I have the following question on ring theory, I would like somebody to help me:

Let $R =\mathbb Z_2[x]$, and consider the ideal $I$ of $R$ generated by the irreducible polynomial $f(x)=x^2+x+1\in R$. Show that the factor ring $R/I$ is a field.

Here is the solution:

$R/I =\{f(x)+I\mid f(x)\in R\}$ so by division algorithm $f(x)=q(x)( x^2+x+1)+r(x)$, then $f(x)= q(x)( x^2+x+1)+ax+b$. Hence in $R/I$: $$f(x)+I=q(x)(x^2+x+1)+ax+b+I=ax+b+I$$ $$R/I=\{ax+b+I\mid a,b\in\mathbb Z_2\}=\{0+I,1+I,x+I,1+x+I\}$$ Every element in $R/I$ has a multiplicative inverse, hence is a field.

I'm not clear with this - somebody help me, thanks.

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    $\begingroup$ I have not gone through the question thoroughly! But, if every element in $R/I$ has an inverse, $0_{R/I}=1_{R/I}$ and hence, $R/I$ will be a zero ring! So, you probably wanted to say, every non-zero element has an inverse! $\endgroup$ – user21436 Jan 29 '12 at 7:15
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    $\begingroup$ Hint: $\mathbb{Z}_2[x]/(x^2 + x + 1) = \mathbb{Z}/(2, x^2 + x + 1)$. To show that this is a field is equivalent to showing that $(2,x^2 + x + 1)$ is a maximal ideal in thepolynomial ring $\mathbb{Z}[x]$. $\endgroup$ – user38268 Jan 29 '12 at 7:19
  • $\begingroup$ is the last line the part you don't follow? $\endgroup$ – Tim kinsella Jan 29 '12 at 7:20
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    $\begingroup$ $f\;$ is irreducible $\implies$ $I=(f)$ is maximal $\implies$ $R/I$ is a field... $\endgroup$ – anon Jan 29 '12 at 7:24
  • 1
    $\begingroup$ Sorry bad mistake it should have been $\mathbb{Z}[x]/(2,x^2 + x + 1)$. $\endgroup$ – user38268 Jan 29 '12 at 7:51

At a more abstract elevation, we can view this particular case as a corollary to more general facts from commutative algebra that I expect can be found in most introductory notes or texts.

Lemma 1. If $R$ is an integral domain, then the polynomial ring $R[x]$ is also an integral domain.

Proof. Let $ab=0$ in $R[x]$. Write $a=a'x^n+\cdots$ and $b=b'x^m+\cdots$. Then we have

$$ab=a'b'x^{n+m}+\cdots=0\implies a'b'=0\implies a'\text{ or }b'=0.$$

This however contradicts the implicit supposition that $\deg a=n$ and $\deg b=m$.

Lemma 2. If $K$ is a field, then the polynomial ring $K[x]$ is a principal ideal domain.

Proof. Let $G=(g_1,\dots,g_n)$ be an ideal of $K[x]$. By induction on Bezout's identity, there must exist polynomials $a_1,\dots, a_n$ such that $a_1g_1+\cdots a_ng_n=d=\gcd(g_1,\dots,g_n)$. Since each $g_i$ is a multiple of $d$, every $K[x]$-linear combination of the $g_i$'s is a multiple of $d$ and conversely, since $d\in G$ and $G$ is an ideal, every multiple of $d$ is in $G$. Therefore $G=(d)$, which is principal. By Lemma 1, we also know that $K[x]$ is a domain, so it is a principal ideal domain.

Lemma 3. If $K$ is a field and $f\in K[x]$ is irreducible, the ideal $(f)$ is maximal.

Proof. Suppose $(f)\subsetneq G\subsetneq K[x]$. By Lemma 2, $G=(g)$ for some $g$. Now $f\in(g)$ implies $f=ag$ for some $a$; if $a$ is a unit then $g=a^{-1}f\in(f)$ implies $(g)\subseteq(f)$, contradicting our hypothesis, or else $f$ factors nontrivially, which contradicts our supposition that $f$ is irreducible.

Lemma 4. If $R$ is an integral domain and $I$ a maximal ideal, then the factor ring $R/I$ is a field.

Proof. Let $a + I\in R/I$ be nonzero. Since $\{i+ra:i\in I,r\in R\}$ is an ideal in $R$ containing $I$ and $a$, it must be all of $R$, whence $i+ra=1$ for some $i\in I,r\in R$, ergo $(r + I)(a+I)=1+I$. Since the nonzero cosets have inverses, the factor ring must be a field.

Now $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$ is a field, hence $\mathbb{F}_2[x]$ is a PID. The polynomial $f=x^2+x+1$ is irreducible, so $(f)$ is maximal, from which we conclude $\mathbb{F}_2[x]/(f)$ is a field. For more on commutative algebra, I recommend the Algebra Handouts in the Number Theory I course or the Commutative Algebra notes from Pete L. Clark (the latter is much more advanced).

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  • $\begingroup$ In fact, as you no doubt know, in Lemma 4 you don't need $R$ to be an integral domain, you just need it to be commutative with unity. $\endgroup$ – Arturo Magidin Feb 2 '12 at 4:18

At a more "naive" level, you can just show directly that $R/I$ is a field. To do this, let's "rename" the elements of $R/I$:

$0+I \to 0$, $1+I \to 1$, $x+I \to a$, $1+x+I \to b$.

$(R/I,+,\cdot)$ is a commutative ring (being a quotient ring of a commutative ring). It only remains to be shown that $\{1,a,b\}$ is a group (which is necessarily abelian). We already have associativity of multiplication, so we only need to show closure, which we can do explicitly:

$1 \cdot 1 = 1$, $1 \cdot a = a$, $1 \cdot b = b$, $a \cdot a = b$ because:

$(x + I)(x + I) = (x^2 + I) = x^2 + 2x + 2 + I = (x + 1) + (x^2 + x + 1) + I = x +1 + I$

$a \cdot b = 1$:

$(x + I)(1 + x + I) = x^2 + x + I = x^2 + x + 2 + I = 1 + x^2 + x + 1 + I = 1 + I$

$b \cdot b = a$:

$(x + 1 + I)(x + 1 + I) = x^2 + 2x + 1 + I = x + x^2 + x + 1 + I = x + I$

and we have the other 3 possible products by commutativity. The above calculations also provide us with the inverses explicitly (although we already know that we have them, since our set is finite):

$1^{-1} = 1, a^{-1} = b, b^{-1} = a$.

Congratulations, you're the proud parent of a 4-element field!

(Note to other answerers: showing $(f)$ is maximal (which is equivalent to showing $x^2 + x + 1$ is irreducible in $\mathbb{Z}_2[x]$, which is equivalent to showing neither 0 nor 1 is a root of $f$, since $\text{deg}(f) = 2$, so if $f$ is reducible it has a linear factor) is probably the route the instructor is hoping the original poster will take, and is certainly "higher-level" and perhaps more enlightening. Nevertheless, why use a sledgehammer to swat a fly?)

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