# Uniqueness of prime ideal

Let $$K$$ be a field and $$R$$ be the Ring $$K[x]/(X²)$$ then prove that $$R$$ has exactly one prime ideal.

Here,R is a commutative ring with identity, and that it is a quotient of the polynomial ring K[x], which is itself a domain. Hence, R is also a domain.

Let I be an ideal of R. Then, I corresponds to an ideal J of K[x] that contains the ideal (x²). By the correspondence theorem for ideals, J/(x²) is an ideal of R. Moreover, we have a natural isomorphism

R/(J/(x²)) ≅ (K[x]/J)/(x²)

By the third isomorphism theorem, we have

(K[x]/J)/(x²) ≅ K[x]/(x², J)

Thus, we need to find all ideals J of K[x] such that (x², J) is a prime ideal.

Suppose J = (f) for some non-constant polynomial f in K[x]. Then, (x², J) = (x², f). Note that x² is irreducible in K[x], since it has no roots in K. Hence, (x², f) is a prime ideal if and only if one of the following holds:

1.x² is in (f), i.e., f is a multiple of x².

2.f is in (x²), i.e., f is a polynomial of the form g(x)x² for some polynomial g(x) in K[x].

In the first case, (f) contains (x²), and hence I = J/(x²) contains the zero element of R. Since R is a domain, this means that I = {0}. In the second case, (x², f) contains the element x in R, since x = f(x) mod (x²). Hence, I contains x + J/(x²). Since R is a domain, this means that I is a prime ideal.

Therefore, the prime ideals of R are precisely the ideals of the form J/(x²), where J is a non-constant polynomial in K[x] that is divisible by x². There is exactly one such polynomial up to multiplication by a unit in K[x], namely x² itself. Hence, R has exactly one prime ideal, namely (x)/(x²).

Also Examine whether $$R$$ is isomorphic as a ring to the product ring $$k × k$$ (with coordinate wise addition and multiplication).

$k[x]/(x^2)$ has only one prime ideal here is the answer of my first questions,but please have a look of this approach and please help by giving your arguments...

• As for the later question: the ring's local and therefore can't be a product of two nonzero rings. Commented Mar 29, 2023 at 14:35

False: $$0$$ is a root of $$X^2$$. Furthermore, a polynomial needs not have a root in K to be reducible: for example $$(X^2+1)^2$$ is obviously reducible in $$\mathbb R$$, but has not root in $$\mathbb R$$, and your polynomial $$X^2= X\cdot X$$ is even more obviously reducible.
Note: this also implies that $$R$$ is not a domain, since $$X\cdot X = 0$$ in $$R$$. But $$R$$ is still a commutative ring and this suffices.
The proof was globally OK up to this point. So, I continue from this stage. You have shown that you need to find all ideals $$J$$ such that $$(X^2, J)$$ is a prime ideal of $$K[X]$$. But $$K[X]$$ is a principal ideal domain, hence the generator $$f$$ of $$J$$ must divide $$X^2$$ and by prime. This implies $$f = X$$. Consequently, $$J = (X)$$, and there can be only one prime ideal in $$K[X]/(X^2)$$.