Abstract algebra true/false?
(a) The polynomial a_n x^n +· · · +a_1 x+a_0 ∈ R[x] is equal to 0 if and only if a_i = 0 for i = 0, 1, . . . , n.
(b) If R is any ring and f (x), g (x) ∈ R[x] are both of degree 2, then f (x)g (x) is of degree 4.
(c) Given any ring R, x is never a divisor of zero in R[x].
(d) 4x − 8 is irreducible over Q.
(e) If F is a field and if f (x) ∈ F [x] satisfies f (α) = 0 for all α ∈ F , then f (x) = 0.
(f ) If a polynomial f (x) ∈ F [x] has no zeros then it is irreducible.
(g) Every subring of Z is an ideal in Z.
(h) Every subring of Q is an ideal in Q.
(i) An ideal N in a ring R with unity 1 is all of R if and only if 1 ∈ N .
(j) If N is an ideal in R and φ : R → R′ is a ring homomorphism, then φ[N ] is an ideal in R′.
(k) If N′ is an ideal in R′ and φ : R → R′ is a ring homomorphism, then φ^−1 [N′] is an ideal in R.
(l) If R has no divisors of 0, then every ideal of R is a prime ideal.
(m) If R is a finite commutative ring with unity then every prime ideal in R is a maximal ideal.
Comments
(a) if by "0" you mean the 0-polynomial, then yes. if by "0" you mean the RING element 0, then no.
(b) hint: consider 2x^2 and 3x^2 in Z6[x]
(c) true. let the leading coefficient of f(x) be non-zero (which is the case when f(x) = 0). then the leading coefficient of xf(x) is non-zero, so xf(x) ≠ 0.
(d) yes. note that 4x - 8 = 4(x - 2), x -2 is irreducible over Z, and 4 is a unit in Q.
(e) rediculous. consider x^2 + x in Z2[x].
(f) again, rediculous. consider x^4 + 2x^2 + 1 = (x^2 + 1)^2 in Q[x], which is clearly reducible over Q, but has no roots in Q (or even R, for that matter).
(g) every subring in Z is of the form nZ, for some n in Z, which in turn is the ideal generated by n.
(h) Z is a subring of Q, but it is not an ideal of Q (the element (1/2)k is not in Z if k is an odd integer).
(i) if 1 ∈ N, then clearly r = r1 ∈ N, for all r in R, so N = R. on the other hand, if N = R, and R is a ring with unity, then 1 ∈ N since 1 ∈ R.
(j) clearly φ(N) is an additive subgroup of R'. but the trouble is, φ may not be surjective. suppose S is some other ring, and we take N = R = S[x], and R' = S[x,y], with φ(f(x)) = f(x).
then y is in S, and x is in φ( N), but yx is not in φ(N).
(k) again we have φ^-1(N') an additive subgroup of R. but here, we DO have an ideal:
let r be any element of R. and let x be in φ^-1(N'), so that φ(x) is in N'. then φ(rx) = φ(r)φ(x), which is in N', since N' is an ideal of R'. thus rx is in φ^-1(N'), which shows it is an ideal (use xr as well if the ring is non-commutative).
(l) consider (6) which is an ideal of Z, which has no zero divisors. is this a prime ideal?
(m) consider a prime ideal P, and the ring R/P. now suppose (a+P)(b+P) = ab + P = P.
this means that ab is in P, so either a is in P, or b is in P. so R/P is an integral domain. since R is finite, R/P is also finite. but a finite integral domain D is a field (to see this consider the map:
x→dx for any non-zero d in D. suppose dx = dy. then dx - dy = d(x - y) = 0. since D is an integral domain, and d is non-zero, x-y must be 0, so x = y. thus this map is injective, and because D is finite it must be surjective as well. so there is some x with dx = 1. thus any non-zero element of D must be a unit, so D is in fact, a field).
but if R/P is a field, then P is a maximal ideal.
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