I have a question concerning page 43 of this book. In Corollary 2.7 it says that the map $\mathfrak{p}\mapsto \overline{\{\mathfrak{p}\}}$ is a bijection from Spec($A$) onto the sets of closed irreducible subsets of Spec($A$). It further says that minimal prime ideals of $A$ correspond to the irreducible components of Spec($A$). I don't understand this point. By the first part $\overline{\{0\}}=Spec(A)$ should be irreducible. So $Spec(A)$ has only one irreducible component. And this component corresponds to the zero ideal and not the minimal prime ideals. So what did I get wrong?
[Math] Irreducible components in the spectrum of a ring
algebraic-geometrycommutative-algebra
Related Solutions
In order to find an ideal which doesn't have a primary decomposition, the following construction is useful. Let $R$ be a commutative ring and $M$ an $R$-module. On the set $A=R\times M$ one defines the following two algebraic operations:
$$(a,x)+(b,y)=(a+b,x+y)$$
$$(a,x)(b,y)=(ab,ay+bx).$$
With these two operations $A$ becomes a commutative ring with $(1,0)$ as unit element. ($A$ is called the idealization of the $R$-module $M$ or the trivial extension of $R$ by $M$). Let's list some important properties of this ring:
$\{0\}\times M$ is an ideal of $A$ isomorphic to $M$ (as $R$-modules) and there is a ono-to-one correspondence between the ideals of $R$ and the ideals of $A$ containing $\{0\}\times M$.
$A$ is a Noetherian ring if and only if $R$ is Noetherian and $M$ is finitely generated.
All prime (maximal) ideals of $A$ have the form $P\times M$, where $P$ is a prime (maximal) ideal of $R$.
If $R$ is an integral domain and $M$ is divisible, then all the ideals of $A$ have the form $I\times M$ with $I$ ideal of $R$, or $\{0\}\times N$ with $N$ submodule of $M$.
Now I suggest to consider $R=\mathbb{Z}_{(2)}$ (the localization of $\mathbb{Z}$ at the prime ideal $2\mathbb{Z}$), $M=\mathbb{Q}$, $A=R\times M$ (as before) and $\mathfrak{a}=\{0\}\times H$ with $H$ a proper $\mathbb{Z}_{(2)}$-submodule of $\mathbb{Q}$. There are only two prime ideals of $A$ containing $\mathfrak{a}$ (and one of them is minimal over $\mathfrak{a}$), and $\mathfrak{a}$ has no finite primary decomposition because the primary ideals of $A$ have the following form: $\{(0,0)\}$, $\{0\}\times\mathbb{Q}$ and $2^n\mathbb{Z}_{(2)}\times\mathbb{Q}$, $n\in\mathbb{N}^*$.
Yes, your scheme $X$ has exactly one irreducible component - i.e., it is an irreducible scheme. Note that an affine scheme $\operatorname{Spec} R$ is irreducible if and only if the nilradical of $R$ is prime. Here, the nilradical of $k[X,Y]/(X^2,XY)$ is $(\overline{X})$ (where $\overline{X}$ denotes the image of $X$ in $k[X,Y]/(X^2,XY)$) and this is a prime ideal since $(k[X,Y]/(X^2,XY))/(\overline{X}) \cong k[Y]$. However, you should be careful to understand your claim that "the subscheme $V(X)$" is the irreducible component of $X$ correctly.
Let me explain. There is a natural way to view $X$ as a closed subscheme of $\mathbb{A}_k^2 = \operatorname{Spec} k[X,Y]$. The underlying set of this subscheme contains precisely those prime ideals $\mathfrak{p}$ of $k[X,Y]$ such that $\mathfrak{p} \supset (X^2,XY)$. As A.P.'s comment to your question shows, we have $$ \mathfrak{p} \supset (X^2,XY) \Leftrightarrow \mathfrak{p} \supset (X) .$$ It is customary to express this by writing $$ V(X) = V(X^2,XY), $$ which is perfectly fine as long as you understand this as an equality of sets only (of course, the subspace topology induced by the Zariski topology on $\mathbb{A}_k^2$ then agrees as well, which is why you could also view the above equality as an equality of topological spaces).
However, sometimes $V(X)$ is not used to denote a subset, but to denote a subscheme (and writing "the subscheme $V(X)$" suggests just that) - here, that would be the scheme $\operatorname{Spec} k[X,Y]/(X)$. But $\operatorname{Spec} k[X,Y]/(X^2,XY)$ and $\operatorname{Spec} k[X,Y]/(X)$ are not isomorphic as schemes. In fact, there is a bijective correspondence between the closed subschemes of any affine scheme $\operatorname{Spec} R$ and the ideals of $R$ - but clearly, $(X^2, XY) \neq (X)$.
(Well, $\operatorname{Spec} k[X,Y]/(X)$ is [naturally isomorphic to] a closed subscheme of $\operatorname{Spec} k[X,Y]/(X^2,XY)$ and, strictly speaking, it is true that $\operatorname{Spec} k[X,Y]/(X)$ is the irreducible component of $\operatorname{Spec} k[X,Y]/(X^2,XY)$ because the underlying topological spaces of those two schemes agree. But, at least to me, this seems to be a rather strange way of asserting that $\operatorname{Spec} k[X,Y]/(X^2,XY)$ is irreducible.)
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I now feel a bit silly for writing this pedantic and probably superfluous remark so let me add another comment which might be of more interest to you: While the scheme $X$ is irreducible, it has an embedded prime. For this reason, it might help you to understand the "geometric meaning" of associated primes and primary decompositions - in particular, in what sense they yield more refined information than the decomposition into irreducible components.
Best Answer
$(0)$ is not always in $\operatorname {Spec} A$ is where your reasoning went wrong. One can show rather straightforwardly that $\operatorname {Spec} A$ is irreducible iff the the nilradical is prime (Exercise 19 in Ch. 1 of Atiyah and Macdonald's commutative algebra book) Note there is a natural homeomorphism induced by the projection map between $\operatorname {Spec} A/I$ and the closed subset of $\operatorname {Spec} A$, $V(I)= \{p\in \operatorname {Spec} A\ | \ I\subseteq p\}$, so this deals with the irreducible components question as well.
Just friendly advice, I would recommend you do most of the exercises in a book like Atiyah and Macdonald before seriously attempting to learn about schemes.