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(xiii) For every positive integer n , there exists a ﬁeld with exactly 10n
elements.
Solution. False.
(xiv) For every positive integer n , there exists a ﬁeld with exactly 9n
elements.
Solution. True.
(xv) There exists a ﬁeld E of characteristic 2 such that x 4 + x + 1 is a
product of linear factors in E [x ].
Solution. True.
3.93 For every commutative ring R , prove that R [x ]/(x ) ∼ R .
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Solution. Deﬁne ϕ : R [x ] → R by
f (x ) → constant term of f (x ).
It is easy to check that ϕ is a surjective homomorphism with ker ϕ = (x ).
3.94 (Chinese Remainder Theorem in k [x ])
(i) Prove that if k is a ﬁeld and f (x ), f (x ) ∈ k [x ] are relatively
prime, then given b(x ), b (x ) ∈ k [x ], there exists c(x ) ∈ k [x ]
with
c − b ∈ ( f ) and c − b ∈ ( f );
moreover, if d (x ) is another common solution, then c − d ∈ ( f f ).
Solution. Every solution of c − b ∈ ( f ) has the form c = b + g f ,
where g (x ) ∈ k [x ]. Hence, we must ﬁnd g such that
g f − (b − b ) ∈ ( f ).
Since ( f , f ) = 1, there are polynomials s , t ∈ k [x ] with s f +
t f = 1. Deﬁne g = s (b − b). Then
g f = s f (b − b )
= (1 − t f )(b − b )
= (b − b ) − t (b − b ) f ∈ ( f ).
If d is another common solution, then both f and f divide c − d .
By Exercise 3.94, f f  c − d , and so c − d ∈ ( f f ).
(ii) Prove that if k is a ﬁeld and f (x ), g (x ) ∈ k [x ] are relatively prime,
then
k [x ]/( f (x )g (x )) ∼ k [x ]/( f (x )) × k [x ]/(g (x )).
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Solution. We adapt the proof of Theorem 2.131. ...
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 Fall '11
 KeithCornell
 Prime number, positive integer, Integral domain, common solution

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