Prove that the ideal <x^2+1> is prime in Z[x] but not maximal in Z[x]. Solution

Question

Prove that the ideal <x^2+1> is prime in Z[x] but not maximal in Z[x].

Explanation / Answer

Let R be a commutative ring and I an ideal of R. (i) I is prime R/I is an integral domain. (ii) I is maximal R/I is a field. --------------------------------------… In order to use these facts, we first observe that Z[x]/ = Z[i]. (The "=" stands for isomorphism.) To show this, define a map (phi): Z[x] --> Z[i] via (phi)(f(x)) = f(i). It is easy to see that this is a ring homomorphism. (phi) is onto, since phi(a+bx) = a+bi for any a,b in Z. Moreover, ker(f) = {f(x)| f(i) = 0}: Since f is in Z[x], if i is a root, so is -i. Thus, (x^2+1) | f(x). ==> ker f = . Thus, by the first isomorphism theorem, Z[x]/ = Z[i]. --------------------------------------… Now, this is easy! (i) Z[i] is an integral domain, because Z[i] is a subring of C, and C enjoys the zero product property. Since Z[i] is an integral domain, and Z[x]/ = Z[i], we have that is prime. (ii) Z[i] is not a field, because (1+i)^(-1) = (1/2) + (1/2)i is not in Z[i]. This, combined with Z[x]/ = Z[i], shows that is not maximal.

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