Consider $\mathbb Z [x]$ is th polynomial ring over $\mathbb Z$. Why $\mathbb Z[x]/(2x-1)$ is not a field?
Why is $\mathbb Z[x]/(2x-1)$ not a field?
0
$\begingroup$
abstract-algebra
ring-theory
-
3Intuitively you are adding $1/2$ to the integers. But this is not enough to invert all elements. i.e, we still cant invert $3,5,7,\cdots$. The important property being that $\mathbb{Z}[X]$ is not a PID and $(2x-1)$ is contained in for example $(2x-1,3x-1)$ and so on. – 2017-02-21
2 Answers
7
If $ [3] $ denotes the coset $ 3 + (2x - 1) $, then $ [3] $ has no inverse in this quotient. Indeed, if it did, there would be $ a(X), b(X) \in \mathbf Z[X] $ such that
$$ 3 a(X) + (2x - 1)b(X) = 1 $$
Now, evaluate at $ X = 1/2 $ (we can do this because this equality is also an equality in $ \mathbf Q[X] $) to get a contradiction.
2
Hopefully you're aware of the following theorem:
Theorem: Let $R$ be a commutative ring with unity and $I \subset R$ an ideal. Then $R/I$ is a field if and only if $I$ is maximal.
So your question reduces to whether $(2x-1)$ generates a maximal ideal. To show that it isn't maximal, we just need to find an ideal properly containing it.
EDIT: Removed suggestion of ideal generated by $x,1$. As pointed out in the comment below, this is not a proper ideal.