Question #4996

Prove that 5Z is both a prime and maximal ideal of Z

Expert's answer

Let& R be a ring. A two sided ideal I of R is called Maximal Ideal if I is not equal to R and no proper ideal of R properly contains I.

For suppose that I is an ideal of Z properly containing& 5I, then there exists some m ε I but m does not in 5I, i.e. 5 does not divide m. Then gcd(5,m)=1 , since& 5 is prime.

Therefore, we can write

1= 5s + m t

for integers s and t. Since 5s& in I& and mt is also in I, this means 1 in I& and hence I = Z.

Therefore, 5Z is a maximal ideal.

Let R be a commutative ring. An ideal I of R is called prime if I not eqal to R. And whenever ab in I for some a, b in R, then either a in I or b in I.

The ideal 5Z is prime in Z, since the product of two integers is a multiple of 5 only if at least one of the two is a multiple of 5.

For suppose that I is an ideal of Z properly containing& 5I, then there exists some m ε I but m does not in 5I, i.e. 5 does not divide m. Then gcd(5,m)=1 , since& 5 is prime.

Therefore, we can write

1= 5s + m t

for integers s and t. Since 5s& in I& and mt is also in I, this means 1 in I& and hence I = Z.

Therefore, 5Z is a maximal ideal.

Let R be a commutative ring. An ideal I of R is called prime if I not eqal to R. And whenever ab in I for some a, b in R, then either a in I or b in I.

The ideal 5Z is prime in Z, since the product of two integers is a multiple of 5 only if at least one of the two is a multiple of 5.

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