# Ring Isomorphism Proof

Homework Helper

## Homework Statement

The question : http://gyazo.com/5372336302b5ef289b305172bcd16a2a

## Homework Equations

First Isomorphism theorem.

## The Attempt at a Solution

Define $\phi : \mathbb{Q}[x]/<x^2-2> → Q[ \sqrt{2} ] \space | \space \phi (f(x)) = f( \sqrt{2})$

So showing phi is a homomorphism is quite easy so I'll skip those details.

My question lies in my argument for phi being a bijection. I COULD show it's 1-1 and onto which would mean it's an isomorphism, but I need practice with the first isomorphism theorem so I'm going to try it and hope I know what's going on.

So, I believe I need to show phi is onto and then argue about the kernel.

To show phi is onto, suppose (a+b√2) is in Q[√2] and f(x) is in Q[x]/<x2-2> so that f(x) = ax+b.

Now, $\phi (f(x)) = f( \sqrt{2}) = a + b \sqrt{2}$. Hence for every a+b√2 in Q[√2] there exists f(x) in Q[x]/<x2-2> such that phi(f(x)) = a+b√2.

Hence phi is onto.

Now consider that ker(phi) = { f(x) in Q[x]/<x2-2> | f(√2) = 0 }. Since x2-2 is in ker(phi) and x2-2 is of the smallest degree, we conclude that ker(phi) = x2-2 and hence by the first isomorphism theorem, Q[x]/<x2-2> is isomorphic to Q[√2].

This is my first time trying to apply the theorem rather than doing it the long way of showing 1-1 and onto correspondence.

If anyone could tell me where I may have went wrong or if it looks good, it would be much appreciated.

Related Calculus and Beyond Homework Help News on Phys.org
jbunniii
Homework Helper
Gold Member
Your domain for ##\phi## is incorrect. It should be
$$\phi : \mathbb{Q}[x] \rightarrow Q[\sqrt{2}]$$
The polynomial ##f(x)## is in ##\mathbb{Q}[x]##, not ##\mathbb{Q}[x]/\langle x^2 - 2\rangle##.

Homework Helper
Your domain for ##\phi## is incorrect. It should be
$$\phi : \mathbb{Q}[x] \rightarrow Q[\sqrt{2}]$$
The polynomial ##f(x)## is in ##\mathbb{Q}[x]##, not ##\mathbb{Q}[x]/\langle x^2 - 2\rangle##.
Ah this makes sense to me.

It wouldn't make sense for my f(x) to be inside a set of cosets. So supposing I switch all instances of Q[x]/<x2-2> to Q[x]. Does the argument still apply?

jbunniii
Homework Helper
Gold Member
Ah this makes sense to me.

It wouldn't make sense for my f(x) to be inside a set of cosets. So supposing I switch all instances of Q[x]/<x2-2> to Q[x]. Does the argument still apply?
Yes, your argument is fine. However, I suggest being a bit more explicit in your argument about the kernel of ##\phi##. Perhaps something along the following lines: "the kernel of ##\phi## is an ideal, and ##\mathbb{Q}[x]## is a principal ideal domain, so we know the kernel is generated by a single element of ##\mathbb{Q}[x]##, namely the minimal polynomial of ##\sqrt{2}##. The minimal polynomial of ##\sqrt{2}## is ##x^2 - 2## because..."

Homework Helper
Yes, your argument is fine. However, I suggest being a bit more explicit in your argument about the kernel of ##\phi##. Perhaps something along the following lines: "the kernel of ##\phi## is an ideal, and ##\mathbb{Q}[x]## is a principal ideal domain, so we know the kernel is generated by a single element of ##\mathbb{Q}[x]##, namely the minimal polynomial of ##\sqrt{2}##. The minimal polynomial of ##\sqrt{2}## is ##x^2 - 2## because..."
because (√2)2 - 2 = 0 so that x2-2 = ker(phi) (i.e x2-2 generates ker(phi)).

jbunniii
Homework Helper
Gold Member
because (√2)2 - 2 = 0 so that x2-2 = ker(phi) (i.e x2-2 generates ker(phi)).
To be more precise, ##x^2 - 2## satisfies the definition of the minimal polynomial for ##\sqrt{2}##: (1) it is monic, (2) ##\sqrt{2}## is a root, and (3) if ##p(x)## is a monic polynomial of smaller degree, then ##\sqrt{2}## cannot be a root of ##p(x)## because...

(This may seem obvious, but if it's one of your first homework assignments on this material, being pedantic is never a bad idea.)

Homework Helper
To be more precise, ##x^2 - 2## satisfies the definition of the minimal polynomial for ##\sqrt{2}##: (1) it is monic, (2) ##\sqrt{2}## is a root, and (3) if ##p(x)## is a monic polynomial of smaller degree, then ##\sqrt{2}## cannot be a root of ##p(x)## because...

(This may seem obvious, but if it's one of your first homework assignments on this material, being pedantic is never a bad idea.)
Woah, now this was never mentioned in the book, but a quick read up on wiki has informed me of what you mean.

(1) Is satsified because the coefficient of the highest powered term is 1.
(2) (√2)2 - 2 = 0 so √2 is a root of x2-2.
(3) If p(x) is a monic polynomial of smaller degree, then √2 cannot be a root of p(x) because the only polynomials of smaller degree are of degree 1 and the constant polynomial. i.e either p(x) = ax + b or p(x) = a for some constant a.

Thus p(√2) = a√2 + b or p(√2) = a which tells us that x2-2 is indeed the minimal polynomial.

jbunniii
Homework Helper
Gold Member
Thus p(√2) = a√2 + b or p(√2) = a which tells us that x2-2 is indeed the minimal polynomial.
Right, if ##p(x) \in \mathbb{Q}[x]## is a monic polynomial of degree 1, then it is of the form ##p(x) = x + a##. If we had ##p(\sqrt{2}) = 0## then that means ##\sqrt{2} + a = 0##, or ##\sqrt{2} = -a##, a contradiction because ##a## is rational and ##\sqrt{2}## is not.

And the only monic polynomial of degree 0 is ##p(x) = 1##, which clearly does not satisfy ##p(\sqrt{2}) = 0##, so this tells us that ##x^2 - 2## is indeed the minimal polynomial.

By the way, if this is all new to you, you may want to take a few minutes to convince yourself why we say "the" minimal polynomial, i.e., why is it unique? Can't there be some other monic polynomial of degree 2 with ##\sqrt{2}## as a root?

Homework Helper
Right, if ##p(x) \in \mathbb{Q}[x]## is a monic polynomial of degree 1, then it is of the form ##p(x) = x + a##. If we had ##p(\sqrt{2}) = 0## then that means ##\sqrt{2} + a = 0##, or ##\sqrt{2} = -a##, a contradiction because ##a## is rational and ##\sqrt{2}## is not.

And the only monic polynomial of degree 0 is ##p(x) = 1##, which clearly does not satisfy ##p(\sqrt{2}) = 0##, so this tells us that ##x^2 - 2## is indeed the minimal polynomial.

By the way, if this is all new to you, you may want to take a few minutes to convince yourself why we say "the" minimal polynomial, i.e., why is it unique? Can't there be some other monic polynomial of degree 2 with ##\sqrt{2}## as a root?
I had a look over the proof for uniqueness and this line jumped at me :

"there is at least one polynomial in ker(subα) that generates ker(subα). Such a polynomial will have least degree among all non-zero polynomials in ker(subα), and a(x) is taken to be the unique monic polynomial among these."

Source : http://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory)

jbunniii
Homework Helper
Gold Member
Here's a simple proof of uniqueness. If ##p(x)## and ##q(x)## both satisfy the requirements of a minimal polynomial for ##\sqrt{2}##, then ##r(x) = p(x) - q(x)## is also a polynomial satisfying ##r(\sqrt{2}) = 0##, and the degree of ##r(x)## is strictly lower than that of ##p(x)## and ##q(x)## because the leading coefficients of ##p(x)## and ##q(x)## are both 1, so the leading term cancels when you subtract them.

If ##r(x)## is not the zero polynomial, then we can normalize it by dividing by its leading coefficient, thereby obtaining a monic polynomial satisfying ##r(\sqrt{2}) = 0##. This contradicts the minimality of ##p(x)## and ##q(x)##. We conclude that ##r(x)## must be the zero polynomial, so ##p(x) = q(x)##.

Homework Helper
Here's a simple proof of uniqueness. If ##p(x)## and ##q(x)## both satisfy the requirements of a minimal polynomial for ##\sqrt{2}##, then ##r(x) = p(x) - q(x)## is also a polynomial satisfying ##r(\sqrt{2}) = 0##, and the degree of ##r(x)## is strictly lower than that of ##p(x)## and ##q(x)## because the leading coefficients of ##p(x)## and ##q(x)## are both 1, so the leading term cancels when you subtract them.

If ##r(x)## is not the zero polynomial, then we can normalize it by dividing by its leading coefficient, thereby obtaining a monic polynomial satisfying ##r(\sqrt{2}) = 0##. This contradicts the minimality of ##p(x)## and ##q(x)##. We conclude that ##r(x)## must be the zero polynomial, so ##p(x) = q(x)##.
That's a lot better and easier to follow for sure. So since p = q, the minimal polynomial is monic and unique.

jbunniii
Homework Helper
Gold Member
By the way, we can also make a direct argument for why the kernel is equal to ##\langle x^2 - 2\rangle##, without mentioning minimal polynomials. We just use the fact that we already established, namely that if ##p(x)## is a polynomial of degree less than 2, such that ##p(\sqrt{2}) = 0##, then ##p(x)## must be the zero polynomial.

First, note that the kernel consists precisely of those polynomials ##p(x)## for which ##p(\sqrt{2}) = 0##.

If ##p(x)## is an element of ##\langle x^2 - 2\rangle##, that means ##p(x) = (x^2 - 2)q(x)##, where ##q(x)## is a polynomial in ##Q[x]##, and evaluating at ##x = \sqrt{2}## shows that ##p(\sqrt{2}) = 0##, so ##p(x)## is in the kernel. Therefore ##\langle x^2 - 2 \rangle \subset \textrm{ker}(\phi)##.

Conversely, if ##p(x) \in \textrm{ker}(\phi)##, then by the division algorithm, we may write ##p(x) = q(x)(x^2 - 2) + r(x)##, where the degree of ##r(x)## is strictly less than the degree of ##x^2 - 2##. But ##r(x) = p(x) - q(x)(x^2 - 2)##, so ##r(\sqrt{2}) = 0##, which means ##r(x) \in \textrm{ker}(\phi)##. We already established that this forces ##r(x)## to be the zero polynomial. Thus ##p(x) = q(x)(x^2 - 2)##, which shows that ##p(x) \in \langle x^2 - 2 \rangle##. Thus ##\textrm{ker}(\phi) \subset \langle x^2 - 2 \rangle## and we're done.