Discussion
Let \(x\) be the image of X in the quotient ring \(A := \mathbf{R}[X, Y]/\langle X^{2} + Y^{2} - 1\rangle\). Then the element \(x\) is :
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.
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