title: Irreducible ideal
text: # Irreducible ideal
  
  In mathematics, a proper ideal of a commutative ring is said to be irreducible if it cannot be written as the intersection of two strictly larger ideals.
  
  ## Examples
  
  * Every prime ideal is irreducible. Let $J$ and $K$ be ideals of a commutative ring $R$, with neither one contained in the other. Then there exist $a\in J\setminus K$ and $b\in K\setminus J$, where neither is in $J\cap K$ but the product is. This proves that a reducible ideal is not prime. A concrete example of this are the ideals $2\mathbb {Z}$ and $3\mathbb {Z}$ contained in $\mathbb {Z}$. The intersection is $6\mathbb {Z}$, and $6\mathbb {Z}$ is not a prime ideal.
  * Every irreducible ideal of a Noetherian ring is a primary ideal, and consequently for Noetherian rings an irreducible decomposition is a primary decomposition.
  * Every primary ideal of a principal ideal domain is an irreducible ideal.
  * Every irreducible ideal is primal.
  
  ## Properties
  
  An element of an integral domain is prime if and only if the ideal generated by it is a non-zero prime ideal. This is not true for irreducible ideals; an irreducible ideal may be generated by an element that is not an irreducible element, as is the case in $\mathbb {Z}$ for the ideal $4\mathbb {Z}$ since it is not the intersection of two strictly greater ideals.
  
  An ideal I of a ring R can be irreducible only if the algebraic set it defines is irreducible (that is, any open subset is dense) for the Zariski topology, or equivalently if the closed space of spec R consisting of prime ideals containing I is irreducible for the spectral topology. The converse does not hold; for example the ideal of polynomials in two variables with vanishing terms of first and second order is not irreducible.
  
  If k is an algebraically closed field, choosing the radical of an irreducible ideal of a polynomial ring over k is exactly the same as choosing an embedding of the affine variety of its Nullstelle in the affine space.
  
  