Rings and Ideals Questions and Answers

Answer: B

\(I\) is a prime ideal but not a maximal ideal of \(R\).

The projection \((a, b, c) \mapsto c\) is a surjective ring homomorphism from \(R\) onto \(\mathbb{Z}\) whose kernel is exactly \(I = \mathbb{Z} \times \mathbb{Z} \times \{0\}\). So \(R/I \cong \mathbb{Z}\).

An ideal is prime exactly when the quotient is an integral domain, and maximal exactly when the quotient is a field. \(\mathbb{Z}\) is an integral domain but not a field, so \(I\) is prime and not maximal. Concretely, \(I \subsetneq \mathbb{Z} \times \mathbb{Z} \times 2\mathbb{Z} \subsetneq R\), so \(I\) is not maximal. That rules out the three other options.

Answer: A

\(R\) has exactly 1 maximal ideal.

Let \(\mathfrak{m}\) be any maximal ideal. It is proper, so it contains no unit. Every element outside \(I\) is a unit, so every element of \(\mathfrak{m}\) lies in \(I\): \(\mathfrak{m} \subseteq I\). Since \(I\) is proper and \(\mathfrak{m}\) is maximal, \(\mathfrak{m} = I\).

So every maximal ideal equals \(I\), and \(I\) is the only one: \(R\) is a local ring. Such an \(I\) exists, for example \(R = \mathbb{Z}/4\mathbb{Z}\) with \(I = \{0, 2\}\). The condition that \(R\) is not a field only ensures \(I \neq 0\); it does not create more maximal ideals, so 2, 3 and infinite are wrong.

Answer: A

The answer is P implies Q but Q does not imply P.

P implies Q: take \(r_{1} = 1\) and \(r_{2} = 0\). P gives an r with \(r - 1 \in J_{1}\) and \(r \in J_{2}\). Then \(1 = r - (r - 1) \in J_{2} + J_{1}\), so \(J_{1} + J_{2} = R\).

Q does not imply P: Q gives existence through the Chinese remainder theorem, but not uniqueness. In \(\mathbb{Z}\) with \(J_{1} = 2\mathbb{Z}\) and \(J_{2} = 3\mathbb{Z}\), the ideals are comaximal, yet if r solves the two congruences so does r + 6. Uniqueness in P actually forces \(J_{1} \cap J_{2} = 0\), which Q does not give.

So the options saying Q implies P, or that neither implies the other, are wrong.

Answer: D

The ring that need not have an identity is the ring of polynomials \(P_{0}[\mathbf{R}]\) vanishing at the origin.

The polynomials with zero constant term are closed under addition and multiplication, so they form a ring. Suppose \(e(x)\) were an identity. Then \(e(x)\cdot x = x\) for the polynomial \(x\), so \(e(x) = 1\), and the constant 1 does not vanish at the origin. So no identity exists in this ring.

The other rings all have one. Homomorphisms of a vector space to itself (linear maps) form a ring whose identity is the identity map. Upper triangular \(n \times n\) matrices have the identity matrix \(I\), which is upper triangular. The automorphisms contain the identity map too; as printed this option is loosely worded, because invertible maps are not closed under addition and so are not a ring in the strict sense, but whichever way it is read, the identity map belongs to it.

Answer: A

The possible value is \(n = 32\).

In a Boolean ring every element satisfies \(x^{2} = x\). Expanding \((x + x)^{2} = x + x\) gives \(4x = 2x\), so \(2x = 0\) and the ring has characteristic 2 and its additive group is a vector space over \(\mathbf{Z}_{2}\). A finite one therefore has \(2^{k}\) elements; in fact a finite Boolean ring with identity is isomorphic to \(\mathbf{Z}_{2}^{k}\).

\(32 = 2^{5}\) is a power of 2. 23 is prime, 27 is \(3^{3}\) and 15 is \(3 \times 5\), none of them a power of 2.

Answer: C

The true statement is wherever R is finite, \(m \ne 0\).

In a finite ring the elements \(1, 1 + 1, 1 + 1 + 1, \ldots\) cannot all be distinct, so \(k \cdot 1 = 0\) for some positive \(k\), and the characteristic is positive.

The other statements fail. \(\mathbf{Z}_{p}[x]\) is infinite with characteristic \(p\), which refutes A (infinite forces \(m = 0\)) and B (\(m \ne 0\) forces finite). \(\mathbf{Z}_{p}\) itself is finite with \(m = p \ne 0\), which refutes D (\(m \ne 0\) forces infinite).

Answer: A

The ring that is not Noetherian is \(\mathbf{R}[x_{1}, x_{2}, \ldots]\).

In the polynomial ring in infinitely many variables, the chain of ideals

\[ \begin{aligned} (x_{1}) &\subset (x_{1}, x_{2}) \\ &\subset (x_{1}, x_{2}, x_{3}) \subset \cdots \end{aligned} \]

is strictly increasing and never stops, so the ascending chain condition fails. The ideal generated by all the variables is not finitely generated.

\(\mathbf{Z}\) and \(\mathbf{Z}[i]\) are PIDs, hence Noetherian, and \(\mathbf{Z}_{205}[x]\) is Noetherian by Hilbert's basis theorem, since \(\mathbf{Z}_{205}\) is a finite ring and so Noetherian.

Answer: B

\(x\) is irreducible, but not prime in A.

Not prime: in A, \(x^{2} = 1 - y^{2} = (1 - y)(1 + y)\), so \(x\) divides the product \((1 - y)(1 + y)\). But \(x\) divides neither factor, since \(A/(x) \cong \mathbf{R}[Y]/(Y^{2} - 1)\), in which \(1 - y\) and \(1 + y\) are non-zero. That quotient is \(\mathbf{R} \times \mathbf{R}\), not a domain, which is another way of saying \((x)\) is not prime.

Irreducible: putting \(x = \cos\theta\), \(y = \sin\theta\) identifies A with the ring of real trigonometric polynomials, in which the degree of a product is the sum of the degrees and the units are the non-zero constants (degree 0). \(x\) has degree 1, so any factorisation \(x = gh\) has a factor of degree 0, a unit. This is the textbook example showing that A is not a UFD.

So 'not irreducible', 'prime' and 'a unit' are all wrong.

Answer: D

The principal ideal is \((2x, 3x, x^{2})\).

Since \(3x - 2x = x\), the ideal contains \(x\); and each generator \(2x, 3x, x^{2}\) is a multiple of \(x\). So \((2x, 3x, x^{2}) = (x)\), which is principal.

The others are not principal. \((2, x)\) consists of polynomials with even constant term; a single generator would have to divide both 2 and \(x\), so it would be \(\pm 1\), but 1 is not in the ideal. The same argument works for \((3, x)\). And \((x, x^{2} - x, x + 4) = (x, 4)\), which fails for the same reason.

Answer: D

The quotient has 4 maximal ideals.

The roots of \(X^{8} + 1\) are the eight primitive 16th roots of unity, none of them real. They form four conjugate pairs, so over \(\mathbf{R}\) the polynomial factors into four distinct irreducible quadratics \(q_{1}q_{2}q_{3}q_{4}\).

Maximal ideals of \(\mathbf{R}[X]/\langle X^{8} + 1 \rangle\) correspond to maximal ideals of \(\mathbf{R}[X]\) containing \(X^{8} + 1\), that is, to its distinct irreducible factors. By the Chinese remainder theorem the ring is \(\mathbf{C}^{4}\), with exactly 4 maximal ideals.