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I want to see if the three algebraic varieties are irreducible or not. We use the notation

$V(I)=\{P\in\mathbb{A}_k^n:f(P)=0\text{ for all }f\in I\}$ for an ideal $I\subset k[x_1,...,x_n]$.

  1. $V((xy^3))$
  2. $V((x^2+y^3+xy))$
  3. $V((x^2+y^3+xy,xy^3))$
  4. $V((xy+yz+zx+xyz))$

$1.$ is reducible since we have $V((xy^3))=V((x)\cap (y^3))=V((x))\cup V((y^3))$.

But I can't say something about the other sets. I have drawn them, so I saw that they are reducible. But how can I see this without using maple?

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    Is the polynomial $x^2+y^3+xy$ irreducible? – 2017-02-25
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    Yes, it is! So? – 2017-02-25
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    Well, there is a certain relationship between a variety being irreducible and the polynomial defining it being irreducible... – 2017-02-25
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    Ok, so the variety 2 is irreducible! Thank you. But what about 3 and 4? – 2017-02-25
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    Are you going to make me ask you if $xy+yz+zx+xyz$ is irreducible? :| – 2017-02-25
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    And for 3: what are the points of that agebraic set? – 2017-02-25
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    So: 1 is reducible, 2 and 4 are irreducible and 3 consists only of the point $(0,0)$. So it is irreducible, isn't it? – 2017-02-25
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    Please write a complete answer explaining your reasoning. – 2017-02-25
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    To see that 1 is reducible, see my question. 2 is irreducible since the associated polynomial is irreducible. Same argument works for 4. A point $(a,b)$ in the 3rd variety must statisfy $ab^3=0$ and $a^2+b^3+ab=0$, so $(a,b)=(0,0)$. So 3 is the same as $V(x,y)$, which is irreducible. – 2017-02-25
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    No. Please use the answer box to write down a complete answer —not a comment. – 2017-02-25
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    Okay, but is my answer right? A false answer wouldn't be helpful :) – 2017-02-25

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