Show Irreducibility of $K^n$ Algebraic Set $V$ iff $I(V)$ is a Prime Ideal

[evinda]

Hello! (Wave)I want to show that the algebraic set of $K^n$, $V$, is irreducible iff $I(V)$ is a prime ideal.

That's what I have tried so far:

We know that the algebraic set $V$ is irreducible iff $V$ cannot be written as $V=V_1 \cup V_2$, where $V_1, V_2$ are algebraic sets of $K^n$ and $V_1 \subset V$, $V_2 \subset V$.

We suppose that $V$ is reducible.

Then, there are the algebraic sets $V_1, V_2 \subset V$ such that $V=V_1 \cup V_2$.

So, $I(V)=I(V_1 \cup V_2)=I(V_1) \cap I(V_2)$.

We have that $V_1 \subset V \Rightarrow I(V) \subset I(V_1)$ and that $V_2 \subset V \Rightarrow I(V) \subset I(V_2)$.

We want to show that $R/I$ is not an integral domain.

$R/I=\{a+I | a \in R\}$

$(a+I)(b+I)=a \cdot b+a \cdot I+b \cdot I+I \cdot I=a \cdot b$

We take $a \in V_1(K^n)$ and $b \in V_2 (K^n)$.

That means that $f(a)=0, \forall f \in K^n$ and $g(b)=0, \forall g \in K^n$.

Is it right? If so, how could we continue? (Thinking)

 

[Fallen Angel]

Hi evinda,

Left you some comments in red

Hello! (Wave)I want to show that the algebraic set of $K^n$, $V$, is irreducible iff $I(V)$ is a prime ideal.

That's what I have tried so far:

We know that the algebraic set $V$ is irreducible iff $V$ cannot be written as $V=V_1 \cup V_2$, where $V_1, V_2$ are algebraic sets of $K^n$ and $V_1 \subset V$, $V_2 \subset V$.

We suppose that $V$ is reducible.

Then, there are the algebraic sets $V_1, V_2 \subset V$ such that $V=V_1 \cup V_2$.

So, $I(V)=I(V_1 \cup V_2)=I(V_1) \cap I(V_2)$.

We have that $V_1 \subset V \Rightarrow I(V) \subset I(V_1)$ and that $V_2 \subset V \Rightarrow I(V) \subset I(V_2)$.Everything OK until this point.

We want to show that $R/I$ is not an integral domain. $I$ is not defined, I know it's $I(V)$, but you need to be careful.
The same happens with R

$R/I=\{a+I | a \in R\}$

$(a+I)(b+I)=a \cdot b+a \cdot I+b \cdot I+I \cdot I=a \cdot b$$+I$

The next sentences makes no sense, or at least I can't see it.
We take $a \in V_1(K^n)$ and $b \in V_2 (K^n)$.

That means that $f(a)=0, \forall f \in K^n$ and $g(b)=0, \forall g \in K^n$.

Is it right? If so, how could we continue? (Thinking)

First, the operator \(\displaystyle V\) is applied over \(\displaystyle K[x_{1},\ldots,x_{m}]\), not over $K^{n}$

What you are supposed to do it's find two polynomials $f,g\in K[x_{1},\ldots, x_{m}]$ such that $f\cdot g \in I(V)$ with $f\notin I(V), g\notin I(V)$

In order to do that try to use the inclusions you wrote at first steps.

 

[evinda]

What you are supposed to do it's find two polynomials $f,g\in K[x_{1},\ldots, x_{m}]$ such that $f\cdot g \in I(V)$ with $f\notin I(V), g\notin I(V)$

Why do we have to find two polynomials $f,g\in K[x_{1},\ldots, x_{m}]$ such that $f\cdot g \in I(V)$ with $f\notin I(V), g\notin I(V)$? (Thinking)

Does this mean that $I$ is not a prime ideal? If so, why is it like that?

 

[Fallen Angel]

Sorry for not writing maths, I am with the mobile phone.

Being in I(V) is the same as being zero in the quotient, so finding that polynomials is proving that the quotient is not an integral domain and then I(V) is not prime.

 

[evinda]

Sorry for not writing maths, I am with the mobile phone.

Being in I(V) is the same as being zero in the quotient, so finding that polynomials is proving that the quotient is not an integral domain and then I(V) is not prime.

Could you explain it further to me? (Worried)

 

[Fallen Angel]

Hi,

You want to prove that an ideal \(\displaystyle I=I(V)\) of a ring \(\displaystyle R=K[x_{1},\ldots, x_{m}]\) is not a prime ideal.

Or, equivalently, you want to prove that $R/I$ is not an integral domain (¿You know this equivalence or do I need to prove it?)

For proving that some ring is not an integral domain, the natural way is find two non zero elements such that their product is zero.

In a quotient, the elements you got are equivalence classes modulo the relation \(\displaystyle a+I=b+I \Leftrightarrow a-b \in I\), so \(\displaystyle a+I=0+I \Leftrightarrow a\in I\).

Now, traslating the paragraph above, you want to find two elements $a,b\in R$, such that $a,b\notin I$(i.e., they are nonzero in the quotient $R/I$) and $a\cdot b \in I$(i.e, their product is zero in the quotient $R/I$)

For doing that you got the inclusions $I(V)\subset I(V_{1})$ and $I(V)\subset I(V_{2})$, so you can take $f\in I(V_{1})\setminus I(V)$, $g\in I(V_{2})\setminus I(V)$

Hence, $f\cdot g\in I(V)$ (i.e They are zero in the quotient) with $f,g\notin I(V)$(i.e. They are non zero in the quotient).If you have understood this exercise, you should be able to explain me why $f\cdot g \in I(V)$ :D

 

[evinda]

If you have understood this exercise, you should be able to explain me why $f\cdot g \in I(V)$ :D

Since $f \in I(V_1) \setminus I(V)$ and $g \in I(V_2) \setminus I(V)$, is their product $f \cdot g$ in the union of $I(V_1) \setminus I(V)$ and $I(V_2) \setminus I(V)$? (Thinking)

 

[Fallen Angel]

No, $f\cdot g \in I(V)$.

$f\in I(V_{1})$ means that for every point $x\in V_{1}$, $f(x)=0$.

$g\in I(V_{2})$ means that for every point $y\in V_{2}$, $g(y)=0$.

And now $V=V_{1}\cup V_{2}$ so if we got $x\in I(V)$, so $x\in V_{1}$ or $x\in V_{2}$, hence $(f\cdot g)(x)=0$

 

[evinda]

Hi,

You want to prove that an ideal \(\displaystyle I=I(V)\) of a ring \(\displaystyle R=K[x_{1},\ldots, x_{m}]\) is not a prime ideal.

Or, equivalently, you want to prove that $R/I$ is not an integral domain (¿You know this equivalence or do I need to prove it?)

For proving that some ring is not an integral domain, the natural way is find two non zero elements such that their product is zero.

In a quotient, the elements you got are equivalence classes modulo the relation \(\displaystyle a+I=b+I \Leftrightarrow a-b \in I\), so \(\displaystyle a+I=0+I \Leftrightarrow a\in I\).

Now, traslating the paragraph above, you want to find two elements $a,b\in R$, such that $a,b\notin I$(i.e., they are nonzero in the quotient $R/I$) and $a\cdot b \in I$(i.e, their product is zero in the quotient $R/I$)

For doing that you got the inclusions $I(V)\subset I(V_{1})$ and $I(V)\subset I(V_{2})$, so you can take $f\in I(V_{1})\setminus I(V)$, $g\in I(V_{2})\setminus I(V)$

Hence, $f\cdot g\in I(V)$ (i.e They are zero in the quotient) with $f,g\notin I(V)$(i.e. They are non zero in the quotient).If you have understood this exercise, you should be able to explain me why $f\cdot g \in I(V)$ :D

Could you explain me why we take $f\in I(V_{1})\setminus I(V)$ and $g\in I(V_{2})\setminus I(V)$? (Worried)Also, we want to show that the algebraic set of $K^n$, $V$, is irreducible iff $I(V)$ is a prime ideal.

How can we show the other direction? (Thinking)

 

[Fallen Angel]

We take $f,g$ in this way because we want two polynomials who's product is zero in the quotient but both being nonzero in the quotient.

For the other implication assume $I(V)$ is not prime, so there exists, $F_{1}\cdot F_{2}\in I(V)$ with $F_{i}\notin I(V)$ ($i$ can be 1 or 2, it doesn't matter).
Then $V=(V\cap V(F_{1}))\cup (V\cap V(F_{2}))$ and $V\cap V(F_{i})\underset{\neq}{\subset}V$, so $V$ is reducible

 

[evinda]

No, $f\cdot g \in I(V)$.

$f\in I(V_{1})$ means that for every point $x\in V_{1}$, $f(x)=0$.

$g\in I(V_{2})$ means that for every point $y\in V_{2}$, $g(y)=0$.

And now $V=V_{1}\cup V_{2}$ so if we got $x\in I(V)$, so $x\in V_{1}$ or $x\in V_{2}$, hence $(f\cdot g)(x)=0$

So, we have that $x \in V_1$ or $x \in V_2$?

Does it stand that $V_1=V_1(I(V_1))$, or why do we conclude that $\forall f \in I(V_1): f(x)=0$ ? (Thinking)

 

[evinda]

We take $f,g$ in this way because we want two polynomials who's product is zero in the quotient but both being nonzero in the quotient.

For the other implication assume $I(V)$ is not prime, so there exists, $F_{1}\cdot F_{2}\in I(V)$ with $F_{i}\notin I(V)$ ($i$ can be 1 or 2, it doesn't matter).
Then $V=(V\cap V(F_{1}))\cup (V\cap V(F_{2}))$ and $V\cap V(F_{i})\underset{\neq}{\subset}V$, so $V$ is reducible

How do we get that $V=(V\cap V(F_{1}))\cup (V\cap V(F_{2}))$ and $V\cap V(F_{i})\underset{\neq}{\subset}V$? (Thinking)

 

[Fallen Angel]

So, we have that $x \in V_1$ or $x \in V_2$?

Does it stand that $V_1=V_1(I(V_1))$, or why do we conclude that $\forall f \in I(V_1): f(x)=0$ ? (Thinking)

I don't understand this question, $V_1=V_1(I(V_1))$ make no sense (probably a typo,$V_1=V(I(V_1))$ )and I haven't concluded that.For the other post, we have $F_{1}\cdot F_{2} \in I(V)$, so $(F_{1}\cdot F_{2})(x)=0, \ \forall x\in V$, then $\forall x\in V, \ F_{1}(x)=0$ or $F_{2}(x)=0$.

So every point in $V$ is a zero of $F_{1}$ or a zero of $F_{2}$.

By definition, $V(F_{1})$ is the set of zeros of $F_{1}$.

Hence, $V$ can be expressed as the points that are in $V$ and vanishes $F_{1}$ , this is $V\cap V(F_{1})$ and the points that are in $V$ and vanishes $F_{2}$, this is $V\cap V(F_{2})$.

$V\cap V(F_{i})\subset V$ is obvious and the non equality follows from the fact that $F_{i}\notin I(V)$.PS: In the post you quoted me there is a typo, I will correct it now

 

[Fallen Angel]

Correction

No, $f\cdot g \in I(V)$.

$f\in I(V_{1})$ means that for every point $x\in V_{1}$, $f(x)=0$.

$g\in I(V_{2})$ means that for every point $y\in V_{2}$, $g(y)=0$.

And now $V=V_{1}\cup V_{2}$ so if we got $\color{red}x\in V\color{black}$, so $x\in V_{1}$ or $x\in V_{2}$, hence $(f\cdot g)(x)=0$