Proving Ring Isomorphism using the First Isomorphism Theorem

[STEMucator]

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.

 

[jbunniii]

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 $\mathbb{Q}[x]/\langle x^2 - 2\rangle$.

 

[STEMucator]

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]

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..."

 

[STEMucator]

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]

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.)

 

[STEMucator]

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]

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?

 

[STEMucator]

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

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

 

[jbunniii]

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 $p(x) = q(x)$.

 

[STEMucator]

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]

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.