Answer to Question #165500 in Abstract Algebra for AbdulAleem

Question #165500

Let ℤ[𝑥] be the polynomial ring with coefficients in ℤ. Prove or disprove 

that the ideal 𝐼 = (4, 𝑥) is a principal ideal in ℤ[𝑥]. Is the ideal 𝐼 a 

maximal ideal in ℤ[𝑥]? Explain your answer.


1
Expert's answer
2021-02-24T06:00:00-0500

The ideal 𝐼 = (4, 𝑥), by the definition, consists of those polynomials over Z, which can be expressed as a sum "4\\cdot u(x) + x\\cdot v(x)".

Every such polynomial f(x) satisfies the condition "f(0)\\equiv 0\\ (mod\\ 4)". Therefore, "1\\notin I, 2\\notin I".

If we assume that I is a principal ideal in ℤ[𝑥], then there exists a polynomial d(x) in I, which divides all elements in I.

d(x) is a divisor of 4. All divisors of 4 are: 1, 2, 4, -1, -2, -4. Only 4 and -4 are elements of I, but no one of them is a divisor of x.

The contradiction proves that I is not a principal ideal.

Since "2\\notin I" , the ideal I' = (2, x) is greater than I. Moreover, "1\\notin I'", because of each polynomial f(x) from I' satisfies the condition "f(0)\\equiv 0\\ (mod\\ 2)".

Therefore, I is not a maximal ideal.


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