Computations/Applications 1. Find the inverse of x + 1 in the ring F 5 [x]/(x 3 1). Solution: We use the Euclidean Algorithm: x 3 1 (x + 1)(x + 4x + 1) + 3 (x + 1) 3(x + ) + 0. Thus 3 (x 3 1) + (x + 1)(4x + x + 4) and after multiplying by 3 1 we get 1 (x 3 1) + (x + 1)(3x + x + 3) showing that (x + 1)(3x + x + 3) 1 (mod x 3 1) and hence the inverse is 3x + x + 3.. Compute gcd(8 + 5, 3 + ) in the ring Z[ ]. Solution: Again, we use Euclidean algorithm. To find the first quotient, we look at 8 + 5 3 + (8 + 5 )(3 ) (3 + )(3 ) 4 4 1 So in fact 3 + divides 8 + 5 and hence gcd(8 + 5, 3 + ) 3 +. Interestingly, 3 + is a unit in Z[ ] since it has norm 1. This means we didn t even have to do the division to recognize that 3 + divides 8 + 5. 3. Find all ideals in Z[i] containing the element 33 + 9i. Solution: Since Z[i] is a PID, it suffices to find all divisors of 33 + 9i. We can do this by factoring 33 + 9i into irreducible elements. The easiest way to do this is to factor N( 33 + 9i) 1170 9 5 13. We ve classified all the irreducible elements of Z[i]. They are, up to associates, elements with prime norm equal to or a prime congruent to 1 (mod 4), and primes in Z congruent to 3 (mod 4). Since 3 and 1 + i are the only two irreducibles (up to multiplication by a unit) of norms 9 and, we see that 3 + 3i divides 33 + 9i. Dividing yields 33 + 9i 3 + 3i 11 + 3i 1 + i ( 11 + 3i)(1 i) 8 + 14i 4 + 7i.
Up to multiplication by units, the only elements of norm 5 are 1 + i and 1 i. Only one of these elements will divide 4 + 7i, and you can see that it is 1 + i since 4 + 7i 1 + i Thus we have the prime factorization: ( 4 + 7i)(1 i) 5 10 + 15i 5 33 + 9i (1 + i) 3 (1 + i) ( + 3i). + 3i. This means that there are 16 ideals containing 33 + 9i, given by multiplying these factors in all possible ways. For your convenience, they are (ordered by norm) (1) (1 + i) (1 + i) (3) ( 1 + 3i) ( + 3i) (3 + 3i) ( 1 + 5i) (3 + 6i) ( 4 + 7i) ( 3 + 9i) (6 + 9i) ( 11 + 3i) ( 3 + 15i) ( 1 + 1i) ( 33 + 9i). 4. Find all the units in the polynomial ring (Z[ 5])[x, y, z]. Solution: For R an integral domain, we ve seen that (R[x]) R. By induction, this means that (R[x, y, z]) R. Thus the units in (Z[ 5])[x, y, z] are precisely the units in Z[ 5]. We know that α Z[ 5] is a unit if and only if N(α) ±1. Since the norm of a + b 5 is a + 5b, we see that the only units in this ring are ±1. 5. Show that the polynomial x 3 + x y + xy + xy + y Q[x, y] is irreducible. Solution: Thinking about this as a polynomial in (Q[y])[x], this element is x 3 + yx + (y + y)x + y. Notice that all the coefficients of this polynomial are divisible by y, and the constant term is not divisible by y. Since y is irreducible, hence prime in the UFD Q[y], Eisenstein tells us that the above polynomial is irreducible. 6. How many irreducible polynomials of degree are there in F 3 [x]? Solution: It s slightly easier to count the number of reducible polynomials of degree. To do this, we will only count reducible monic polynomials. Each must factor into the form (x a)(x b) for some a, b F 3. Since order doesn t matter, when a b, there are ( 3 ) 3 such polynomials with a b, and an additional 3 when a b. Therefore there are 6 reducible monic polynomials of degree, and since F 3 [x] has 3 9 monic quadratic polynomials, we see that there are 9 6 3 irreducible monic quadratic polynomials. Since F, we see that in total there are quadratic irreducible polynomials. 3 6 Page
7. How many elements are there in (F 7 [x]/(x 3 + x + 4x + 1))? Solution: By considering roots, one can factor: x 3 + x + 4x + 1 (x + 4)(x + 5)(x + 6) in F 7 [x]. Thus the Chinese Remainder Theorem gives (F 7 [x]/(x 3 + x + 4x + 1)) (F 7 [x]/(x + 4)) (F 7 [x]/(x + 5)) (F 7 [x]/(x + 6)). Because the three rings on the right hand side are fields of size 7, they each have 6 units. Thus the ring on the left has 6 3 16 units. Proof Problems 1. Let R be a Euclidean Domain with respect to a norm N satisfying N(a) N(ab) whenever a, b are non-zero elements of R. (a) Prove that N(a) N(1) for every a R {0}. Solution: This follows immediately from the definition. We have for any a 0 in R. N(1) N(1 a) N(a) (b) Let m N(1). Show that R {r R {0} N(r) m}. Solution: Let u R. By part (a), we know that N(u) N(1) m. Since u is invertible, there exists v R with uv 1. But then N(u) N(uv) N(1) m and so indeed we get equality. This proves that R {r R {0} N(r) m}. For the opposite inclusion, let x {r R {0} N(r) m}. The element x is non-zero, so there exists q, r R with 1 xq + r where r 0 or N(r) < N(x). By part (a), the norm of x is minimal among non-zero elements in R. This proves that r 0 and so 1 xq and hence x R. REMARK: Given a Euclidean Domain R with norm N, the norm N might not satisfy N(a) N(ab). There is, however, a trick to use the fact that R is a Euclidean Domain to construct a different norm Ñ which still turns R into a Euclidean Domain and also satisfies Ñ(a) Ñ(ab).. Let F be a field. Prove that the ideal (x, y) F [x, y] is not principal. Page 3
Solution: Assume for the sake of contradiction that (x, y) is a principal ideal. The ring F [x, y] is a UFD, so if (x, y) is principal then it must be generated by the GCD of x and y. The elements x and y are both irreducible and non-associate so their GCD is 1. Thus we get (x, y) (1). But then there exist f(x, y), g(x, y) F [x, y] with 1 xf(x, y) + yg(x, y). Plugging in x y 0 yields 1 0 which is a contradiction. 3. Let D { n n 0} Z. Prove that D 1 Z Z[x]/(x 1). Solution: Define a map ϕ: Z[x] Q via f(x) f Since x n 1 n, we see that ϕ(z[x]) D 1 Z. Now ( ) 1. ϕ(x 1) 1 1 0 so (x 1) ker(ϕ) and hence ϕ induces a surjective homomorphism ϕ: Z[x]/(x 1) D 1 Z. To get an inverse to ϕ, consider the injection ι: Z Z[x] and the quotient map The composition gives a ring homomorphism π : Z[x] Z[x]/(x 1). π ι: Z Z[x]/(x 1). Under this map, π(ι()) is a unit since it s inverse is x in Z[x]/(x 1). By the universal property of rings of fractions, this map induces a map ψ : D 1 Z Z[x]/(x 1). By their definitions, the maps ϕ and ψ are inverses of one another, and hence we get an isomorphism D 1 Z Z[x]/(x 1). 4. Let R Z[ 5] and let F Q( 5) be the field of fractions of R. (a) Prove that x + x 1 is irreducible in R[x] yet reducible in F [x]. Page 4
Solution: There are several ways to check that x + x 1 is irreducible in R[x]. Here is one nice way using modular arithmetic. Let I (, 1 + 5) Z[ 5]. I claim that I is a proper ideal of Z[ 5]. To prove this, assume for the sake of contradiction that 1 z + (1 + 5)y for some z, y Z[ 5]. Multiplying through by (1 5) gives 1 5 ((1 5)z) 4y which is absurd since doesn t divides 1 5. Now modulo I, we have 5 1 and 0, so we get that a + b 5 0, 1 (mod I). Since I is proper, we thus see that Z[ 5]/I Z/Z. The polynomial x + x 1 is irreducible in Z/Z, and since this polynomial is monic, we see it must be irreducible in R[x] as well. As for F [x], we can use the quadratic formula. The roots of x + x 1 are x 1 ± 1 + 4 1 ± 5. Since both of these are elements of Q( 5), we see that x + x 1 factors into linears in F [x]. (b) Explain why part (a) guarantees that R is not a UFD. Solution: If R were a UFD, then the content of x + x 1 would be 1 since it is monic. Thus x + x 1 would be irreducible in R[x] if and only if it were irreducible in F [x]. Since this clearly doesn t happen, we see that R cannot be a UFD. 5. Let R be a commutative ring with 1 0 with the property that for every x R, there exists n with x n x. Prove that every prime ideal of R is maximal. Solution: Let P R be a prime ideal of R. Since R is commutative with 1 0, we know that R/P is an integral domain. To prove that P is maximal, it suffices to show that R/P is actually a field. Thus it suffices to prove that if x R P then x has a multiplicative inverse in R/P. To this end, let x R P. Then there exists n > 1 with x n x. This is equivalent to 0 x n x x(x n 1 1). Since 0 P, and P is a prime ideal, this says that either x P or x n 1 1 P. By assumption on x we have x P so in fact 0 x n 1 1 implying that and hence x is invertible in R/P. 1 x x n 1 6. Let R be a finite commutative ring with 1 0. Prove that every non-zero element of R is either a unit or a zero divisor. Page 5
Solution: Let 0 r R. Since R in finite, there must exist positive integers n < m with r n r m. We can rewrite this expression as 0 r m r n r n (r m n 1). If r is a zero-divisor then there is nothing to prove. Otherwise, r is not a zero-divisor and hence r n is also not a zero-divisor. Therefore we can cancel to obtain which shows that r has a multiplicative inverse. r m n 1 0 7. Consider the function ϕ: Z[ ] Z/Z given by ϕ(a + b ) a (mod ). (a) Prove that ϕ is a ring homomorphism. Solution: Let α a + b and β c + d. We have ϕ(α + β) ϕ((a + c) + (b + d) ) a + c (mod ) ϕ(α) + ϕ(β). Similarly, ϕ(αβ) ϕ((ac + bd) + (ad + bc) ) ac + bd (mod ) ac (mod ) ϕ(α)ϕ(β). This proves that ϕ is a homomorphism. (b) Find generators for the ideal ker(ϕ). Solution: The ring Z[ ] is a Euclidean Domain with respect to N(a + b ) a b. Thus ker(ϕ) is a principal ideal generated by a non-zero element in ker(ϕ) whose norm is as small as possible. If a + b ker(ϕ) then a is even, and so N(a + b ) must also be even, and hence N(a + b ). Of course ker(ϕ) and N( ) and so ker(ϕ) ( ). Page 6