Using First Isomorphism Theorem with quotient rings

In summary, the conversation discusses using the First Isomorphism Theorem to prove the isomorphism between \mathbb{Z}_2[x]/I and \mathbb{Z}_2[x]/J. The suggested approach is to use the given homomorphism \varphi: \mathbb{Z}_2[x] \rightarrow \mathbb{Z}_2[x] given by x \rightarrow x+1, and show that it is a surjective homomorphism with kernel I. This will then prove the isomorphism between \mathbb{Z}_2[x]/I and \mathbb{Z}_2[x]/J.
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PirateFan308
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Homework Statement


Try to apply the First Isomorphism Theorem by starting with a homomorphism from a polynomial ring [itex]R[x][/itex] to some other ring [itex]S[/itex].

Let [itex]I = \mathbb{Z}_2[x]x^2[/itex] and [itex]J = \mathbb{Z}_2[x](x^2+1)[/itex]. Prove that [itex]\mathbb{Z}_2[x]/I[/itex] is isomorphic to [itex]\mathbb{Z}/J[/itex] by using the homomorphism [itex]\mathbb{Z}_2[x] \rightarrow \mathbb{Z}_2[x][/itex] given by [itex]x \rightarrow x+1[/itex].


Homework Equations


First isomorphism theorem states that if [itex]\varphi: R \rightarrow S[/itex] is a homomorphism then [itex]R / ker \varphi[/itex] is isomorphic to Image[itex]\varphi[/itex]


The Attempt at a Solution


The only thing that I was able to think of was to use the First Isomorphism Theorem twice to find some ring S that both [itex]\mathbb{Z}/I[/itex] and [itex]\mathbb{Z}/J[/itex] are isomorphic (hence making them isomorphic to each other), but I am completely stumped about which ring S I should use. I'm also confused about how the homomorphism given will help?

I attempted to find [itex]kernel (\varphi)[/itex], but had no luck as I'm confused about how which functions would be s.t. [itex]f(x+1)=0[/itex] in [itex]\mathbb{Z}_2[x][/itex]?

I did notice that [itex]\varphi (x^2+1) = (x+1)^2+1 = x^2 + 2x + 1 + 1 = x^2[/itex] in [itex]\mathbb{Z}_2[x][/itex]. Similarly, [itex]\varphi (x^2) = (x+1)^2 = x^2 + 2x + 1 = x^2 + 1[/itex], but I'm not sure how exactly this could help.

Any hint on where to get started would be greatly appreciated! Thanks!
 
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  • #2




Thank you for your question. It seems like you are on the right track with using the First Isomorphism Theorem to prove the isomorphism between \mathbb{Z}_2[x]/I and \mathbb{Z}_2[x]/J. However, you do not need to find a specific ring S that both \mathbb{Z}/I and \mathbb{Z}/J are isomorphic to. Instead, you can use the given homomorphism \varphi: \mathbb{Z}_2[x] \rightarrow \mathbb{Z}_2[x] given by x \rightarrow x+1 to show that \mathbb{Z}_2[x]/I and \mathbb{Z}_2[x]/J are isomorphic to each other.

To do this, you can start by showing that \varphi is a surjective homomorphism with kernel I. Since \varphi is surjective, by the First Isomorphism Theorem, \mathbb{Z}_2[x]/I is isomorphic to \varphi(\mathbb{Z}_2[x]) = \{f(x+1) \mid f(x) \in \mathbb{Z}_2[x]\}. Can you see how this set is isomorphic to \mathbb{Z}_2[x]/J?

To show that \varphi has kernel I, you can use the fact that \varphi(x^2) = (x+1)^2 = x^2 + 2x + 1 = x^2 + 1 in \mathbb{Z}_2[x], as you mentioned. This means that x^2 \in ker \varphi. Can you think of any other elements in I that would also be in ker \varphi?

I hope this helps you get started on your proof. Good luck!
 

What is the First Isomorphism Theorem?

The First Isomorphism Theorem is a fundamental result in abstract algebra that relates the structure of a group or ring to its quotient structure. It states that if G and H are groups (or rings) and φ: G → H is a surjective homomorphism, then G/ker(φ) is isomorphic to H. In other words, the quotient group/ring is structurally equivalent to the target group/ring.

What is a quotient ring?

A quotient ring is a mathematical structure that is obtained by "quotienting out" a ring by a certain subset. This means that we are essentially dividing the ring into smaller pieces based on some criteria. For example, we can quotient out a ring by its ideal, which results in a new ring with fewer elements, but with the same underlying structure.

How is the First Isomorphism Theorem used with quotient rings?

The First Isomorphism Theorem can be used to show that the quotient ring obtained by dividing a ring by its ideal is isomorphic to another ring. This is because the homomorphism used in the theorem maps elements in the original ring to elements in the quotient ring, preserving the underlying structure. This allows us to identify the quotient ring with a different ring, making it easier to study and understand.

What is the importance of the First Isomorphism Theorem in abstract algebra?

The First Isomorphism Theorem is an important result in abstract algebra because it allows us to analyze and understand complicated algebraic structures by breaking them down into simpler pieces. It also helps us to identify when two seemingly different structures are actually equivalent, which can be useful in solving problems and making connections between different areas of mathematics.

Are there any limitations to using the First Isomorphism Theorem with quotient rings?

Yes, there are some limitations to using the First Isomorphism Theorem with quotient rings. One limitation is that the homomorphism used must be surjective, meaning that every element in the target ring must have a corresponding element in the original ring. Additionally, the theorem only applies to quotient rings that are obtained by dividing out by an ideal, so it cannot be used in other types of quotient structures. Finally, the theorem does not provide any information about the specific elements or structure of the quotient ring, it only guarantees that it is isomorphic to the target ring.

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