
CM222, Linear Algebra Mock Test 3 Solutions 1. Let P2 denote the
... Solution: a), c) Explanation: a) is the definition. c) is the same as a) since the eigenvalues with eigenvalue λ are precisely the non-zero vectors in the null space of A − λI. b) is incorrect, since for example the geometric multiplicity of the eigenvalue 1 for the 2 × 2 identity matrix is 2, but t ...
... Solution: a), c) Explanation: a) is the definition. c) is the same as a) since the eigenvalues with eigenvalue λ are precisely the non-zero vectors in the null space of A − λI. b) is incorrect, since for example the geometric multiplicity of the eigenvalue 1 for the 2 × 2 identity matrix is 2, but t ...
TRUE/FALSE. Write `T` if the statement is true and `F` if the
... 10) A field is a set in which we can do addition, subtraction, multiplication and division without leaving the set. ...
... 10) A field is a set in which we can do addition, subtraction, multiplication and division without leaving the set. ...
enumerating polynomials over finite fields
... A polynomial is irreducible if it is of positive degree and cannot be factored into polynomials of strictly smaller degree. So for instance every polynomial of degree one is irreducible. In fact, every polynomial can be uniquely factored into irreducible polynomials (possibly repeated). More precise ...
... A polynomial is irreducible if it is of positive degree and cannot be factored into polynomials of strictly smaller degree. So for instance every polynomial of degree one is irreducible. In fact, every polynomial can be uniquely factored into irreducible polynomials (possibly repeated). More precise ...
CHAP11 Z2 Polynomials
... Cast your mind back to the time when you first learnt about complex numbers. Your whole world of numbers was the field of real numbers. There were many polynomial equations which had no solutions such as x2 + 1 = 0. What we did was to invent solutions for this polynomial. A new “imaginary” number “i ...
... Cast your mind back to the time when you first learnt about complex numbers. Your whole world of numbers was the field of real numbers. There were many polynomial equations which had no solutions such as x2 + 1 = 0. What we did was to invent solutions for this polynomial. A new “imaginary” number “i ...
1 PROBLEM SET 9 DUE: May 5 Problem 1(algebraic integers) Let K
... (3). Let m be a maximal ideal of OK , and let k = OK /m be the residue field. Verify there exists a natural ring epimorphism OK [x] → k[x]. (4). (Eisenstein’s criterion) Let f (x) = an xn +an−1 xn−1 +....+a0 ∈ Z[x], satisfying an 6≡ 0 mod p, ai ≡ 0 mod p f or i 6= n, a0 6≡ 0 mod p2 , where p is a pr ...
... (3). Let m be a maximal ideal of OK , and let k = OK /m be the residue field. Verify there exists a natural ring epimorphism OK [x] → k[x]. (4). (Eisenstein’s criterion) Let f (x) = an xn +an−1 xn−1 +....+a0 ∈ Z[x], satisfying an 6≡ 0 mod p, ai ≡ 0 mod p f or i 6= n, a0 6≡ 0 mod p2 , where p is a pr ...
MatlabTutorial
... functions. To use these tools, the polynomial should be represented as a vector with the leftmost number being the highest power and the rightmost number being the constant. For example, x² + 2x + 1 would be represented as [1 2 1]. • The roots function gives the roots of the polynomial and polyval e ...
... functions. To use these tools, the polynomial should be represented as a vector with the leftmost number being the highest power and the rightmost number being the constant. For example, x² + 2x + 1 would be represented as [1 2 1]. • The roots function gives the roots of the polynomial and polyval e ...
Review Problems
... 4. Factor x5 − 2x4 − 2x3 + 12x2 − 15x − 2 over Q. Solution: The possible rational roots are ±1, ±2, and since 2 is a root we have the factorization x5 − 2x4 − 2x3 + 12x2 − 15x − 2 = (x − 2)(x4 − 2x2 + 8x + 1). The only possible rational roots of the second factor are 1 and −1, and these do not work. ...
... 4. Factor x5 − 2x4 − 2x3 + 12x2 − 15x − 2 over Q. Solution: The possible rational roots are ±1, ±2, and since 2 is a root we have the factorization x5 − 2x4 − 2x3 + 12x2 − 15x − 2 = (x − 2)(x4 − 2x2 + 8x + 1). The only possible rational roots of the second factor are 1 and −1, and these do not work. ...
Final Exam conceptual review
... (b) Goals: Determine the number of elements of F [x]/hpi, find all units and all zero divisors of F [x]/hpi (c) Homework problems: #10 (d) Additional practice problems: #9, 11 Proof-based problems. Here is a collection of additional (some more challenging) proofs to practice with that require knowle ...
... (b) Goals: Determine the number of elements of F [x]/hpi, find all units and all zero divisors of F [x]/hpi (c) Homework problems: #10 (d) Additional practice problems: #9, 11 Proof-based problems. Here is a collection of additional (some more challenging) proofs to practice with that require knowle ...
43. Here is the picture: • • • • • • • • • • • • •
... 44. Because the polynomial is primitive, it is irreducible over Z if and only if it is irreducible over Q. As it has degree 3, it is irreducible over Q if and only if it has no roots in Q. Moreover, if a = n/d ∈ Q is a root, then n|4 and d|1, so a ∈ {±1, ±2, ±4}. Checking these one by one, we find t ...
... 44. Because the polynomial is primitive, it is irreducible over Z if and only if it is irreducible over Q. As it has degree 3, it is irreducible over Q if and only if it has no roots in Q. Moreover, if a = n/d ∈ Q is a root, then n|4 and d|1, so a ∈ {±1, ±2, ±4}. Checking these one by one, we find t ...
Wedderburn`s Theorem on Division Rings: A finite division ring is a
... Φd (x) where either d|mi or d = n; so the quotient is a product of the Φd (x)’s where d is a proper divisor of n that does not divide mi ; hence the quotient is a polynomial with integer coefficients. Substituting the integer q for the variable x, we see that the integer Φn (q) divides the integer ( ...
... Φd (x) where either d|mi or d = n; so the quotient is a product of the Φd (x)’s where d is a proper divisor of n that does not divide mi ; hence the quotient is a polynomial with integer coefficients. Substituting the integer q for the variable x, we see that the integer Φn (q) divides the integer ( ...
Principal Ideal Domains
... Proof. The result follows from Theorems 57 and 62. Theorem 63. Every nonzero prime ideal in a PID is a maximal ideal. Corollary 64. If R is a commutative ring such that the polynomial ring R[x] is a PID, then R is necessarily a field. Example 65. Here are a few quick examples. (1) We already know th ...
... Proof. The result follows from Theorems 57 and 62. Theorem 63. Every nonzero prime ideal in a PID is a maximal ideal. Corollary 64. If R is a commutative ring such that the polynomial ring R[x] is a PID, then R is necessarily a field. Example 65. Here are a few quick examples. (1) We already know th ...
PDF
... The polynomial ring over R in two variables X, Y is defined to be R[X, Y ] := R[X][Y ] ∼ = R[Y ][X]. Elements of R[X, Y ] are called polynomials in the indeterminates X and Y with coefficients in R. A monomial in R[X, Y ] is a polynomial which is simultaneously a monomial in both X and Y , when cons ...
... The polynomial ring over R in two variables X, Y is defined to be R[X, Y ] := R[X][Y ] ∼ = R[Y ][X]. Elements of R[X, Y ] are called polynomials in the indeterminates X and Y with coefficients in R. A monomial in R[X, Y ] is a polynomial which is simultaneously a monomial in both X and Y , when cons ...