Proving Intersection of Ideals in $K[x_1,x_2,...,x_n]$

[evinda]

Hi! (Smile)

Let $(I_a)_{a \in A}$ be a family of ideals of $K[x_1,x_2, \dots, x_n]$.
I want to prove that:

$$V \left ( \sum_{a \in A} I_a\right )=\bigcap_{a \in A} V(I_a)$$

Do we have to use the definition:

$$V(S)=\{ (a_1,a_2, \dots, a_n) \in K^n| f_a(a_1,a_2, \dots, a_n)=0 \forall a \in A\}$$

If so, how could use it? (Thinking)

 

[Fallen Angel]

Hi evinda,

From the inclusion \(\displaystyle I_{a}\subset \displaystyle\sum_{a\in A}I_{a}\)
you got that \(\displaystyle V(\displaystyle\sum_{a\in A}I_{a})\subset V(I_{a})\).

Now a point in the intersection means that every polynomial that belongs to one of the ideals vanishes at this point, and every polynomial in the sum is a linear combination of polynomials in this ideals, hence you got the other inclusion. (Just need a better writing :p)

 

[evinda]

Could we prove it like that? (Thinking)

$$\supseteq:$$

We assume that $x \in \cap_{a} V(I_a)$. By definition of intersection, that means that $x \in V(I_a), \forall a$.
By definition of $V$, we have that $f(x)=0, \forall f \in \sum_{a} I_a$. That means that $x \in V(\sum_a I_a)$.

Therefore, $V(\sum_a I_a) \subset \cap_a V(I_a)$.

$$\subseteq:$$

Let $x \in V(\sum_a I_a)$. By definition of $V$, we have that $f(x)=0, \forall f \in I_a$. Does this mean that $f(x)=0, \forall f \in I_a, \forall a$ ?

 

[Fallen Angel]

Hi,

Could we prove it like that? (Thinking)

$$\supseteq:$$

We assume that $x \in \cap_{a} V(I_a)$. By definition of intersection, that means that $x \in V(I_a), \forall a$.
By definition of $V$, we have that $f(x)=0, \forall f \in \color{red} I_a, \forall a\in A\color{black}$. That means that $x \in V(\sum_a I_a)$.

Therefore, $V(\sum_a I_a) \color{red} \supset \color{black} \cap_a V(I_a)$.

$$\subseteq:$$

Let $x \in V(\sum_a I_a)$. By definition of $V$, we have that $f(x)=0, \forall f \in \color{red}\displaystyle\sum_{a\in A}I_a\color{black}$. Does this mean that $f(x)=0, \forall f \in I_a, \forall a$ ?

The parts in red were mistakes, for the second inclusion, just need to know that the \(\displaystyle V\) operator inverts the inclusions, so \(\displaystyle J\subset I \Rightarrow V(I) \subset V(J)\).
Hence you got \(\displaystyle V(\displaystyle\sum_{a\in A}I_{a})\subset V(I_{a}), \ \forall a\in A\).

 

[evinda]

Hi,
The parts in red were mistakes, for the second inclusion, just need to know that the \(\displaystyle V\) operator inverts the inclusions, so \(\displaystyle J\subset I \Rightarrow V(I) \subset V(J)\).
Hence you got \(\displaystyle V(\displaystyle\sum_{a\in A}I_{a})\subset V(I_{a}), \ \forall a\in A\).

So, you mean that it is like that?

$$I_a \subset \sum_{a \in A} I_a \Rightarrow V(I_a) \supset V \left ( \sum_{a \in A} I_a \right )$$

If so, how can we prove that $I_a \subset \sum_{a \in A} I_a$ ? (Thinking)

Also, how can we show the other inclusion? (Thinking)

 

[Fallen Angel]

Hi,

\(\displaystyle \displaystyle\sum_{a\in A}I_{a}=\{\displaystyle\sum_{i\in I}a_{i}\alpha_{i} \ : \ \alpha_{i}\in I_{a}, \ I\subset A, \ I \ finite\}\)

So the inclusion is obvious just by taking \(\displaystyle I=\{a\}\).You got thje other inclusion in my post above, given \(\displaystyle g\in \displaystyle\sum_{a\in A}I_{a}\), then \(\displaystyle g(x)=\displaystyle\sum_{a\in A}k_{a}f_{a}(x)\) and using the condition \(\displaystyle x\in \displaystyle\cap_{a\in A}V(I_{a})\) you got \(\displaystyle f_{a}(x)=0, \ \forall a\in A\)

 

[evinda]

Hi,

\(\displaystyle \displaystyle\sum_{a\in A}I_{a}=\{\displaystyle\sum_{i\in I}a_{i}\alpha_{i} \ : \ \alpha_{i}\in I_{a}, \ I\subset A, \ I \ finite\}\)

So the inclusion is obvious just by taking \(\displaystyle I=\{a\}\).You got thje other inclusion in my post above, given \(\displaystyle g\in \displaystyle\sum_{a\in A}I_{a}\), then \(\displaystyle g(x)=\displaystyle\sum_{a\in A}k_{a}f_{a}(x)\) and using the condition \(\displaystyle x\in \displaystyle\cap_{a\in A}V(I_{a})\) you got \(\displaystyle f_{a}(x)=0, \ \forall a\in A\)

We have shown that:

$I_a \subset \sum_{a \in A} I_a \Rightarrow V \left ( \sum_{a \in A} I_a \right ) \subset V(I_a)$

Does this imply that $V \left ( \sum_{a \in A} I_a \right ) \subset \bigcap_{a \in A} V(I_a)$? Or do we have to prove this? If so, how could we do this? (Thinking)

Could you explain me further what we have to show at the other inclusion?

 

[Fallen Angel]

Hi,

It implies the inclusion because \(\displaystyle V(\displaystyle\sum_{a\in A}I_{a})\subset V(I_{a})\) holds for every $a\in A$. It doesn't need a prove, it's just de definition of intersection.In the other inclusion, you have to take a point $x\in \displaystyle\cap_{a\in A} V(I_{a})$, i.e. $x$ is a point such that given a polynomial $p$, if there exists $a\in A$ such that $p\in I_{a}$, then $p(x)=0$.

And now we want to show that $x\in V(\displaystyle\sum_{a\in A}I_{a})$.
So we take a polynomial $g\in \displaystyle\sum_{a\in A}I_{a}$ and we want to show that $g(x)=0$.Then if $g\in \displaystyle\sum_{a\in A}I_{a}$ we can write $g=\displaystyle\sum_{j\in J\subset A}k_{j}f_{j}$, where $J$ is finite, $k_{j}\in K$ and $f_{j}\in I_{j}$ for every $j$.

Hence $f_{j}(x)=0 \ \forall j\in J$ (see above) and then $g(x)=0$, what finishes the proof.

 

[evinda]

It implies the inclusion because \(\displaystyle V(\displaystyle\sum_{a\in A}I_{a})\subset V(I_{a})\) holds for every $a\in A$. It doesn't need a prove, it's just de definition of intersection.

Do you mean that we use the definition of the generalized intersection:
$$\{x : (\forall b \in A ) x \in b\}$$ ? (Thinking)

 

[Fallen Angel]

$A$ is just an index set so this is not the intersection.

$\displaystyle\cap_{a\in A}V(I_{a})=\{x\in K^{n} \ : \ x\in V(I_{a}) \ \forall a\in A\}$

 

[evinda]

In the other inclusion, you have to take a point $x\in \displaystyle\cap_{a\in A} V(I_{a})$, i.e. $x$ is a point such that given a polynomial $p$, if there exists $a\in A$ such that $p\in I_{a}$, then $p(x)=0$.

And now we want to show that $x\in V(\displaystyle\sum_{a\in A}I_{a})$.
So we take a polynomial $g\in \displaystyle\sum_{a\in A}I_{a}$ and we want to show that $g(x)=0$.Then if $g\in \displaystyle\sum_{a\in A}I_{a}$ we can write $g=\displaystyle\sum_{j\in J\subset A}k_{j}f_{j}$, where $J$ is finite, $k_{j}\in K$ and $f_{j}\in I_{j}$ for every $j$.

Hence $f_{j}(x)=0 \ \forall j\in J$ (see above) and then $g(x)=0$, what finishes the proof.

In my notes, there is this definition $$\sum_{a \in A} I_a =\{\alpha_{i1}+\alpha_{i2}+ \dots +\alpha_{ij} | \alpha_{ij} \in I_{a_j}\}$$

So, should I write it like that? (Thinking)

If $g\in \displaystyle\sum_{a\in A}I_{a}$ we can write $g=\displaystyle\sum_{i,j}f_{ij}$, where $f_{ij} \in I_{a_j}$

So, $f_{ij}(x)=0 \ \forall i,j$.

So, $g(x)=0$.

 

[Fallen Angel]

There is an abuse of notation in your definition, you are not using the subscript $i$, so in the sum you are only summing over $j$.

But in essence both definitions are two ways of writing the same thing, and you can write the proof in this way if you see it better. (Yes)

 

[evinda]

There is an abuse of notation in your definition, you are not using the subscript $i$, so in the sum you are only summing over $j$.

But in essence both definitions are two ways of writing the same thing, and you can write the proof in this way if you see it better. (Yes)

Is this definition:

$$\sum_{a \in A} I_a =\{\alpha_{i1}+\alpha_{i2}+ \dots +\alpha_{ij} | \alpha_{ij} \in I_{a_j}\}$$

right or is there a typo? (Worried)

 

[Fallen Angel]

Is right.