Show that $F= Y^2 +X^2(X-1)^2 \in \mathbb{R}[X,Y]$ is an irrdeucible polynomial , but $V(F)$ is reducible

I have found the solution here

It is written that

$V(F)= \{(0,0),(1,0) \} $ is reducible i,e $V(F)=V(Y^2 +X^2) \cup V(Y^2 +(X-1)^2).$

My confusion: Take$ f =Y^2+X^2$ and $g=Y^2+(X-1)^2$

$V(F)= V(f) \cup V(g) \implies V(F)= V(fg) \implies F= fg$

This led to a contradiction because $Y^2 +X^2(X-1)^2 \neq(Y^2+X^2) (Y^2+(X-1)^2)$

Actually , $F= Y^2 +X^2(X-1)^2=\left((Y+iX)(X-1)(Y-iX)(X-1)\right)$

My question:why is $V(F)= \{(0,0),(1,0) \} =V(Y^2 +X^2) \cup V(Y^2 +(X-1)^2).?$

My thinking: $V(F)= \{(0,0),(1,0)\}= V(Y^2+X^2, Y^2+(X-1)^2)= V(Y^2+X^2) \cap V(Y^2+(X-1)^2)$

Also,$V(Y^2+X^2) \cap V(Y^2+(X-1)^2) \neq V(Y^2 +X^2) \cup V(Y^2 +(X-1)^2)$

  • 2
    $\begingroup$ In general, $V(F) = V(G) \nRightarrow F = G$. $\endgroup$
    – Brian Shin
    May 15 at 16:02
  • $\begingroup$ Have you tried applying the definition of $V(F)$ $\endgroup$
    – ronno
    May 17 at 13:57


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