# Can't understand this statement about factor rings

1. Mar 20, 2013

### Silversonic

Claim: If $p \in R$ is irreducible and non-zero, then p is irreducible and prime in R[X]

Proof: Let I be the ideal generated by p in R[X]. Clearly I consists of polynomials where all the coefficients are divisible by p. Therefore the factor ring R[X]/I is the same as (R/<p>)[X].

The proof goes on and I'm able to understand the rest. But I cannot make physical sense of how these last two rings could be the same or how I would even go about understanding why.

R[X]/I is just a factor ring, whose elements of the cosets of I in R[X].

(R/<p>)[X] is the ring of polynomials whose coefficients are cosets of <p> in I.

So even in the definition I'm confused, R[X]/I is a set of sets of polynomials with coefficients in R. (R/<p>)[X] is just a set of polynomials with coefficients in (R/<p>). How could the two possibly be the same if this last statement is the case?

My lecturer has put "Sketch of proof" to begin with (instead of just "proof") and, since I'm completely able to understand the rest, this leads me to believe the bolded statement is actually much more in depth than he's gone in to. There's no justifaction for "these are the same". Is there an isomorphism between the rings and thus they are classified as the same up to isomorphism?

Any help appreciated.

2. Mar 20, 2013

### jbunniii

You are correct that the rings are not the same. However, they are isomorphic via the map $\phi : R[x]/I \rightarrow (R/\langle p\rangle)[x]$ defined by
$$\phi((a_n x^n + \ldots + a_0) + I) = (a_n + \langle p\rangle)x^n + \ldots + (a_0 + \langle p\rangle)$$
Of course you have to show that this is well defined, a bijection, and a homomorphism. To show that it is well defined, suppose that
$$(a_n x^n + \ldots + a_0) + I = (b_n x^n + \ldots + b_0) + I$$
Therefore,
$$(a_n x^n + \ldots + a_0) - (b_n x^n + \ldots + b_0) = (a_n - b_n)x^n + \ldots + (a_0 - b_0) \in I$$
which means that $a_n - b_n, \ldots, a_0 - b_0$ are all divisible by $p$. So $a_i - b_i \in \langle p \rangle$ for each $i$, whence $a_i + \langle p \rangle = b_i + \langle p \rangle$.

I will leave the proof that $\phi$ is a bijection and a homomorphism to you.