What is the connection between ideals and fields in Z[x]?

[STEMucator]

Homework Statement

Part b and c.

http://gyazo.com/821bceafd1c49adc366c63208066bd05

Homework Equations

Z[x]/I is a field ⇔ I is maximal.

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.

 

[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].

 

[micromass]

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

 

[STEMucator]

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

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.

 

[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.

 

[STEMucator]

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.

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.

 

[Kreizhn]

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.

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.