Showcasing $V(I \cap J)=V(I) \cup V(J)$ in Polynomials

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In summary, the conversation discusses the relationship between ideals $I$ and $J$ in the polynomial ring $K[x_1, x_2, \dots, x_n]$. The participants consider whether it is necessary to show that $V(IJ)=V(I\cap J)$ and then $V(IJ)=V(I)\cup V(J)$ in order to prove that $V(I \cap J)=V(I) \cup V(J)$. They present a proof that $V(IJ)=V(I)\cup V(J)$ and discuss alternative methods for showing the equality $V(I \cap J)=V(I) \cup V(J)$.
  • #1
evinda
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Hello! (Wave)

I want to show that if $I,J$ ideals of $K[x_1, x_2, \dots , x_n]$, then $V(I \cap J)=V(I) \cup V(J)$.

Do I have to show that $ V(IJ)=V(I\cap J)$ and then $V(IJ)=V(I)\cup V(J)$? (Thinking)

If so, that's what I have tried:

$x \in V(IJ) \leftrightarrow (f_i \cdot g_j)(x)=0$, where $f_i \in I$ and $g_j \in J$

$\leftrightarrow f_i(x) \cdot g_j(x)=0 \leftrightarrow f_i(x)=0 \text{ OR } g_j(x)=0 \leftrightarrow x \in V(I) \text{ OR } x \in V(J)$ $\leftrightarrow x \in V(I)\cup V(J)$

So, $V(IJ)= V(I)\cup V(J)$.

Is it right so far? Or have I done something wrong? (Thinking)

How could I continue? :confused:
 
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  • #2
Is there also an other way, to show that $V(I \cap J)=V(I) \cup V(J)$? (Thinking)
 

FAQ: Showcasing $V(I \cap J)=V(I) \cup V(J)$ in Polynomials

What is the significance of showcasing $V(I \cap J)=V(I) \cup V(J)$ in polynomials?

Showcasing $V(I \cap J)=V(I) \cup V(J)$ in polynomials is important because it demonstrates the relationship between the intersection of two polynomial ideals and the union of their varieties. This concept is crucial in algebraic geometry and has many applications in solving systems of polynomial equations.

How does the equation $V(I \cap J)=V(I) \cup V(J)$ relate to the Nullstellensatz theorem?

The equation $V(I \cap J)=V(I) \cup V(J)$ is a direct consequence of the Nullstellensatz theorem, which states that the variety of an ideal is equal to the radical of the ideal. In the case of polynomial ideals, the radical is the intersection of all prime ideals containing it, and the variety is the set of points that satisfy all the polynomials in the ideal. Therefore, the equation showcases the connection between the geometric and algebraic properties of polynomials.

Can you provide an example of how $V(I \cap J)=V(I) \cup V(J)$ works in practice?

One example of how $V(I \cap J)=V(I) \cup V(J)$ works is with the ideals $I=(x^2+y^2-1)$ and $J=(x-1)$. The intersection of these two ideals is $I \cap J=(x-1)$, and the varieties are $V(I)=\{(x,y)|x^2+y^2=1\}$ and $V(J)=\{(1,y)|y \in \mathbb{R}\}$. Thus, the equation holds as $V(I \cap J)=V(x-1)=\{(1,y)|y \in \mathbb{R}\}=V(J)=V(I) \cup V(J)$.

How does showcasing $V(I \cap J)=V(I) \cup V(J)$ relate to solving systems of polynomial equations?

Showcasing $V(I \cap J)=V(I) \cup V(J)$ is essential in solving systems of polynomial equations because it allows us to simplify the system by finding the intersection of the individual polynomial ideals. This reduces the number of equations to solve and can provide a more efficient method for finding solutions. Additionally, it helps us understand the geometric interpretation of the solution set of a system of polynomial equations.

Are there any limitations to $V(I \cap J)=V(I) \cup V(J)$ in polynomials?

While the equation $V(I \cap J)=V(I) \cup V(J)$ holds true for polynomial ideals, it may not be applicable in other contexts. For example, it may not hold for ideals in different rings or for non-polynomial functions. Additionally, it is important to note that the equality may not always hold when considering varieties over an algebraically closed field.

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