# Math Help - Extension Fields - Gallian Ch 20 - Example 1

1. ## Extension Fields - Gallian Ch 20 - Example 1

I am working through Gallian's book "Contemporary Abstract Algebra" (Fifth Edition) and I am on Chapter 20 - Extension Fields.

I need help with Example 1 on page 345 - see attached pdf. [My text below is on the pdf - I am trying to understand the equations]

Example 1 states:

Let $f(x) = x^2 + 1 \in Q[x]$

Then in $E = Q[x]/< x^2 + 1>$ we have

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

$= x^2 + + 1$

Could you also explain how the last two eqations of the 4 are equal - see attached sheet.

Bernhard

2. Originally Posted by Bernhard
I am working through Gallian's book "Contemporary Abstract Algebra" (Fifth Edition) and I am on Chapter 20 - Extension Fields.

I need help with Example 1 on page 345 - see attached pdf. [My text below is on the pdf - I am trying to understand the equations]

Example 1 states:

Let $f(x) = x^2 + 1 \in Q[x]$

Then in $E = Q[x]/< x^2 + 1>$ we have

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

$= x^2 + + 1$

Could you also explain how the last two eqations of the 4 are equal - see attached sheet.

Bernhard

You´re supposed to know elementary ring theory BEFORE getting into fields extensions, and thus you must know that

if R is a ring and I and ideal in it, then for any element $a\in R\,,\,\,(a+I)^2:=a^2+I\in R/I$ , since we define $(a+I)(b+I):=ab+I$ ...

I don't understand what "the last equations of the 4..." can possibly mean, but remember

that $a+I=I\iff a\in I$ , with $I$ being, of course, the zero element of

the quotient ring $R/I$

Tonio

3. one way some authors approach this is the following:

let α = x + <x^2 + 1> in Q[x]/<x^2 + 1>. note that the mapping q ---> q + <x^2 + 1> is an monomorphism (when q is in Q), so we can identify

Q with its image under this mapping.

then α^2 + 1 = (x + <x^2 + 1>)(x + <x^2 + 1>) + 1 + <x^2 + 1> = x^2 + 1 + <x^2 + 1> = <x^2 + 1> = 0 + <x^2 + 1> = 0.

(where this last 0 is the additive identity of Q[x]/<x^2 + 1>, which we identify with 0 (in Q)).

in other words: in the larger ring (which is actually a field) Q[x]/<x^2+1>, α^2 = -1, that is, α is a square root of -1 (actually, -1 + <x^2 + 1>,

but this "acts just like the rational number -1").

4. Thanks Deveno

Just a clarifying question.

If we let a = x + < $x^2$ +1> in Q[x]/ < $x^2$ +1>

then surely

$a^2$ + 1 = (x + < $x^2$ +1>) (x + < $x^2$ +1>) +1

But you write:

$a^2$ + 1 = (x + < $x^2$ +1>) (x + < $x^2$ +1>) +1 + < $x^2$ +1>

Why is this? Is it to formally acknowledge or indicate that the product is a coset?

Bernhard

5. because the "1" in Q[x]/<x^2 + 1> is actually the coset 1 + <x^2 + 1>. however, because <x^2 + 1> is an ideal of Q[x], we can regard an element

q + <x^2 + 1>, where q is a rational number (that is, a constant polynomial in Q[x]), as being just "the number q".

this is similar to how we regard "real numbers" as being "complex numbers". we have the isomorphism a --> a+0i = (a,0).

note that these aren't quite the same thing. a real number a, is just one number. the complex number a+0i = (a,0) is a PAIR of numbers,

but the second coordinate "doesn't do anything", it's always 0.

the same thing happens in a quotient ring R/I. the ideal I in R/I is the additive identity, the "0" of that ring.

think about what happens in the ring Z/nZ. we take all the multiples of n in Z, and "call them all 0" (that is, they are 0 mod n).

in the case of Q[x]/<x^2 + 1>, we call any polynomial having x^2 + 1 as a factor, "0".

in other words, we are creating a structure where x^2 + 1 has a solution (by setting it "equal to 0").

that what ideals DO. an ideal is a subring that can "take the place of 0". specifically, these are the properties of 0 we keep:

0 + 0 = 0
a0 = 0a = 0. note that if we replace "0" with "I", we get the rules an ideal obeys.

the short answer to your question is "yes". in abstract algebra, there is a tendency to replace "equality" with the broader equivalence of "isomorphism".

why? because we are concerned with algebraic properties of structures, not the specific "names" or definitions.

instead of Q, one might say: form the field of fractions of Z, Frac(Z). and certainly ZxZ*/~ where (a,b) ~ (c,d) iff ad = bc isn't quite

the same thing as Q (we think of the rationals as magnitudes on a line, generally, whereas we think of the fraction a/b as a ratio of two numbers).

but they are isomorphic, so the differences are mainly a matter of convention. the algebra works out the same.