# Abstract Alg. - maximal ideals in Z[x]

1. Jan 12, 2010

### WannaBe22

1. The problem statement, all variables and given/known data
Can someone tell me how can I prove that every ideal in Z[x] generated by (p,f(x)) where f(x) is a polynomial that is irreducible in Zp is maximal??

Thanks!

2. Relevant equations

3. The attempt at a solution

2. Jan 12, 2010

### Hurkyl

Staff Emeritus
What have you tried?

I assume you've tried at least one obvious way to get started? Where did you get stuck?

Also, how many ways can you fill in the blank in the following sentence?
If I can prove ______, then (p, f(x)) is maximal​
One answer might suggest to you a line of attack....

3. Jan 13, 2010

### WannaBe22

Well..There are 2 possible ways:
1. If I can prove that Z[x]/I where I=(p,f(x)) is a field, then (p,f(x)) is maximal...
2. If I can prove that if I is contained in an ideal J that is contained in Z[x] then J=I or J=Z[x] thjen (p,f(x)) is maximal...

The first way doesn't help me at all... It seems like the second way is the one we need to use... But I can't figure out what is the contradiction we get by assuming that there is an element k that exists in J but not in I...How does it help us?

THanks a lot

4. Jan 13, 2010

### Hurkyl

Staff Emeritus
That's a shame -- it's the way I would have done it, but the method I had in mind requires knowledge of finite fields.

Maybe restating it more algebraically would help? For some other element a, You're looking to show the existence of a linear combination (coefficients in Z[x])
ua + vp + wf = 1​

5. Jan 13, 2010

### WannaBe22

When talking about the element "1" , you mean the unit element of Z[x] right? It doesn't have to be 1 ...
Hmmmm When adding up an element a such as a isn't in (p,f(x)) , that means that:
gcd(p,f(x),a)=1 so we have u,v,w such as ua+vp+wf=Unit element of Z[x]...
Am I right?

If you'll be able to verify my answer it will be great... Afterwards I'll be delighted if you'll be able to answer the folowing questions:
1. Lets take p=5, f(x)=x^3+x+1, a=2 ...
By the statement above, in the ideal that contains a,p and f(x) there are all the polynomials over Z... But how can we create x^2+1 for example by linear combinations of the 3?
2. What is the unit element in Z[x]? If it's the same as the one in Z, So how come it "spans" the whole ring? I mean, if 1 is the unit element in Z[x], so how can we produce every element in Z[x] by multiplications of 1?

Thanks a lot man!

6. Jan 13, 2010

### Hurkyl

Staff Emeritus
The coefficients of linear combinations come from the ring. So, the element f(x) can be produced by f(x) * 1.

7. Jan 13, 2010

### WannaBe22

Well, so how can we show that ua + vp + wf = 1 ?

Thanks :(

8. Jan 13, 2010

### Hurkyl

Staff Emeritus
Well, you brought up GCD's. What can you tell me about those? (You may need to break into cases on what kind of element a is)

9. Jan 13, 2010

### WannaBe22

Well...If we take an element "a" that is not in (p,f(x)) ... That means that we can't produce this element by final sums of p & f(x) ...I don't that the gcd of p,f(x)&a must be 1... I'm not even sure if gcd(p,f(x)) must be 1... We only know that f(x) has no roots in Zp, I don't think that their gcd must be 1...

Can you give me a more detailed guidance please?
I become pretty desperate...

Thanks