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If $V$ is an algebraic set of $K^n$, I want to show that $V$ is irreducible exactly when $I(V)$ is a prime ideal.

That's what I have tried:

We suppose that $V$ is not irreducible. Then, it can be written in the form $V=V_1 \cup V_2$, where $V_1,V_2$ algebraic sets of $K^n$, $V_1 \subseteq V, V_2 \subset V$.

$$I(V)=I(V_1 \cup V_2)=I(V_1) \cap I(V_2)$$

How can I continue?

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Here is an equivalent definition of a prime ideal, which might help here:

Let $R$ be a (commutative) ring and let $P\le R$ be an ideal. $P$ is a prime ideal if and only if the following holds - for any two ideals $I,J\le R$, if $P \supseteq IJ$ then $P \supseteq I$ or $P \supseteq J$ (If you are not familiar with it, try Proving it, it's a nice exercise in basics of ring theory)

Edit: I'll give a slightly different proof to avoid using $I(V_1 \cup V_2)=I(V_1)I(V_2)$. I'll denote by $V(S)$ ,for a subset $S\subseteq K[x_1,...,x_n]$, the subset of $K^n$ on which all of the elements of $S$ vanish simultaneously.

For the direction you started proving, we write $V=V(J)$ when $J$ is radical, and similarly $V_1=V(J_1), V_2=V(J_2)$. This is useful as for any ideal S, $I(V(S))=\sqrt{S}$ (Nullstellensatz). The strict inclusion $V_i \subset V$ implies that $J_i\supsetneq J$, but: $$ V(J_1)\cup V(J_2)=V(J_1 J_2) \tag{1} $$

This is easier to understand geometrically - if all elements of J_1 vanish somewhere, the anything times them will also vanish there (and same goes for $J_2$). On the contrary, if $p\in J_1$ and $q\in J_2$ do not vanish at some point, so does $pq\in J_1 J_2$.

(1) now gives $V(J_1 J_2)=V(J)$. Taking $I$ of both sides gives: $$ I(V)=J=\sqrt{J}=I(V(J))=I(V(J_1 J_2))=\sqrt{J_1 J_2}\supseteq J_1 J_2 $$ To conclude, we found two ideals $J_1,J_2$, both not inside $J=I(V)$, whose product is inside $I(V)$. Thus $I(V)$ is not prime

The other direction is pretty much the same thing reversed - If $I(V)$ is not prime, we find some polynomials $p,q\in K[x_1,...,x_n]$ which do not lie in $I(V)$ but their product does lie there. We define: $$ V_1=\{y\in V:\ p(y)=0\}, V_2=\{y\in V: q(y)=0\} $$

Then $V_1,V_2$ are algebraic subsets of $V$ which are not everything (because $p,q\notin I(V)$, i.e. $p$ and $q$ both do not vanish on all of $V$), but their union is all of $V$.

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    $\begingroup$ Miel Sharf Do you mean $I(V_1) \cap I(V_2)=I(V_1)I(V_2)$ instead of $I(V_1 \cap V_2)=I(V_1)I(V_2)$ ? If so, why does it stand that $$I(V_1) \cap I(V_2)=I(V_1)I(V_2)$$ ? $\endgroup$
    – evinda
    Nov 20 '14 at 21:19
  • $\begingroup$ Miel Sharf Could you explain me why $I(V_1) \cap I(V_2)=I(V_1)I(V_2)$ stands? $\endgroup$
    – evinda
    Nov 20 '14 at 21:34
  • $\begingroup$ I changed the proof above to avoid using it, but instead using a "dual" version of it which is easier to understand geometrically (and I explained it) $\endgroup$
    – Miel Sharf
    Nov 20 '14 at 21:54
  • $\begingroup$ Miel Sharf Is there also an other way to prove this? Not geometrically? $\endgroup$
    – evinda
    Nov 21 '14 at 0:53
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Let $I(V)$ is a prime ideal. Suppose $V$ is reducible. Then there exists two non-empty proper closed subset $V_1, V_2$ of $V$ such that $V = V_1 \cup V_2.$ Then $I(V) = I(V_1 \cup V_2) = I(V_1) \cap I(V_2).$ Since $I(V)$ is a prime ideal, either $I(V) = I(V_1)$ or $I(V) = I(V_2) \Rightarrow V = V_1$ or $V = V_2.$

Now assume $V$ is irreducible. Suppose $I(V)$ is not a prime ideal. Then there exist two polynomials $f, g \in k[x_1, x_2, \cdots, x_n]$ such that $f \notin I(V), g \notin I(V)$ but $fg \in I(V).$ Set $V_1 := \{ x \in V | f(x) = 0\}$ and $V_2 := \{ x \in V | g(x) = 0\}.$ Then both $V_1$ and $V_2$ are non-empty and proper closed subsets of $V$ with $V = V_1 \cup V_2.$

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  • $\begingroup$ At the first case, why when $I(V)=I(V_1)\cap I(V_2)$ is a prime ideal, then $I(V)=I(V_1)$ or $I(V)=I(V_2)$? How does this imply that $V=V_1$ or $V=V_2$? $$$$ At the second case, when an ideal is not prime, does this mean that there two polynomials $f,g$ such that $f \notin I(V), g \notin I(V)$ but $fg \in I(V)$? $\endgroup$
    – evinda
    Nov 21 '14 at 1:16
  • $\begingroup$ Let $A$ be a commutative ring with identity and let $I, J, P$ be three ideals of $A$ where $P$ is prime and $IJ \subseteq P.$ Then either $I \subseteq P$ or $J \subseteq P.$ Also note that $IJ \subseteq I \cap J.$ From this we get that $I(V_1) \subseteq I(V)$ or $I(V_2) \subset I(V).$ The other side inclusion is obvious since $I(V) = I(V_1) \cap I(V_2).$ It is a fact that, for any subset $X$ of $k^n$, $Z(I(X)) = \overline{X}$, where $\overline{X}$ denotes the closure of $X$ in Zariski topology. (See, R. Hartshorne, Algebraic Geometry, Chap. I, Prop. 1.2(e)) $\endgroup$
    – Krish
    Nov 21 '14 at 1:33
  • $\begingroup$ For the second case, it is a negation of the definition of prime ideal. If $I \subsetneq A$ is not a prime ideal then $A/I$ is not a domain. So we can find $a, b \in A$ such that $\overline{a}, \overline{b}$ are both non-zero in $A/I$, but $\overline{a} \overline{b} = \overline{0}$, i.e. $a \notin I, b \notin I$ but $ab \in I.$ $\endgroup$
    – Krish
    Nov 21 '14 at 1:46
  • $\begingroup$ Could you explain me further, why when we have $I(V)=I(V_1) \cap I(V_2)$ and I(V) is a prime ideal, then $I(V)=I(V_1)$ or $I(V)=I(V_2)$ ? $\endgroup$
    – evinda
    Nov 21 '14 at 8:27
  • $\begingroup$ @evinda: set $I = I(V_1), J = I(V_2), P = I(V)$. then $IJ \subseteq I \cap J \subseteq P \Rightarrow I \subseteq P$ or $J \subseteq P.$ On the other hand, $P = I \cap J \Rightarrow P \subseteq I$ or $P \subseteq J.$ $\endgroup$
    – Krish
    Nov 21 '14 at 12:48
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Take $F_1,F_2 \in K[X_1,\ldots,X_n]$ such that $F_1\cdot F_2 \in I(V)$. Now $$V = V\cap V(F_1 \cdot F_2) = V\cap \left(V(F_1) \cup V(F_2)\right) = \left(V\cap V(F_1)\right) \cup \left(V\cap V(F_1)\right)$$

Intuitively, we know that $F_1\cdot F_2$ vanishes on $V$. Therefore, for all $x\in V$ either $F_1(x) = 0$ or $F_2(x) = 0$. This suggests we can divide $V$ in two sets:

  • $W_1$: The points in $V$ such that $F_1$ vanishes, $W_1 = V \cap V(F_1)$.
  • $W_2$: The points in $V$ such that $F_2$ vanishes, $W_2 = V \cap V(F_2)$.

Now, we can write $V$ as the union of two algebraic sets: $$V= W_1 \cup W_2$$

  • If $V$ is irreducible, one of this sets must be equal to $V$. Say $W_1=V$, then $V \subset V(F_1)$ and therefore $F_1\in I(V)$. Thus, $I(V)$ is a prime ideal.
  • If $I(V)$ is a prime ideal, we can say (for example) $F_1\in I(V)$. In this case $V\subset V(F_1)$ and therefore $W_1 = V$. Thus, $V$ is irreducible.
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  • $\begingroup$ When we have that $x \in V \Rightarrow x \in V_1 \cup V_2 \Rightarrow x \in V_1 \text{ or } x \in V_2$, do we know that $V_1=V_1(I(V_1))$ ? $\endgroup$
    – evinda
    Nov 27 '14 at 22:30
  • $\begingroup$ Could you explain me how you found the following equalities? $$V = V\cap V(F_1 \cdot F_2) = V\cap \left(V(F_1) \cup V(F_2)\right) = \left(V\cap V(F_1)\right) \cup \left(V\cap V(F_1)\right)$$ $\endgroup$
    – evinda
    Nov 27 '14 at 22:36
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    $\begingroup$ For the second question: Since $F_1 F_2$ vanishes on $V$, the vanishing set of $F_1 F_2$ is bigger than $V$. Thus $V \subset V(F_1 F_2)$. Now it should be clear that $V = V \cap V(F_1 F_2)$. Then I use the property $V(I_1 I_2) = V(I_1) \cup V(I_2)$. $\endgroup$ Nov 27 '14 at 22:45
  • $\begingroup$ Raul When we have that $F_1 \cdot F_2$ vanishes on $V$, why does it stand that for all $x \in V$ either $F_1(x)=0$ or $F_2(x)=0$ ? $\endgroup$
    – evinda
    Nov 27 '14 at 23:22
  • $\begingroup$ Because $0 = F_1F_2(x) = F_1(x)F_2(x)$ $\endgroup$ Nov 28 '14 at 0:21

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