Quotient Ring of a Field: Is it Trivial or Isomorphic to the Field?

In summary, the conversation discusses the quotient ring of a field and how it can either be the trivial one or isomorphic to the field. To prove this, it is shown that if the ideal $N$ is not equal to {0}, then it is actually equal to the entire field $F$. This is proven by showing that any non-zero element of $N$ can be multiplied by the inverse of any element in $F$ to obtain 1, thus showing that $N$ contains all elements of $F$. It is also mentioned that the conditions for the homomorphism $\phi$ to be bijective are that $N$ must be equal to {0}.
  • #1
Fantini
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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)
 
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  • #2
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}.
 
  • #3
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!
 

1. What is a quotient ring of a field?

A quotient ring of a field is a mathematical structure that is created by dividing a field by a non-zero ideal. It is a set of equivalence classes of elements of the field with addition and multiplication operations defined on them.

2. How is a quotient ring of a field different from a field?

A quotient ring of a field is a modification of a field, where the elements of the field are grouped into equivalence classes. It is not a field itself, but it inherits some properties of a field such as closure under addition and multiplication.

3. What is the significance of quotient rings of fields in mathematics?

Quotient rings of fields are important in abstract algebra as they provide a way to study the structure of a field by breaking it down into smaller, simpler parts. They also have applications in other areas of mathematics, such as algebraic geometry and number theory.

4. How are quotient rings of fields used in practical applications?

Quotient rings of fields have practical applications in coding theory, where they are used in the construction of error-correcting codes. They are also used in cryptography to encrypt and decrypt messages using algebraic operations.

5. Are there any limitations to quotient rings of fields?

Yes, there are some limitations to quotient rings of fields. For example, not all fields have a non-zero ideal, which means that not all fields can have a quotient ring. Additionally, quotient rings of fields do not always have the same properties as the original field, so they cannot always be used as a substitute for the field itself.

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