# Thread: Proving that an ideal is prime

1. ## Proving that an ideal is prime

Let R= ℚ[X,Y] and let I=<X + 1>

I am trying to prove that I is a prime ideal of R.

I have tried arguing the following way (but i am unsure whether it is correct because the ring has two variables X,Y):

Suppose that f,g ∈ℚ[X,Y] and that fg∈ <X + 1>

That is, (x+1)|fg.

Hence, fg(-1)=0, that is f(-1)g(-1)=0 .

Therefore either f(-1)=0 , that is (X+1)|f, or g(-1)=0, that is (x+1)|g.

Thus either f or g is in the ideal <X+1>.

Therefore the ideal is prime.

Is this a correct argument?

How would i show that I is not a maximal ideal of R?

Thanks

2. Originally Posted by Siknature
Let R= ℚ[X,Y] and let I=<X + 1>

I am trying to prove that I is a prime ideal of R.

I have tried arguing the following way (but i am unsure whether it is correct because the ring has two variables X,Y):

Suppose that f,g ∈ℚ[X,Y] and that fg∈ <X + 1>

That is, (x+1)|fg.

Hence, fg(-1)=0, that is f(-1)g(-1)=0 .

Therefore either f(-1)=0 , that is (X+1)|f, or g(-1)=0, that is (x+1)|g.

Thus either f or g is in the ideal <X+1>.

Therefore the ideal is prime.

Is this a correct argument?

How would i show that I is not a maximal ideal of R?

Thanks

The argument is correct though you must be sure that here applies the same as with one unknown...

As for maximality: $\,$ ...

Tonio

3. Originally Posted by tonio

As for maximality: $\,$ ...

Tonio
Are you saying that <x +1> is not maximal because <x+1,y> is an ideal which is greater than <x+1> ?

4. Originally Posted by Siknature
Are you saying that <x +1> is not maximal because <x+1,y> is an ideal which is greater than <x+1> ?

Yes...somehow I ommited the "<" sign between those two ideals.

Tonio