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... It follows from Tychonoff’s theorem that R[[X]] is compact if and only if R is finite. The topology on R[[X]] can also be seen as the I-adic topology, where I = (X) is the ideal generated by X (whose elements are precisely the formal power series with zero constant coefficient). If R = K is a field, ...
... It follows from Tychonoff’s theorem that R[[X]] is compact if and only if R is finite. The topology on R[[X]] can also be seen as the I-adic topology, where I = (X) is the ideal generated by X (whose elements are precisely the formal power series with zero constant coefficient). If R = K is a field, ...
Math 614, Fall 2015 Problem Set #1: Solutions 1. (a) Since every
... 2x2 = 6. The quotient is isomorphic as an abelian group with Z/6Z⊕Z/2Z with respective generators that are the images of 1 and x. It has 12 elements. (b) Since Z ⊆ A, A = Z[x2 , 2x] which is spanned over Z by the even powers of x, (x2 )k , and the even multiples of the odd powers of x, 2x(x2 )k . Th ...
... 2x2 = 6. The quotient is isomorphic as an abelian group with Z/6Z⊕Z/2Z with respective generators that are the images of 1 and x. It has 12 elements. (b) Since Z ⊆ A, A = Z[x2 , 2x] which is spanned over Z by the even powers of x, (x2 )k , and the even multiples of the odd powers of x, 2x(x2 )k . Th ...
Solutions - UBC Math
... both 3 − 2i and 3 + 2i are irreducible, since their norms are equal to 13, which is prime; and we know that (a) Norm is multiplicative, so if we had 3 − 2i = αβ, then N (3 − 2i) = N (α)N (β), so this forces N (α) or N (β) to be 1; and (b) Any element of norm 1 is a unit. (b) Prove that any ideal in ...
... both 3 − 2i and 3 + 2i are irreducible, since their norms are equal to 13, which is prime; and we know that (a) Norm is multiplicative, so if we had 3 − 2i = αβ, then N (3 − 2i) = N (α)N (β), so this forces N (α) or N (β) to be 1; and (b) Any element of norm 1 is a unit. (b) Prove that any ideal in ...
Solutions — Ark 1
... Oppgave 1. Show that the principal ideal (P (X1 , . . . , Xn )) in the polynomial ring k[X1 , . . . , Xn ] over the field k is prime if and only if P (X1 , . . . , Xn ) is irreducible. (Hint: Use that k[X1 , . . . , Xn ] is UFD.) Solution: In fact, we are going to show that in any ring A being a UFD ...
... Oppgave 1. Show that the principal ideal (P (X1 , . . . , Xn )) in the polynomial ring k[X1 , . . . , Xn ] over the field k is prime if and only if P (X1 , . . . , Xn ) is irreducible. (Hint: Use that k[X1 , . . . , Xn ] is UFD.) Solution: In fact, we are going to show that in any ring A being a UFD ...
LECTURES MATH370-08C 1. Groups 1.1. Abstract groups versus
... Q, R, C and Z/(pZ) for p prime are fields; H is a skew-field. The notion of a subring of a ring R is defined naturally: it is a subset of R, closed under both ring operations. A subring I ⊂ R is called a left ideal, if I · R ⊂ I; a right ideal, if R · I ⊂ I; a two-sided ideal, if I · R ⊂ I & R · I ⊂ ...
... Q, R, C and Z/(pZ) for p prime are fields; H is a skew-field. The notion of a subring of a ring R is defined naturally: it is a subset of R, closed under both ring operations. A subring I ⊂ R is called a left ideal, if I · R ⊂ I; a right ideal, if R · I ⊂ I; a two-sided ideal, if I · R ⊂ I & R · I ⊂ ...
