Algebra quotient rings

Stuck on iv.

a, b and c can be three possible numbers so is the answer simply 3^3?

Posted from TSR Mobile

It is. You've shown that every member of the quotient is of the form $[a+bx+cx^2]+I$; conversely, it's obvious that every polynomial of the form $a+bx+c x^2$ induces a different member $[a+bx+cx^2]+I$ in the quotient, so we have a bijection taking $(a,b,c) \mapsto [a+bx+c x^2]+I$.

There are, as you say, 27 elements of the left-hand set, so there must be 27 of the right-hand set.

The question has constructed a field extension $\mathbb{F}_3(\theta)$, where $\theta^3 + \theta + 1 = 0$.

Thanks so much. Would you be able to help me on c too.

It doesn't specifically say what I is so I'm a little confused on that. I'm trying to show that I is a subset of (a) and vise versa.

I also am considering the definition of an ideal too. But how do I know that a natural number exists in I...

Posted from TSR Mobile

Thanks so much. Would you be able to help me on c too.

It doesn't specifically say what I is so I'm a little confused on that. I'm trying to show that I is a subset of (a) and vise versa.

I also am considering the definition of an ideal too. But how do I know that a natural number exists in I...

Posted from TSR Mobile

$I$ is an ideal of $\mathbb{Z}$, and therefore it can only contain integers.

You're going to have trouble showing that $I \subset \langle a \rangle$ if you don't know what $a$ is. Suppose I gave you the ideal $I = \langle 6, 15 \rangle \subset \mathbb{Z}$. The question is asking you to show that there is some $a$ such that $I = \langle a \rangle$. Can you tell me what $a$ is in this case?

Im not understanding what the comma is all about. <6, 15> what does the comma mean in this? I have never seen notation like this before...

$I$ is an ideal of $\mathbb{Z}$, and therefore it can only contain integers.

You're going to have trouble showing that $I \subset \langle a \rangle$ if you don't know what $a$ is. Suppose I gave you the ideal $I = \langle 6, 15 \rangle \subset \mathbb{Z}$. The question is asking you to show that there is some $a$ such that $I = \langle a \rangle$. Can you tell me what $a$ is in this case?

Just like <a> is the ideal generated by a, so elements of the form na where n is in Z, <a, b> is the ideal generated by a and b so elements of the form na + mb for n and m in Z.

So <6, 15> is elements of the form 6n + 15m for n and m in Z.

Hi very very sorry that I didnt reply. Been very busy. I had my exam today and the same question came up but m is and integer and not a natural number.

This is whay I did. Please can someone check it. if this was out of 7 marks, how many do you think I would get. Really not sure if what I did made sense...

Posted from TSR Mobile

Hi very very sorry that I didnt reply. Been very busy. I had my exam today and the same question came up but m is and integer and not a natural number.

This is whay I did. Please can someone check it. if this was out of 7 marks, how many do you think I would get. Really not sure if what I did made sense...

$\langle -m \rangle = \langle m \rangle$, of course, so WLOG $m$ is greater than 0.


To be honest, it doesn't really. You need to show that $I$ is principal: that there exists $m$ such that $I = \langle m \rangle$.

For 1), you've written "maximal" instead of "minimal". Nonzero ideals in $\mathbb{Z}$ don't have maximal elements.
It's all a bit unclear for me, to be honest, and I'm not really sure how you're trying to prove it. Your final line doesn't make sense anyway: "$I = \langle m \rangle$ for all $m \in \mathbb{Z}$" is not true of any ideal.

I think what you have written is more convincing as a proof that sets of the form $\langle m \rangle$ are ideals of $\mathbb{Z}$. I see some of the right answer in there, but also some bits that aren't right.

To answer the question as stated in a "model answer" kind of way (bearing in mind that solutions anyone comes up with on the spot are usually much less well-structured than this):

OK I see. Thanks for the model answer.

Do you think Id even get one mark lol...

Thank god this was the ONLY thing I struggled on.