# Ideals, prime & maximal

1. Apr 16, 2013

### Zondrina

1. The problem statement, all variables and given/known data

Part b and c.

http://gyazo.com/821bceafd1c49adc366c63208066bd05

2. Relevant equations

Z[x]/I is a field ⇔ I is maximal.

3. The attempt at a solution

b. So do I need to show Z[x]/<x,2> = { f(x) + <x,2> | f(x) in Z[x] } is a field? That would show that I is maximal and hence it is also prime. Is there an easier way to do this?

c. Two elements I'm pretty sure, but I'll be concerned with this one after I get through b.

2. Apr 17, 2013

### Kreizhn

If I recall correctly, the following result holds:

Let R be a ring and $a,b \in R$. If $\bar b$ is the equivalence class of b in R/(a), then
$$R/(a,b) = [R/(a)]/(\bar b).$$

This essentially just says that if you are careful about the book keeping, you can quotient by (2,x) by first quotienting by (x), then quotienting by (2) [or vice versa if you prefer].

3. Apr 17, 2013

### micromass

Staff Emeritus
For (b), use the isomorphism theorems to show that $\mathbb{Z}[X]/I \cong \mathbb{Z}/2\mathbb{Z}$.

4. Apr 17, 2013

### Zondrina

Would it not be better to consider the map going to $\mathbb{Z_2}$?

Consider the map $\phi : \mathbb{Z}[X]/I → \mathbb{Z_2} \space | \space \phi(f(x)+I) = n, n \in \mathbb{Z_2}$

I need to show this is surjective so I need to find elements in $\mathbb{Z}[X]/I$ which map to either 0 or 1. Having some trouble seeing this.

After that I would claim that $ker(\phi) = I$ since ideals are kernels so that $\mathbb{Z}[X]/I \cong \mathbb{Z_2}$ by the first isomorphism theorem.

Since $\mathbb{Z_2}$ is a field, we know that $\mathbb{Z}[X]/I$ is a field which implies that I is maximal which implies that it is also prime.

5. Apr 17, 2013

### Kreizhn

If you take the isomorphism route, I think it would be better to define the map $\phi: \mathbb Z[x] \to \mathbb Z_2$ by $p(x) \mapsto [p(0)]$. Namely, each element is mapped to its constant term mod 2. Your map is certainly surjective, so just check that the kernel is I, which isn't too bad.

Alternatively, my statement is just the third isomorphism theorem. If you are familiar with the fact that $R[x]/(x) \cong R$ then you get that $\mathbb Z[x]/(x,2) \cong [\mathbb Z[x]/(x)]/(\bar 2) \cong \mathbb Z/(2) \cong \mathbb Z_2$ pretty quickly.

6. Apr 17, 2013

### Zondrina

We aren't allowed to use the T.I.T, only the F.I.T. I took the isomorphism route and it cleaned up quite nicely.

Now as for part (c) which is the number of elements.

Either we have $f \in I$ when $f(0)$ is even or $f \in 1+I$ if $f(0)$ is odd
which implies $\mathbb{Z}[X]/I$ is a field with two elements.

7. Apr 17, 2013

### Kreizhn

Sure. Alternatively, isomorphisms preserve cardinality, so as $\mathbb Z[x]/(x,2) \cong \mathbb Z_2$ and the right-hand-side has two elements, you get the same answer.

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