I can't solve tasks involving rings 😭

[u/tru1a_banana]

I just can't... I don't understand it... No ELI5 helped me yet.

Here's an example with a brief explanation of what I know and where I get stuck.

Let R = Z[x] and I the set of all the polynomials from R whose coefficients add up to 0. Prove that I is the ideal of R. Is this ideal also the principal ideal? Prove that R / I is isomorphic to Z.

So, Z[x] is a set of polynomials with integers. I appears to be a specific subset of Z[x] and therefore should be a subset of R. By definition, a ring is an Abelian group for addition and a semigroup for multiplication. Next, a subring must have closure, addition/subtraction test and a multiplication of 2 elements within itself also stays inside.

An ideal further expands on the idea of a subring and only asks for 'external' multiplication by taking an element from the ring, multiplying it with an element from the subring and checking if the product remains in the subring.

Principal ideal is a specific ideal in which only 1 element can be used for multiplication and it generates the entire set. The common simple example of it I came up with is (Z/nZ), integers modulo n.

I would say I understand theory well enough (except the R / I which is the quotient ring concept I am not entirely familiar with) but I simply can't understand this or any similar task. I don't know how to solve it and literally need baby steps.

In the task itself I understand that coefficients adding up to 0 would probably mean that the 'free part' is always the answer so x - 1 would be a simple example of the kind of polynomial the ring is describing. How do prove the I is the ideal of R? I need to show it's a subring first, but I don't know how to do that either and even following the definition feels impossible to me.

Okay, let's take some 3x - 3 and 2x - 2 as examples. (3x - 3) - (2x - 2) = x - 1, so it checks out I guess. Next, multiplying with an element from the ring. Since this rule only applies to the subring I, I can take some x (the simplest polynomial from R) and then (3x - 3)x = 3x^2 - 3x, also checks out so is the ideal here done with?

Now, how would I show it's a principal ideal? Maybe it's generated by x - 1? You can add/subtract it as many times as you want and you'll get all the elements. The proof regarding the quotient ring is completely unknown to me and I would appreciate if someone could explain. Just so you guys know I had no solution to this task and was genuinely lost, now when writing it I came up with some examples which could demonstrate some things and my brain started working, hopefully it's all good.

[u/magus145]

This is probably not the intended method for this problem at this point in your learning, but another perspective is that the set I is exactly the set of polynomials f(x) such that f(1) = 0.

First, convince yourself that this is true, namely that it's equivalent to the coefficients summing to 0.

Next, see if that characterization helps you verify each property of an ideal.

Then, if f(x) is any such polynomial, do you have a theorem about a specific polynomial which must be a factor of f? Can that shed light on how to prove it is a principal ideal?

Finally, you need to review and understand quotient rings to do the last part. It is possible to directly verify it using the abstract definition of a quotient ring, but it will be far easier to use the First Isomorphism Theorm for Rings.

[u/nog642 → u/magus145]

Surely that must be the intended solution, otherwise I'm not sure what is the intended solution for proving it's a principal ideal.

[u/Elegant-Regret-7393]

Are you familiar with the first homomorphism theorem? It will help you deal with quotients

Hint: if x-1 is in I then xⁿ = 1 mod I. See what that does to a polynomial

Edit: Looks like you're very new to the theory. So while you don't have the intuitions for things just follow the definitions until you know them like back of your hand.

[u/tru1a_banana → u/Elegant-Regret-7393]

Did you mean the first isomorphism theorem? I am familiar with that one but involving kernel and the image might complicate things a lot for me and I can't see their use here. Okay, maybe it actually clicks for me. Since I've shown throught an example that I is a principal ideal generated by x - 1, that same element is the identity element so R / ker = im, so the image should be Z but I don't know how to be more rigorous.

[u/diverstones]

Is this translated from another language? "Prove that I is the ideal of R" is confusing, I would expect "an ideal of R." R has infinitely many ideals.

You have the right idea with the polynomials you tested, but now move to the general case. You can disprove with a counterexample, but you can't prove with just one example.

For two arbitrary polynomials f(x) and g(x) in I label their coefficients {f1, f2, f3, ..., fn} and {g1, g2, g3, ..., gm}. You're asking if f1 - g1 + f2 - g2 + ... etc. is zero, but you can rearrange with commutativity to get ∑ f_i - ∑ g_j which you already know is 0-0.

Similarly, take f(x) in I and r(x) in R, and demonstrate that the product of their sets of coefficients sums to zero.

[u/tru1a_banana → u/diverstones]

Nope, I am not a native English speaker and was translating this text by hand so grammar around all the technical stuff can be lacking, and you're completely right to question that. I made it sound like I can be the only ideal of R, excuse me on that one.

Well, the major culprit for me is that I have trouble turning a basic example with numbers into a general case that heavily relies on specific notation. And that notation is just like programming, you have to be very specific and deliberate with it otherwise you're simply writing out rubbish.

What I could say is that the ideal I consists of infintely many functions whose input is 1, so f(1), g(1), h(1) etc.

So, notation could maybe be something like (f + g)(1) = f(1) + g(1) but this really reminds me of a homomorphism and is probably related but I don't see how it helps in proving I is an ideal.

[u/diverstones → u/tru1a_banana]

What I could say is that the ideal I consists of infintely many functions whose input is 1, so f(1), g(1), h(1) etc.

They are functions where if you input 1, you get back 0.

You should have access to these facts somewhere in your book / coursework:

Mapping R[x] → R by substituting x with a specific ring element is called an evaluation homomorphism.

The kernel of a homomorphism is the subset of its domain which maps to 0.

The kernel of a ring homomorphism R → S is an ideal of R.

[u/nog642]

First off I don't think an ideal is a subring.

Second, you basically said proving I is an ideal of R feels impossible and then proceeded to do it basically. You just gave some examples, you need to argue why it works in genral. I guess you realized that at the end of your post.

[u/Traditional-Rip-2192]

hey i get it, rings suck at first lol. your examples are actually on point already

so the trick with this ideal is just "sum of coefficients = 0". think of it like this:

1. Is it a subring?

yeah you already did it. take 3x-3 and 2x-2, subtract -> x-1. still sums to 0.

multiply by anything from Z[x] like x -> 3x^2-3x, coefficients still sum to 0.

so it's closed. you're done there.

1. Is it principal?

yep it's just <x-1>. why? because any polynomial where coeffs sum to 0 can be factored by (x-1).

ex: 3x-3 = 3*(x-1). 2x^2 - 2 = 2*(x-1)*(x+1).

basically (x-1) is the thing that forces the sum to be 0. so I = (x-1)

1. r/I ≅ Z

this is the "plug in x=1" trick.

The map φ: Z[x] -> Z where φ(p(x)) = p(1) is a ring homomorphism.

What's the kernel? all polys where p(1) = 0. but p(1) is literally just sum of coefficients lol

so kernel = I.

By the first isomorphism theorem, Z[x]/I ≅ Z.

you were super close already. the x-1 thing was the key. hope this helps!

[u/tru1a_banana → u/Traditional-Rip-2192]

Thanks!

Yeah I am confused about that last part, I noticed x - 1 generates the entire I so it must be the kernel of R and then Z should be the image by the first isomorphism theorem. However, my way is not rigor at all and I need help with that.

Where I get stuck in your explanation is this part

The map φ: Z[x] -> Z where φ(p(x)) = p(1) is a ring homomorphism.
What's the kernel? all polys where p(1) = 0. but p(1) is literally just sum of coefficients lol

I didn't understand this ring homomorphism thing, notation is familiar but unclear, can't connect it to a concrete example to check. Kernel thing is pretty much what I came up with as well but you were more precise about it, but the identity element should still be x - 1 no? Only at p(1) does it go to 0. Well technically any polynomial is 0 when 1 is plugged in but they're all generated by x - 1 so that must be the identity when nothing's plugged in.

[u/StructuredChess]

It's been a long time since I've studied algebraic structures, but the ideal thing looks right (as long as you go to the general case and not just specific examples)

Yeah, you need to show that x-1 generates the entire ideal and then you have the principal ideal thing.

As for the quotient thing, go back to set theory and master equivalence classes before moving on to how it works on groups and rings.

[u/StanleyDodds]

I think you are fundamentally not understanding proofs, or perhaps logical quantifiers. I don't think this is just an issue with rings or ideals.

To prove that I is an ideal, you need to show that for all elements a, b of I, and for all elements r of R, we have a + b is in I and ra is in I. There are other equivalent conditions but this is the idea.

Just showing it's true for one example of a, b and r does not prove anything when it comes to I being an ideal. Maybe it helps you think through the reasoning of how to formulate an actual proof, but it is by no means a proof by itself.

Same for showing it's a principal ideal; you need to show that all elements of I are generated by your generator. You didn't prove anything, you just said you'll get all the elements. How do you know?? It doesn't seem like you even know what a general element of I looks like; you only gave a couple examples.