Ring Isomorphism Proof

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Homework Statement



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

Homework Equations



First Isomorphism theorem.

The Attempt at a Solution



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

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, [itex]\phi (f(x)) = f( \sqrt{2}) = a + b \sqrt{2}[/itex]. 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.
 

Answers and Replies

  • #2
jbunniii
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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##.
 
  • #3
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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?
 
  • #4
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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..."
 
  • #5
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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)).
 
  • #6
jbunniii
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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.)
 
  • #7
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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.
 
  • #8
jbunniii
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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?
 
  • #9
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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)
 
  • #10
jbunniii
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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)##.
 
  • #11
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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.
 
  • #12
jbunniii
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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.
 

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