TEST II

1. (34) a. Define: A homomorphism of rings. φ : R → S is a ring homomorphism if for all x, y ∈ R, φ(x + y) = φ(x + φ(y); φ(xy) = φ(x)φ(y) and φ(1R) = 1S. 2 b. Is φ : Z[x] → Z[x] given by φ(f(x)) = f(x ) a ring homomorphism? Why? Yes. It is a ring homomorphism. In fact, it is the unique ring homomorphism that extends the 2 identity on Z to a ring homomorphism on Z[x] by mapping x to x . Since any ring homomorphism must map the multiplicative identity to multiplicative identity, 2 there is only one ring homomorphism, identity map, from Z → Z. Then if we let φ(x) = x ∈ Z[x], this gives a unique ring homomorphism φ that extends identity on Z to φ : Z[x] → Z[x]. This is the map φ(f) = f(x2). c. Using Eisenstein’s criterion or otherwise, prove that for all odd prime numbers p, the polyno- mial 1 + x + x2 + ... + xp−1 is irreducible. 2 p−1 Let f(x) = 1 + x + x + ... + x . φ : Z[x] → Z[x] defined by φ(x) = x + 1 is a ring isomorphism because φ−1(x) = x − 1 is also a ring homomorphism. So, f(x) is irreducible if and only if f(x + 1) = φ(f) is irreducible. p p p−1 Pp−1 p
p−i−1 Now, f(x) = x −1/x−1 and hence φ(f) = f(x+1) = ((x+1) −1)/x = x + i=1 i x . p
Since for any prime p, the binomial coefficients i are all multiples of p, and the constant term of f(x + 1) = p is not a multiple of p2, the monic polynomial f(x + 1) satisfies the Eisenstein’s criterion. So, it is irreducible. Hence f(x) is also irreducible. 2 d. How many elements are there in Z[x]/(5, x + x − 2) ? 2 2 Since 5 is in the ideal, this ring Z[x]/(5, x + x − 2) = (Z/5Z)[x]/(x + x − 2) and hence a vector 2 space over Z/5Z generated by 1 and x, for x = 2 − x and hence is a two dimensional vector space of the field with five elements. So it has 52 = 25 elements. Alternately, every polynomial f(x) satisifies, f(x) = q(x)(x2 +x−2)+r(x) where r(x) has degree 2 2 2 ≤ 1. Hence, Z[x]/(5, x + x − 2) = {f(x) + (5, x + x − 2)|f ∈ Z[x]} = {a + bx + (5, x + x − 1)|0 ≤ 2 1 ≤ 4, 0 ≤ b ≤ 4}. Hence the number of elements in Z[x]/(5, x + x − 2) is 5 × 5 = 25.

II (50) Prove or Disprove: If true prove it. If not, give an example to show why it is not true. You should then justify your example. i If R is an infinite ring which is not a domain, then it has infinitely many zero divisors. True: Since R is not a domain, there are two non zero elements x, y ∈ R such that xy = 0. Every element of Rx = {rx|r ∈ R} is a zero divisor since rxy = r0 = 0 by associativity of multiplication. t If Rx is finite, with Rx = {r1, . . . rt}, then R = ∪i=1{r|rx = ri}. Since R is infinite, one of these t sets, say the first one, must be infinite. So, {r|rx = r1} is infinite. So, {r − r1|r − r1(x) = 0} is an infinite set of left zero divisors in R. ii. If a commutative ring R contains only one prime ideal, then every element of R that is not a unit must be nilpotent. True: 1 2 TEST II

Let x be an element of R that is neither a unit nor nilpotent. Then the ideal generated by x which is not the whole ring must be contained in a maximal ideal, which is prime p. Now, since x is not nilpotent, S = {1, x, x2, . . . xn,...} is a multiplicatively closed subset of R which does not contain 0. So by Zorn’s lemma there is an ideal q which is disjoint from S and is maximal with respect to inclusion which is a prime ideal as well. But x ∈ p and x∈ / q. So, R now has two prime ideals! A contradiction. iii. There are no fields with exactly 27 elements. False: Let F3 = {0, 1, 2} be the field with three elements of integers modulo 3. Then the degree 3 2 three polynomial x + x + 2 has no roots in F3 since none of the three elements 0,1,2 satisfy the 3 2 3 2 polynomial. So x + x + 2 is irreducible in F3[x]. So, F3[x]/(x + x + 2) is a field and it has 33 = 27 elements. iv. All fields are simple but not all simple rings are fields. v. In a finite commutative ring every prime ideal is maximal.

True: Let R be a finite commutative ring and p be a prime ideal in R. Let m be an ideal containing p. Then R/p is an integral domain, since p is prime and it is finite since R is finite. Let R/p be an integral domain with n elements. p ⊂ m. If x ∈ m and x 6 inp, then {x + p, x2 + p, . . . , xn+1 + p} cannot all be different. So, xi + p = xj + p for some i < j. Then xj − xi ∈ p. So, xi(1 − xj−i) ∈ p x∈ / p, p prime, and hence 1 − xj−i ∈ p ⊂ m So, 1 = 1 − xj−i + nxj−i ∈ m. Hence m = R. So, p is maximal. III. (16) 1. List all properties P that you know are inherited by R[x] from a ring R. You don’t have to prove any of it. Commutativity, domain, factorial. 2. List any property P that you know is lost going from a ring R to a polynomial ring R[x]. Give an example to justify. PID, field, finite ring.