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Quotient ring of a field

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Good afternoon! Along the same lines as the other, here is the question:

Show that the quotient ring of a field is either the trivial one or is isomorphic to the field.

My answer: Let $N$ be an ideal of the field $F$. Assume that $N \neq \{ 0 \}$. Consider the homomorphism $\phi: F \to F / N$ defined by $\phi(a) = a + N$. If we show that it is one-to-one and onto we are done. It is clearly surjective, thus all that is left is to show injectivity. If $a \neq b$ then we will have $a + N \neq b + N$, but this is none other than $\phi(a) \neq \phi(b)$.

Thanks for all help! (Yes)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
a+N ≠ b+N does not follow from a ≠ b. all we can say from a+N ≠ b+N is that:

a-b is not in N.

what you need here is that if N ≠ {0}, then N = F, so that $\phi$ is the 0-map.

suppose N is a non-trivial ideal of F. since N is non-trivial there exists a ≠ 0 in N.

since a is non-zero, and F is a FIELD, we have 1/a in F.

since N is an IDEAL, we have 1 = (1/a)a in N.

thus, for any x in F, we have x = x(1) is in N, since N is an ideal.

since N contains all of F, N = F, as desired.

you have your conditions backwards, as well, you need to show that $\phi$ is bijective iff N = {0}.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
That's quite a few arguments missing (not to mention mine is wrong). I need to stop and pay more attention whenever I feel uneasy, because at all times it has been proved the uneasiness is justified.

Thanks Deveno!