**Unformatted text preview: **05/15/2002 WED 12:05 FAX 6434330 MOFFITT LIBRARY 001 George M. Bergman Fall 2001, Math 113, Sec. 3 12 Dec., 2001
3108 Etcheverry Hall Final Exam 12:30-3130 1. (20 points, 4 points each.) Find the following.
(a) (1, 2,3,4)30 in S5. (b) The product [a + bx][c +dx] in the ﬁeld Q[x]/( x2 — x — 1), where a, b, c, deQ.
The answer should be given in the form [p + qx] with p, qE Q. (c) A nonzero polynomial over Q having 1J3 l—wsli as a root.
(d) A splitting ﬁeld of x3— 2 over R.
(e) The characteristic of a ﬁeld of 25 elements. 2. (20 points; 4 points each.) For each of the items listed below, either give an example,
or give a brief reason why no example exists. (If you give an example, you do not have to
prove that it has the property stated.) (a) An abelian group which is not cyclic. (b) A ﬁeld which is not a ring.
(c) A ring which is not an integral domain.
(d) A Euclidean domain which is not a unique factorization domain. (e) A unique factorization domain which is not a principal ideal domain. 3. Short proofs (30 points; 6 points each)
(a) Suppose G is a ﬁnite group acting on a set S, and x is an element of S. Recall
that Gx denotes {gx | gE G}, while Gx denotes {g6 G | gx = x}. Prove using only the deﬁnition of action of G on. S (which you are not asked to
give) and facts about groups that |Gx| = [G : Gx]. (b) Suppose R is a commutative ring with identity 1 :5 O in which {0} is a maximal
ideal (i.e., a proper ideal which is not contained in any larger proper ideal). Show that R
is a ﬁeld. Do not assume any result about maximal ideals except the deﬁnition. (0) Show that if a and b are nonzero elements of an integral domain R, and
aR = bR, then a and b are associates (i.e., there exists a unit it such that b m an). 05/15/2002 WED 12:06 FAX 6434330 MOFFITT LIBRARY 002 George M. Bergman, Fall 2001 - 2 - Math 113, Sec.3, Final, 12 Dee, 2001 (C!) Show that x2 — 7 has no root in Q(J§).
(In a similar situation, where the book said “the roots iﬁ of the polynomial x2 -— 3 do not belong to Q(J§) because they cannot be written in the form a + in”, I pointed
out that one needed a proof that iﬁ could not be written in that form, and I showed you
such a proof. The proof you give of the statement here can be the analog of my proof of
the above fact, or a different proof if you have one.) (e) Suppose R is a commutative ring with identity having characteristic m > O and S
is a commutative ring with identity having characteristic n > 0. Show that the
characteristic of the ring R EB S is the least common multiple of m and n. 4. (30 points, 3 points each) Below, a theorem and a corollary are proved. After certain
steps of the proof I have inserted parenthetical questions such as “(El Why?)”. Give your
explanation at the bottom of the page after the boxed number corresponding to each _
question. Your reasons can be results proved in the text (you don’t have to specify their
statement-numbersl), observations about the given situation, or calculations. Do not worry
about giving further reasons to support your reasons; one key fact or calculation is what is
wanted in each case. Note also that if you can’t justify some step, you may still assume it
in justifying later steps. Throughout this question, (a +bi) will denote the ideal generated by a + bi in Z[i],
while (12) will denote the ideal generated by n in Z. Thus, Z/(n) is the ring we have
more often called Zn. When a symbol of the form [x] is used, if it is referred to as a
member of Z/(n), it will denote the congruence class of an integer xEZ by (n), while
if it is referred to as a member of Z[i]/(a + bi) it will denote the congruence class of an
element xEZ[i] by (a+bi). Theorem. If a and b are relatively prime integers, and n = a2+b2, then
Z[i]/(a +bi) E Z/(n), as rings.
Proof. Let (p: Z ——> Z[i]/(a + bi) be the homomorphism deﬁned by @(k) =
[k] eZ[i]/(a + bi) for all ke Z. We shall show this homomorphism is surjective, and has
kernel (11). These facts imply the desired conclusion, Z[i]/(a + bi) E Z/(n). ( Why?)
To prove surjectivity, ﬁrst note that there exist integers x and y such that
ax+by = 1. ( Why?) Hence the product (a+ bi) (y +xi) = (ay -— bx) + (ax+ by)i
equals c + i, where c = ay ~— beZ. Hence in Z[i] we have i2 —-c (mod (a+bi))
( why?), hence in Z[i]/(a+bi), [i] = [—0]. Hence any [p +qi] EZ[i]/(a+bi) can
be written [p]+[q][i] = [p]+[ql[—c] = [p+q(—c)] = tom-c1 c)Eto(Z). So (9 is
SUI‘JCCtiVB. 05/15/2002 WED 12:06 FAX 6434330 MOFFITT LIBRARY 003 George M. Bergman, Fall 2001 - 3 - Math 113, Sec.3, Final, 12 Dec, 2001 Let us now determine ker(go). We note that if meker(go), then m must be a multiple of a + bi in Z[i] ( why?), that is, we can write
(1) m = (a+bi)(u+vi) where it, 1161. Expanding the above equation and looking at its imaginary part, we get 0 = av+bu, hence
(2) av=—bu, so that a| bu. This in turn implies a| u ( why?). Writing u = aw (we Z),
substituting into (2), and cancelling a. we get 11 = mbw. Substituting these formulas for
u and i) into (1) we get m = (a+ bi)(aw—bwi) =2 w(a +bi)(a—bi) = w(a2+b2) =
wne (n). This shows that ker(qo) g (n). On the other hand, n = (a +bi)(a—bi)e(a+ bi), so nekcr(cp). Hence (n) = ker((p) (El why?), completing the
proof. El Corollary. For any integer n > 1, the following conditions are equivalent:
(a) The ring Z/(n) contains a square root of [—1].
(b) There exist relatively prime integers a and 1) such that (12+!)2 = n. Proof. Assuming (a), let c be an integer such that [c]2 = [— 1] in Z/(n). Let us
deﬁne (p: Z[i] —> Z/(n) by @(x+yi) = [x +yc]. This clearly respects addition; a quick
calculation using the equation [c]2 = [a 1] shows that it also respects multiplication, i.e.,
that (p((x +yi)(x’+y’i)) = (p(x+yi) rp(x’+y’i). ( Show this calculation.) Hence (p is a
ring homomorphism. Hence ker(qo) = (a + bi) for some a, beZ ( why?). I claim that a and b are relatively prime. For if they had a nonunit common divisor
d, then every element of (a + bi) would be a multiple of d, hence the real and
imaginary parts of any such element (i.e., the coefﬁcients of 1 and i) would be divisible
by d in Z. However c —- i e (a + bi) (El why?), and its imaginary part, — 1, is not
divisible by any nonunit. ' Note also that rp is surjective (has all of Z/(n) as its image) ( why?). Now by the Fundamental Homomorphism Theorem for Rings, the image of (p is
isomorphic to Z[i]/ker(rp) = Z[i]/(a + bi), which by the preceding theorem is
isomorphic to Z/(a2+ b2). Hence Z/(n) 3 Z/(a2+ b2). The ﬁrst of these rings has n 2+ b2 elements, so n = 02+ b2, proving (b). elements and the second has a
Conversely, assuming (13), the preceding theorem tells us that Z[i] /(a + bi) E Z/(n).
The former ring has a square root of —1 (that is, a square root of the negative of its multiplicative identity element), namely [i], hence so must the latter, proving (a). El ...

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