 , ···  ,f  s , then  < f  1 , ···  ,f  s  > ⊂  J  . C.  Show that  x 2 ∈ < x − y 2 ,xy >  in C [ x,y ] . D.  Show that  < x − y 2 ,xy,y 2 > = < x,y 2 > . E.  Is  < x − y 2 ,xy > = < x 2 ,xy > ? Why or why not? Problem 3. Elimination Ideals 1 We will consider only commutative rings in this course. 2 a · b  is usually abbreviated as simply  ab . 1  Let  I   be an ideal in C [ x 1 , ···  ,x n ] , let  l  ≥  1  be an integer and let  I  l  consist of the elements in  I   that donot depend on the first  l  variables: I  l  =  I   ∩ C [ x l +1 , ···  ,c n ] .I  l  is called the  l th elimination ideal of   I  . A.  For  I   = < x 2 +  y 2 ,x 2 − z  3 > ⊂ C [ x,y,z  ] , show that  y 2 +  z  3 is in the first elimination ideal  I  1 . B.  Prove that  I  l  is an ideal in the ring C [ x l +1 , ···  ,x n ] . Problem 4. Quotient Rings Let  R  be a ring and  I   ⊂  R  an ideal. We define the quotient ring  R /I   in class as follows. First, wedefine a surjective map  π  :  R  →  R /I   between sets  R  and  R /I  . We write  π ( a ) = [ a ]  and  π ( a ) =  π ( b ) if and only if   a − b  ∈  I  . The surjective map  π  and  R  define the set  R /I  , and each element in  R /I   canbe written (not uniquely) as  [ a ]  for some  a  ∈  R . Next, we define the two operations in  R /I  :1.  [ a ] + [ b ] = [ a  +  b ] ,2.  [ a ] · [ b ] = [ ab ] .Stop for a few moments and convince yourself that you understand the definition above. A.  Show that operations defined above are well-defined. That is, if   [ a ] = [ a  ]  and  [ b ] = [ b  ] ,  [ a ] + [ b ] =[ a  ] + [ b  ]  and  [ a ] · [ b ] = [ a  ] · [ b  ] . B.  Show that this makes  R /I   a ring. C.  Show that if   R  has multiplicative identity, so does  R /I  .2

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Jul 23, 2017

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