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...
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 $K[X]/(X^2)$.