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Suppose $N$ is a subvariety of $M$. Furthermore, suppose $N$ is irreducible.

Can I deduce that $N$ is contained in some irreducible component of $M$?

In most examples I thought of, this works ($M \subseteq \mathbb{A}^{2}$ being the union of $x=0$ and $y=0$, $N$ being the point $(0,0)$ or one of the axes).

I am really a beginner in algebraic geometry, so apart from the definition of irreducibility, I don't have an idea. I know that the following is false, though: if $N$ intersects some irreducible component of $M$, it doesn't mean it is contained in it.

I would be glad to be given an explanation, or a counterexample. More than anything, I would like to see the line of thought.

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up vote 4 down vote accepted

If $M = M_1 \cup M_2 \cup \ldots \cup M_r$ is the decomposition of $M$ into its irreducible components, then $N = \bigcup_{i=1}^r (N \cap M_i)$. Since $N$ is irreducible we must have $N \cap M_i = N$ for some $i$, so $N \subseteq M_i$.

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Thanks. I can't believe it was this easy... – Ofir Feb 11 '13 at 20:38

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