Recent content by mathbalarka

The point is unital rings always have maximal ideals, so you can always take intersection amongst them and $J(R)$ would be well-defined. In fact $J(R)$ is always a proper ideal (because maximal ideals are proper and intersection of a bunch of proper things is always going to give you a proper...

Great idea there. You're right that $\lim_{x \to 1} \dfrac{\sin(x^2 - 1)}{(x - 1)} = \lim_{x \to 1} \dfrac{\sin(x^2 - 1)}{(x^2 - 1)} \cdot (x + 1) = 2 \lim_{x \to 1} \dfrac{\sin(x^2 - 1)}{x^2 - 1}$. To proceed, you have to change the variable in the limit. Note that as $x\to 1$ you have $x^2 -...

You should be able to do the first part of the exercise as I have already given hints in the other thread. For the second part, I really don't see how you can do this without some machinery. $\mathcal{Z}(xy - z^2)$ looks like a cone in the affine $3$-space, if you try to draw it. You can try...

Let's talk a bit about algebras first : Given a commutative unital ring $R$, an $R$-algebra $(A, f)$ is simply a ring $A$ with a ring homomorphism $f : R \to A$. Note that you have a natural $R$-module structure on $A$ defined by $ra = f(r)a$. Sometimes we just say, by abuse of notation, that...

Nah, I have just recently finished studying the first few chapters of Atiyah-Macdonald. Really don't know any algebraic geometry, but I'd have to learn it soon.

Yes, that is correct. I have given an example in my answer :

I have no idea what you mean by that. Why should this be impossible? You go through the polynomials in $k[x_1, \cdots, x_n]$ which vanish in all of $A$ identically, and throw out the ones that doesn't. The remaining polynomials is your set. Sorry, that's not valid. If you have a subset $A$ of...

I'll answer the question first : $\Bbb R[x, y]/(xy - 1)$ is the same thing as $\Bbb R[x, 1/x]$. There is nothing much to prove. Note that $\Bbb R[x, 1/x]$ is the ring of formal polynomials of $x$ and $1/x$. That is, it's NOT a polynomial ring. This is a very important note to bear in mind. A...

No, it's completely false that any subset of $\Bbb A^n$ appears as an affine algebraic set (also called affine variety). First, let's review the definition : For some subset $S \in k[x_1,\cdots, x_n]$, the affine variety corresponding to it is the common zero locus of all polynomials in $S$...

Good problem. This is the "physicist's argument" (no offense topsquark) of the reflexivity principle being redundant, as I was told quite a time ago when I was learning algebra. The problem with the argument is that you might not have "enough wiggle room" to apply reflexivity. What if the...

I do agree that Eisenbud is pretty advanced. However, if you know enough ring theory (which I think you do), you can start off with Atiyah-MacDonald and Reid's Undergraduate Commutative Algebra. Both are undergrad books, and complement each another quite well in the sense that A-M's theory is...

You can always factorize. $x^3 + 3x^2 - 4x - 12 = x^3 - 4x + 3x^2 - 12 = x(x^2 - 4) + 3(x^2 - 4) = (x^2 - 4)(x + 3) = (x + 2)(x+3)(x - 2)$ Hence, the roots are $x = 2, -2, -3$.

(fair warning : Dummit-Foote doesn't have much algebraic geometry. I'd recommend Eisenbud's book or Reid's book for learning commutative algebra with a flavor of algebraic geometry) The surjective ring homomorphism $\psi : k[x_1, \cdots, x_n] \to k$ here is taking a polynomial $f(x_1, x_2...

Fallen Angel has answered the question already, but I will elaborate a bit more on this in case it's not clear : A proper ideal of a ring $R$ is an ideal which is not the whole ring $R$. By definition, prime ideals are proper. If a prime ideal $I \subset R$ contained the identity $1$, then it...