# Math Help - Polynomial Rings - Quotient Rings -- D&F Section 7.4 - Exercise 15

1. ## Polynomial Rings - Quotient Rings -- D&F Section 7.4 - Exercise 15

I am beginning to work on Dummit and Foot Section 7.4 Exercise 15 (page 257 - see attachment)

Exercise is as follows:

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"Let $x^2 +x + 1$ be an element of the polynomial ring $E= {\mathbb{F}}_2 [x]$ and use bar notation to denote the passage to the quotient ring $\overline{E} = {\mathbb{F}}_2 [x]/$.

Prove that $\overline{E}$ has four elements $\overline{0}, \overline{1}, \overline{x}, \overline{x+1}$"

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Now from the result in D&F Section 7.4 Exercise 14 (see attachment) we have that every element of $\overline{E}$ is of the form $\overline{p(x)}$ for some polynomial p(x) $\in$ (Z/2Z)[x] of degree less than 2 ( i.e. of degree 1 or 0).

Listing such polynomials in (Z/2Z)[x] we have 0,1, x, x+1.

So the elements of $\overline{E}$ are $\overline{0}, \overline{1}, \overline{x}, \overline{x+1}$ (one element in each coset!)

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But what if we made the quotient ring $\overline{G} = {\mathbb{F}}_2 / $

The same reasoning as above applies ... so (???) the elements of $\overline{G}$ seem to be the same as $\overline{E}$

But this does not seem right .... $\overline{E}$ is the quotient ring modulo $x^2 + x + 1$ and $\overline{G}$ is the quotient ring modulo $x^2 + 1$

Intuitively it seems to me that $\overline{E}$ and $\overline{G}$ should be different! (or maybe only the degree of the polynomial is significant?)

Can someone please clarify this matter for me?

Peter

2. ## Re: Polynomial Rings - Quotient Rings -- D&F Section 7.4 - Exercise 15

they ARE different. note that x2+1 is not irreducible over F2, as it has the factorization:

x2 + 1 = x2 + 0x + 1 = x2 + (1+1)x + 1 = x2 + x + x + 1 = (x + 1)2.

this means that:

$\overline{G} = \mathbb{F}_2[x]/\langle x^2+1 \rangle \cong \mathbb{F}_2[x]/\langle x+1 \rangle \times \mathbb{F}_2[x]/\langle x+1 \rangle \cong \mathbb{F}_2 \times \mathbb{F}_2$.

$\overline{E}$ is a FIELD, whereas $\overline{G}$ isn't even an integral domain (it has the zero divisor $\overline{x+1}$).

however, you are partially right....if the two polynomials we're modding by are both IRREDUCIBLE and of the same degree, we obtain isomorphic fields (when our polynomials rings have coefficients in a finite field).

3. ## Re: Polynomial Rings - Quotient Rings -- D&F Section 7.4 - Exercise 15

Thanks.

Most helpful re my understanding of polynomial rings

Peter

4. ## Re: Polynomial Rings - Quotient Rings -- D&F Section 7.4 - Exercise 15

Reflecting on your post I have concluded that I do not yet fully understand the situation regarding my initial post.

Regarding $\overline{E} = {\mathbb{F}}_2 [x] / < x^2 + x + 1>$ , (following the solution in Project Crazy Project) I drew on the fact that every element of $\overline{E}$ has the form $\overline{p(x)}$ for some polynomial of degree 0 or 1 in E, and all such polynomials represent distinct elements in $\overline{E}$ (as per the theorem/result of D&F Section 7.4 Exercise 14 (a) - see attachment)

BUT ... ... as I argued previously it seems (???) to me that these arguments based on 14(a) apply also to $\overline{G} = {\mathbb{F}}_2 [x] / $ (neglecting for the moment you have shown that these arguments are not sufficient to determine the situation!) ... since there is no mention in 14(a) of irreducibility and/or fields.

Can you please explain how irreducibility comes into this situation?

Also is my determination of the elements of $\overline{E}$ complete and correct?

Peter

5. ## Re: Polynomial Rings - Quotient Rings -- D&F Section 7.4 - Exercise 15

the two (quotient) rings have exactly the same additive structure. their multiplication is different (the generator of the ideal determines how we "reduce higher order polynomial cosets").

your 2 quotient rings have the same exact number of elements, in this case, 4. the additive group of BOTH rings is, in this case, isomorphic to the klein 4-group. so we have an example of two distinct ways we can introduce a ring multiplication onto Z2xZ2.

the kind of ideal we quotient by greatly affects the kind of quotient ring we get. prime ideals (which are maximal in commutatitve rings with identity) give us a quotient ring which is an integral domain. in effect, we "mod the bad stuff out".

if the ideal we are modding by ISN'T maximal, that is, if J is an ideal between I and R, then J/I is an ideal of R/I. in particular, R/I cannot be a field (since fields have no "intermediate ideals").

and that's where irreducibility comes in. in a unique factorization domain (which is the case for R[x], where R is a UFD) prime = irreducible. in particular, all fields are unique factorization domains (rather trivially, since everything except 0 is a unit).

so every polynomial ring F[x], for any field F, is a UFD. also, F[x] is "even better" than a UFD, it's a PID (principal ideal domain). so in this case:

maximal ideal = prime ideal = ideal generated by prime element = ideal generated by irreducible element

(the above chain of concept equality does NOT hold for arbitrary polynomial rings R[x], fields are "special" because their ideal structure is VERY limited).

let's look at the ideal structure of Z2xZ2 (with the direct product multiplication). this set has only 4 elements, so there's only 16 possible subsets:

the empty set, which is not an ideal since an additive subgroup of (R,+) (which every ideal is) must be non-empty.

{(0,0)} <---clearly an ideal

no other one-element set can possibly be an ideal, since an ideal MUST contain the additive identity of R (this eliminates the other 3 1-element sets).

we can similarly eliminate all 2-element sets that do NOT contain (0,0).

this leaves just 3 2-element sets to check out:

{(0,0),(0,1)}, {(0,0),(1,0)} and {(0,0),(1,1)}. each of these is an additive subgroup so we only need to check the multiplicative "absorption" property. since (1,1) is the multiplicative identity, and (0,0) always sends everything to (0,0), this means only checking that for each 2-element set S = {(0,0),(a,b)}:

(1,0)(a,b) and (0,1)(a,b) are in S. i urge you to verify that {(0,0),(0,1)} and {(0,0),(1,0)} are ideals but {(0,0),(1,1)} is not.

no 3-element set is an ideal (violates lagrange's theorem).

the only 4-element set (the whole ring) is also clearly an ideal.

so the ideals of the direct product Z2xZ2 are:

Z2xZ2, Z2x{0},{0}xZ2 and {0}.

note that since we have ideals besides R and (0), this is yet another way to see that Z2xZ2 (with the direct product multiplication (a,b)*(c,d) = (ac,bd)) is not a field.

it is possible to define a *different* multiplication on Z2xZ2:

(a,b)*(c,d) = (ac+bd,ad+bc+bd), which gives a different ideal structure.

6. ## Re: Polynomial Rings - Quotient Rings -- D&F Section 7.4 - Exercise 15

Thanks for the guidance and help in my quest to understand polynomial rings!

Appreciate the help.

Peter