Ring Structure of R

The given ring is \( R = \mathbb{Z}[X]/(x^2 + 1) \). This is a quotient ring formed by taking the polynomial ring \( \mathbb{Z}[X] \) and modding out by the ideal generated by the polynomial \( x^2 + 1 \). The elements of \( R \) are cosets of the form \( f(X) + (x^2 + 1) \), where \( f(X) \in \mathbb{Z}[X] \). The natural quotient map \( \psi : \mathbb{Z}[X] \to R \) sends a polynomial \( f(X) \) to its corresponding coset \( f(X) + (x^2 + 1) \).

We will examine each of the given statements about this ring \( R \).

For convenience, we can use the fact that \( R \) is isomorphic to the ring of Gaussian integers, \( \mathbb{Z}[i] = \{a + bi \mid a, b \in \mathbb{Z}\} \). This isomorphism is given by the map \( \phi: R \to \mathbb{Z}[i] \) where \( \phi(aX + b + (x^2 + 1)) = b + ai \). Let's verify this isomorphism briefly.

Consider the evaluation homomorphism \( \Phi: \mathbb{Z}[X] \to \mathbb{Z}[i] \) defined by \( \Phi(f(X)) = f(i) \). The image of \( \Phi \) is \( \mathbb{Z}[i] \). The kernel of \( \Phi \) is \( \{f(X) \in \mathbb{Z}[X] \mid f(i) = 0\} \). Since \( x^2 + 1 \) is the minimal polynomial of \( i \) over \( \mathbb{Q} \) and is monic in \( \mathbb{Z}[X] \), the kernel is the ideal \( (x^2 + 1) \) in \( \mathbb{Z}[X] \). By the First Isomorphism Theorem, \( \mathbb{Z}[X]/(x^2 + 1) \cong \mathbb{Z}[i] \).

Statement 1: Ring Isomorphic to a Subring of \( \mathbb{C} \)

Statement 1 says that \( R \) is isomorphic to a subring of \( \mathbb{C} \). As shown above, \( R \cong \mathbb{Z}[i] \). The set of Gaussian integers \( \mathbb{Z}[i] \) is a subset of the complex numbers \( \mathbb{C} \), and it forms a ring under the usual addition and multiplication of complex numbers. Thus, \( \mathbb{Z}[i] \) is a subring of \( \mathbb{C} \).

Since \( R \) is isomorphic to \( \mathbb{Z}[i] \) and \( \mathbb{Z}[i] \) is a subring of \( \mathbb{C} \), it follows that \( R \) is isomorphic to a subring of \( \mathbb{C} \).

Statement 1 is true.

Statement 2: Proper Ideal Generated by \( \psi(p) \)

Statement 2 says that for any prime number \( p \in \mathbb{Z} \), the ideal generated by \( \psi(p) \) is a proper ideal of \( R \). The element \( \psi(p) \) in \( R \) is the coset \( p + (x^2 + 1) \). The ideal generated by \( \psi(p) \) is \( (\psi(p)) \).

An ideal \( I \) in a ring \( S \) is proper if \( I \ne S \). The ideal generated by an element \( a \) is the whole ring \( S \) if and only if \( a \) is a unit in \( S \). So, the ideal \( (\psi(p)) \) is proper if and only if \( \psi(p) \) is not a unit in \( R \).

Using the isomorphism \( R \cong \mathbb{Z}[i] \), the element \( \psi(p) = p + (x^2 + 1) \) in \( R \) corresponds to the integer \( p \) in \( \mathbb{Z}[i] \). The ideal \( (\psi(p)) \) in \( R \) corresponds to the ideal generated by \( p \) in \( \mathbb{Z}[i] \), which is \( (p) = \{p \cdot z \mid z \in \mathbb{Z}[i]\} \).

For the ideal \( (p) \) to be proper in \( \mathbb{Z}[i] \), the element \( p \) must not be a unit in \( \mathbb{Z}[i] \). The units in \( \mathbb{Z}[i] \) are the elements with norm 1, which are \( 1, -1, i, -i \).

Given that \( p \) is a prime number in \( \mathbb{Z} \), \( p \ge 2 \). Therefore, \( p \) is not equal to \( 1, -1, i, \) or \( -i \). Thus, \( p \) is not a unit in \( \mathbb{Z}[i] \).

Since \( p \) is not a unit in \( \mathbb{Z}[i] \), the ideal \( (p) \) is a proper ideal in \( \mathbb{Z}[i] \). Consequently, the ideal generated by \( \psi(p) \) is a proper ideal of \( R \).

Statement 2 is true.

Statement 3: Infinitely Many Prime Ideals

Statement 3 claims that \( R \) has infinitely many prime ideals. Since \( R \) is isomorphic to \( \mathbb{Z}[i] \), \( R \) has infinitely many prime ideals if and only if \( \mathbb{Z}[i] \) has infinitely many prime ideals.

\( \mathbb{Z}[i] \) is a Principal Ideal Domain (PID), which implies it is also a Unique Factorization Domain (UFD). The non-zero prime ideals in a PID are precisely the ideals generated by irreducible elements.

The irreducible elements in \( \mathbb{Z}[i] \) (up to associates) are closely related to the prime numbers in \( \mathbb{Z} \):

• The element \( 1+i \) is irreducible in \( \mathbb{Z}[i] \), and the ideal \( (1+i) \) is a prime ideal. It lies over the prime 2 in \( \mathbb{Z} \).
• For any prime number \( p \in \mathbb{Z} \) such that \( p \equiv 3 \pmod 4 \), \( p \) remains irreducible in \( \mathbb{Z}[i] \). The ideal \( (p) \) is a prime ideal in \( \mathbb{Z}[i] \).
• For any prime number \( p \in \mathbb{Z} \) such that \( p \equiv 1 \pmod 4 \), \( p \) factors into two non-associate irreducible elements in \( \mathbb{Z}[i] \), say \( p = \pi \bar{\pi} \). The ideals \( (\pi) \) and \( (\bar{\pi}) \) are distinct prime ideals in \( \mathbb{Z}[i] \).

By Dirichlet's theorem on arithmetic progressions, there are infinitely many prime numbers \( p \) in \( \mathbb{Z} \) such that \( p \equiv 3 \pmod 4 \). Each such prime \( p \) gives rise to a distinct prime ideal \( (p) \) in \( \mathbb{Z}[i] \).

Also, by Dirichlet's theorem, there are infinitely many prime numbers \( p \) in \( \mathbb{Z} \) such that \( p \equiv 1 \pmod 4 \). Each such prime \( p \) gives rise to two distinct prime ideals \( (\pi) \) and \( (\bar{\pi}) \) in \( \mathbb{Z}[i] \).

Therefore, combining these cases, there are infinitely many distinct prime ideals in \( \mathbb{Z}[i] \).

Since \( R \cong \mathbb{Z}[i] \), \( R \) has infinitely many prime ideals.

Statement 3 is true.

Statement 4: Prime Ideal Generated by \( \psi(X) \)

Statement 4 asks if the ideal generated by \( \psi(X) \) is a prime ideal in \( R \). The element \( \psi(X) \) is \( X + (x^2 + 1) \). The ideal generated by \( \psi(X) \) in \( R \) is \( (\psi(X)) \).

An ideal \( I \) in a ring \( S \) is a prime ideal if and only if the quotient ring \( S/I \) is an integral domain. Let's examine the quotient ring \( R/(\psi(X)) \).

Using the Third Isomorphism Theorem, this is isomorphic to \( \mathbb{Z}[X] / ( (x^2 + 1) + (X) ) \), which is \( \mathbb{Z}[X] / (x^2 + 1, X) \).

The ideal \( (x^2 + 1, X) \) in \( \mathbb{Z}[X] \) is the ideal generated by \( x^2 + 1 \) and \( X \). Since \( X \) is in the ideal, any polynomial \( f(X) \) in this ideal must be divisible by \( X \) plus some multiple of \( x^2+1 \). Specifically, since \( X \in (x^2 + 1, X) \), we can write \( x^2 \in (x^2 + 1, X) \). Since \( x^2 + 1 \in (x^2 + 1, X) \), we have \( (x^2 + 1) - x^2 = 1 \in (x^2 + 1, X) \). Since the ideal contains the element \( 1 \), it must be the entire ring \( \mathbb{Z}[X] \).

So, \( (x^2 + 1, X) = \mathbb{Z}[X] \).

Therefore, \( R/(\psi(X)) \cong \mathbb{Z}[X] / \mathbb{Z}[X] \), which is the zero ring, denoted by \( \{0\} \).

The zero ring \( \{0\} \) contains only one element (the zero element). An integral domain must contain at least two distinct elements (the zero element and the multiplicative identity). Since \( R/(\psi(X)) \) is the zero ring, it is not an integral domain. Therefore, the ideal \( (\psi(X)) \) is not a prime ideal.

Alternatively, using the isomorphism \( R \cong \mathbb{Z}[i] \), the element \( \psi(X) = X + (x^2 + 1) \) corresponds to the element \( i \) in \( \mathbb{Z}[i] \). The ideal generated by \( \psi(X) \) in \( R \) corresponds to the ideal generated by \( i \) in \( \mathbb{Z}[i] \), which is \( (i) \).

The element \( i \) is a unit in \( \mathbb{Z}[i] \) because \( i \cdot (-i) = 1 \), and \( -i \) is also in \( \mathbb{Z}[i] \). An ideal generated by a unit is the entire ring. Thus, \( (i) = \mathbb{Z}[i] \).

Since the ideal generated by \( \psi(X) \) is the entire ring \( R \), it is not a proper ideal. By definition, a prime ideal must be a proper ideal. Therefore, the ideal generated by \( \psi(X) \) is not a prime ideal.

Statement 4 is false.

Summary of Statements

• Statement 1: \( R \) is isomorphic to a subring of \( \mathbb{C} \). (True)
• Statement 2: For any prime number \( p \in \mathbb{Z} \), the ideal generated by \( \psi(p) \) is a proper ideal of \( R \). (True)
• Statement 3: \( R \) has infinitely many prime ideals. (True)
• Statement 4: The ideal generated by \( \psi(X) \) is a prime ideal in \( R \). (False)

The statements that are true are 1, 2, and 3.